Selina ICSE Solutions for Class 10 Maths

Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises.

Sales-Tax (VAT) Chapterwise Revision Exercises

Question 1.
A man purchased a pair of shoes for ₹809.60 which includes 8% rebate on the marked price and then 10% sales tax on the remaining price. Find the marked price of the pair of shoes.
Solution:

Question 2.
The catalogue price of an article is ₹36,000. The shopkeeper gives two successive discounts of 10% each. He further gives an off-season discount of 5 % on the balance. If sales-tax at the rate 10% is charged on the remaining amount, find :
(i) the sales tax charged
(ii) the selling price of the article including sales-tax.
Solution:

Question 3.
A sells an article to B for ₹80,000 and charges sales-tax at 8%. B sells the same article to C for ₹1,12,000 and charges sales- tax at the rate of 12%. Find the VAT paid by B in this transaction.
Solution:

Question 4.
The marked price of an article is ₹1,600. Mohan buys this article at 20% discount and sells it at its marked price. If the sales-tax at each stage is 6%; find :
(i) the price at which the article can be bought,
(ii) the VAT (value added tax) paid by Mohan.
Solution:

Question 5.
A sells an old laptop to B for ₹12,600; B sells it to C for ₹14,000 and C sells the same laptop to D for ₹16,000.
If the rate of VAT at each stage is 10%, find the VAT paid by :
(i) B
(ii) C
Solution:

Banking Chapterwise Revision Exercises

Question 6.
Ashok deposits ₹3200 per month in a cumulative account for 3 years at the rate of 9% per annum. Find the maturity value of this account
Solution:

Question 7.
Mrs. Kama has a recurring deposit account in Punjab National Bank for 3 years at 8% p.a. If she gets ₹9,990 as interest at the time of maturity, find:
(i) the monthly instalment.
(ii) the maturity value of the account.
Solution:

Question 8.
A man has a 5 year recurring deposit account in a bank and deposits ₹240 per month. If he receives ₹17,694 at the time of maturity, find the rate of interest.
Solution:

Question 9.
Sheela has a recurring deposit account in a bank of ₹2,000 per month at the rate of 10% per anum. If she gets ₹83,100 at the time of maturity, find the total time (in years) for which the account was held.
Solution:

Question 10.
A man deposits ₹900 per month in a recurring account for 2 years. If he gets 1,800 as interest at the time of maturity, find the rate of interest .
Solution:

Shares And Dividend Chapterwise Revision Exercises

Question 11.
What is the market value of 4 \(\frac { 1 }{ 2 }\) % (₹100) share, when an investment of ₹1,800 produces an income of ₹72 ?
Solution:

Question 12.
By investing ₹10,000 in the shares of a company, a man gets an income of ₹800; the dividend being 10%. If the face-value of each share is ₹100, find :
(i) the market value of each share.
(ii) the rate per cent which the person earns on his investment.
Solution:

Question 13.
A man holds 800 shares of ₹100 each of a company paying 7.5% dividend semiannually.
(i) Calculate his annual dividend.
(ii) If he had bought these shares at 40% premium, what percentage return does he get on his investment ?
Solution:

Question 14.
A man invests ₹10,560 in a company, paying 9% dividend, at the time when its ₹100 shares can be bought at a premium of ₹32. Find:
(i) the number of shares bought by him;
(ii) his annual income from these shares and
(iii) the rate of return on his investment .
Solution:

Question 15.
Find the market value of 12% ₹25 shares of a company which pays a dividend of ₹1,875 on an investment of ₹20,000.
Solution:

Linear Inequations Chapterwise Revision Exercises

Question 16.
The given diagram represents two sets A and B on real number lines.

(i) Write down A and B in set builder notation.
(ii) Represent A ∪ B, A ∩ B, A’ ∩ B, A – B and B – A on separate number lines.
Solution:

Question 17.
Find the value of x, which satisfy the inequation:

Graph the solution set on the real number line .
Solution:

Question 18.
State for each of the following statements whether it is true or false :
(a) If (x – a) (x – b) < 0, then x < a, and x < b.
(b) If a < 0 and b < 0, then (a + b)² > 0.
(c) If a and b are any two integers such that a > b, then a² > b².
(d) If p = q + 2, then p > q.
(e) If a and b are two negative integers such that a < b , then \(\frac { 1 }{ a }\) < \(\frac { 1 }{ b }\)
Solution:

Question 19.
Given 20 – 5x < 5(x + 8), find the smallest value of x when :
(i) x \(\epsilon\) I
(ii) x \(\epsilon\) W
(iii) x \(\epsilon\) N
Solution:

Question 20.
If x \(\epsilon\) Z, solve : 2 + 4x < 2x – 5 < 3x. Also, represent its solution on the real number line.
Solution:

Quadratic Equation Chapterwise Revision Exercises

Question 21.

Solution:

Question 22.

Solution:

Question 23.
Find the value of k for which the roots of the following equation are real and equal k²x² – 2 (2k -1) x + 4 = 0
Solution:

Question 24.

Solution:

Question 25.
If -5 is a root of the quadratic equation 2x² +px -15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.
Solution:

Problems On Quadratic Equations Chapterwise Revision Exercises

Question 26.
x articles are bought at ₹(x – 8) each and (x – 2) some other articles are bought at ₹(x – 3) each. If the total cost of all these articles is ₹76, how many articles of first kind were bought ?
Solution:

Question 27.
In a two digit number, the unit’s digit exceeds its ten’s digit by 2. The product of the given number and the sum of its digits is equal to 144. Find the number.
Solution:

Question 28.
The time taken by a person to cover 150 km was 2.5 hours more than the time taken in return journey. If he returned at a speed of 10 km/hour more than the speed of going, what was the speed per hour in each direction ?
Solution:

Question 29.
A takes 9 days more than B to do a certain piece of work. Together they can do the work in 6 days. How many days will A alone take to do the work ?
Solution:

Question 30.
A man bought a certain number of chairs for ₹10,000. He kept one for his own use and sold the rest at the rate ₹50 more than he gave for one chair. Besides getting his own chair for nothing, he made a profit of ₹450. How many chairs did he buy ?
Solution:

Question 31.
In the given figure; the area of unshaded portion is 75% of the area of the shaded portion. Find the value of x.

Solution:

Ratio And Proportion Chapterwise Revision Exercises

Question 32.

Solution:

Question 33.
If a:b = 2:3,b:c = 4:5 and c: d = 6:7, find :a:b :c :d.
Solution:

Question 34.

Solution:

Question 35.
Find the compound ratio of:
(i) (a-b) : (a+b) and (b²+ab): (a²-ab)
(ii) (x+y): (x-y); (x²+y²): (x+y)² and (x²-y²)²: (x⁴-y⁴)
(iii) (x²– 25): (x²+ 3x – 10); (x²-4): (x²+ 3x+2) and (x + 1): (x² + 2x)
Solution:

Question 36.
The ratio of the prices of two fans was 16: 23. Two years later, when the price of the first fan had risen by 10% and that of the second by Rs. 477, the ratio of their prices became 11: 20. Find the original prices of two fans.
Solution:
The ratio of prices of two fans = 16 : 23
Let the price of first fan = 6x
then price of second fan = 23x

Remainder And Factor Theorems Chapterwise Revision Exercises

Question 37.
Given that x + 2 and x – 3 are the factors of x³ + ax + b, calculate the values of a and b. Also find the remaining factor.
Solution:

Question 38.
Use the remainder theorem to factorise the expression 2x³ + 9x² + 7x – 6 = 0 Hense, solve the equation 2x³ + 9x² + 7 x – 6 = 0
Solution:

Question 39.
When 2x³ + 5x² – 2x + 8 is divided by (x – a) the remainder is 2a³ + 5a². Find the value of a.
Solution:

Question 40.
What number should be added to x³ – 9x² – 2x + 3 so that the remainder may be 5 when divided by (x – 2) ?
Solution:

Question 41.
Let R₁ and R₂ are remainders when the polynomials x³ + 2x² – 5ax – 7 and x³ + ax² – 12x + 6 are divided by (x +1) and (x – 2) respectively. If 2R₁ + R₂ = 6; find the value of a.
Solution:

Matrices Chapterwise Revision Exercises

Question 42.

Solution:

Question 43.

Solution:

Question 44.

Solution:

Question 45.

Solution:

Arithmetic Progression (A.P.) Chapterwise Revision Exercises

Question 46.
Find the 15th term of the A.P. with second term 11 and common difference 9.
Solution:

Question 47.
How many three digit numbers are divisible by 7 ?
Solution:

Question 48.
Find the sum of terms of the A.P.: 4,9,14,…….,89.
Solution:

Question 49.
Daya gets pocket money from his father every day. Out of the pocket money, he saves ₹2.75 on first day, ₹3.00 on second day, ₹3.25 on third day and so on. Find:
(i) the amount saved by Daya on 14th day.
(ii) the amount saved by Daya on 30th day.
(iii) the total amount saved by him in 30 days.
Solution:

Question 50.
If the sum of first m terms of an A.P. is n and sum of first n terms of the same A.P. is m. Show that sum of first (m + n) terms of it is (m + n).
Solution:

Geometric Progression (GP) Chapterwise Revision Exercises

Question 51.
3^rd term of a GP. is 27 and its 6th term is 729; find the product of its first and 7^th terms.
Solution:

Question 52.
Find 5 geometric means between 1 and 27.
Solution:

Question 53.
Find the sum of the sequence 96 – 48 + 24…. upto 10 terms.
Solution:

Question 54.
Find the sum of first n terms of:
(i) 4 + 44 + 444 + …….
(ii) 0.7 + 0.77 + 0.777 + …..
Solution:

Question 55.
Find the value of 0.4.
Solution:

Reflection Chapterwise Revision Exercises

Question 56.
Find the values of m and n in each case if:
(i) (4, -3) on reflection in x-axis gives (-m, n)
(ii) (m, 5) on reflection in y-axis gives (-5, n-2)
(iii) (-6, n+2) on reflection in origin gives (m+3, -4)
Solution:

Question 57.
Points A and B have the co-ordinates (-2,4) and (-4,1) respectively. Find :
(i) The co-ordinates of A’, the image of A in the line x = 0.
(ii) The co-ordinates of B’, the image of Bin y-axis.
(iii) The co-ordinates of A”, the image of A in the line BB’,
Hence, write the angle between, the lines A’A” and B B’. Assign a special name to the figure B’ A’ B A”
Solution:

Question 58.
Triangle OA₁B₁ is the reflection of triangle OAB in origin, where A, (4, -5) is the image of A and B, (-7, 0) is the image of B.
(i) Write down the co-ordinates of A and B and draw a diagram to represent this information.
(ii) Give the special name to the quadrilateral ABA₁ B₁. Give reason.
(iii) Find the co-ordinates of A₂, the image of A under reflection in x-axis followed by reflection in y-axis.
(iv) Find the co-ordinates of B₂, the image of B under reflection in y-axis followed by reflection in origin.
(v) Does the quadrilateral obtained has any line symmetry ? Give reason.
(vi) Does it have any point symmetry ?
Solution:

Section And Mid-Point Formulae Chapterwise Revision Exercises

Question 59.
In what ratio does the point M (P, -1) divide the line segment joining the points A (1,-3) and B (6,2) ? Hence, find the value of p.
Solution:

Question 60.
A (-4,4), B (x, -1) and C (6,y) are the vertices of ∆ABC. If the centroid of this triangle ABC is at the origin, find the values of x and y.
Solution:

Question 61.
A (2,5), B (-1,2) and C (5,8) are the vertices of a triangle ABC. Pand Q are points on AB and AC respectively such that AP: PB=AQ: QC = 1:2.
(a) Find the co-ordinates of points P and Q
(b) Show that BC = 3 x PQ.
Solution:

Question 62.
Show that the points (a, b), (a+3, b+4), (a -1, b + 7) and (a – 4, b + 3) are the vertices of a parallelogram.
Solution:

Equation Of straight Line Chapterwise Revision Exercises

Question 63.
Given points A(l, 5), B (-3,7) and C (15,9).
(i) Find the equation of a line passing through the mid-point of AC and the point B.
(ii) Find the equation of the line through C and parallel to AB.
(iii) The lines obtained in parts (i) and (ii) above, intersect each other at a point P. Find the co-ordinates of the point P.
(iv) Assign, giving reason, a special names of the figure PABC.
Solution:

Question 64.
The line x- 4y=6 is the perpendicular bisector of the line segment AB. If B = (1,3); find the co-ordinates of point A.
Solution:

Question 65.
Find the equation of a line passing through the points (7, -3) and (2, -2). If this line meets x- axis at point P and y-axis at point Q; find the co-ordinates of points P and Q.
Solution:

Question 66.
A (-3,1), B (4,4) and C (1, -2) are the vertices of a triangle ABC. Find:
(i) the equation of median BD,
(ii) the equation of altitude AE.
Solution:

Question 67.
Find the equation of perpendicular bisector of the line segment joining the points (4, -3) and (3,1).
Solution:

Question 68.
(a) If (p +1) x + y = 3 and 3y – (p -1) x = 4 are perpendicular to each other find the value of p.
(b) If y + (2p +1) x + 3 = 0 and 8y – (2p -1) x = 5 are mutually prependicular, find the value of p.
Solution:

Question 69.
The co-ordinates of the vertex A of a square ABCD are (1, 2) and the equation of the diagonal BD is x + 2y = 10. Find the equation of the other diagonal and the coordinates of the centre of the square.
Solution:

Similarity Chapterwise Revision Exercises

Question 70.
M is mid-point of a line segment AB; AXB and MYB are equilateral triangles on opposite sides of AB; XY cuts AB at Z. Prove that AZ = 2ZB.
Solution:

Question 71.
In the given figure, if AC = 3cm and CB = 6 cm, find the length of CR.

Solution:

Question 72.
The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point O. If BO : OD = 4:7; find:
(i) ∆AOD : ∆AOB
(ii) ∆AOB : ∆ACB
(iii) ∆DOC : ∆AOB
(iv) ∆ABD : ∆BOC
Solution:

Question 73.
A model of a ship is made to a scale of 1 : 160. Find :
(i) the length of the ship, if the length of its model is 1.2 m.
(ii) the area of the deck of the ship, if the area of the deck of its model is 1.2 m².
(iii) the volume of the ship, if the volume of its model is 1.2m³.
Solution:

Question 74.
In trapezium ABCD, AB || DC and DC = 2 AB. EF, drawn parallel to AB cuts AD in F and BC in E such that 4 BE = 3 EC. Diagonal DB intersects FE at point G Prove that: 7 EF = 10 AB.

Solution:

Loci Chapterwise Revision Exercises

Question 75.
In triangle ABC, D is mid-point of AB and CD is perpendicular to AB. Bisector of ∠ABC meets CD at E and AC at F. Prove that:
(i) E is equidistant from A and B.
(ii) F is equidistant from AB and BC.
Solution:

Question 76.
Use graph paper for this questions. Take 2 cm = 1 unit on both axes.
(i) Plot the points A (1,1), B (5,3) and C (2,7)
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:

Circles Chapterwise Revision Exercises

Question 77.
In the given figure, ∠ADC = 130° and BC = BE. Find ∠CBE if AB ⊥ CE.

Solution:

Question 78.
In the given figure, ∠OAB=30° and ∠OCB= 57°, find ∠BOC and ∠AOC.

Solution:

Question 79.
In the given figure, O is the centre of the circle. If chord AB = chord AC, OP⊥ AB and OQ⊥ AC; show that: PB=QC.
Solution:

Question 80.
In the given figure, AB and XY are diameters of a circle with centre O. If ∠APX=30°, find:
(i) ∠AOX
(ii) ∠APY
(iii) ∠BPY
(iv) ∠OAX

Solution:

Question 81.
(a) In the adjoining figure; AB = AD, BD = CD and ∠DBC = 2 ∠ABD.
Prove that: ABCD is a cyclic quadrilateral.

(b) AB is a diameter of a circle with centre O, Chord CD is equal to radius OC. AC and BD produced intersect at P. Prove that ∠APB = 60°.

Solution:

Tangents And Intersecting Chords Chapterwise Revision Exercises

Question 82.
In the given figure, AC=AB and ∠ABC=72°.
OA and OB are two tangents. Determine:
(i) ∠AOB
(ii) angle subtended by the chord AB at the centre.

Solution:

Question 83.
In the given figure, PQ, PR and ST are tangents to the same circle. If ∠P = 40° and ∠QRT = 75°, find a, b and c.

Solution:

Question 84.
In the given figure, ∠ABC = 90° and BC is diameter of the given circle. Show that:
(i) AC x AD =AB²
(ii) AC x CD = BC²

Solution:

Question 85.
In the given figure; AB, BC and CA are tangents to the given circle. If AB = 12 cm, BC = 8 cm and AC=10 cm, find the lengths of AD,BE = CF.

Solution:

Question 86.
(a) AB and CD are two chords of a circle intersecting at a point P inside the circle. If:
(i) AB = 24 cm, AP = 4 cm and PD = 8 cm, determine CP.
(ii) AP = 3 cm, PB = 2.5 cm and CD = 6.5 cm determine CP.
(b) AB and CD are two chords of a circle intersecting at a point P outside the circle. If:
(i) PA = 8 cm, PC – 5 cm and PD = 4 cm, determine AB.
(ii) PC = 30 cm, CD = 14 cm and PA = 24 cm, determine AB.
Solution:

Construction Chapterwise Revision Exercises

Question 87.
Construct a triangle ABC in which AC = 5 cm, BC = 7 cm and AB = 6 cm.
(i) Mark D, the mid point of AB.
(ii) Construct a circle which touches BC at C and passes through D.
Solution:

Question 88.
Using ruler and compasses only, draw a circle of radius 4 cm. Produce AB, a diameter of this circle up to point X so that BX = 4cm. Construct a circle to touch AB at X and to touch the circle, drawn earlier externally.
Solution:

Mensuration Chapterwise Revision Exercises

Question 89.
A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.
Solution:

Question 90.
A tent is of the shape of right circular cylinder upto height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner surface of the tent at Rs. 4 per sq. metre, if the radius of the base is 14 metres.
Solution:

Question P.Q.
In the given figure, diameter of the biggest semi-circle is 108cm, and diameter of the smallest circle is 36 cm. Calculate the area of the shaded portion.

Solution:

Question 91.
A copper wire of diameter 6 mm is evenly wrapped on the cylinder of length 18 cm and diameter 49 cm to cover the whole surface. Find:
(i) the length
(ii) the volume of the wire
Solution:

Question 92.
A pool has a uniform circular cross-section of radius 5 m and uniform depth 1.4m. It is filled by a pipe which delivers water at the rate of 20 litres per sec. Calculate, in minutes, the time taken to All the pool. If the pool is emptied in 42 min. by another cylindrical pipe through which water flows at 2 m per sec, calculate the radius of the pipe in cm.
Solution:

Question 93.
A test tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is 2849/3cm³ and 2618/3cm³ of water are required to fill the tube to a level which is 2 cm below the top of the tube. Find the radius of the tube and the length of its cylinderical part.
Solution:

Question 94.
A sphere is placed in an inverted hollow conical vessel of base radius 5 cm and vertical height 12 cm. If the highest point of the sphere is at the level of the base of the cone, find the radius of the sphere. Show that the volume of the sphere and the conical vessel are as 40 : 81.
Solution:

Question 95.
The difference between the outer and the inner curved surface areas of a hollow cylinder, 14cm. long is 88sq. cm. Find the outer and the inner radii of the cylinder given that the volume of metal used is 176 cu. cm.
Solution:

Trigonometry Chapterwise Revision Exercises

Question 96.

Solution:

Question 97.
If tan A = 1 and tan B = √3 ; evaluate :
(i) cos A cos B – sin A sin B
(ii) sin A cos B + cos A sin B
Solution:

Question 98.
As observed from the top of a 100 m high light house, the angles, of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Solution:

Question 99.

Solution:

Question 100.
From the top of a light house, it is observed that a ship is sailing directly towards it and the angle of depression of the ship changes from 30° to 45° in 10 minutes. Assuming that the ship is sailing with uniform speed; calculate in how much more time (in minutes) will the ship reach to the light house.
Solution:
Let LM be the height of light house = h
Angle of depression changes from 30° to 45° in 10 minutes.

Statistics Chapterwise Revision Exercises

Question 101.
Calculate the mean mark in the distribution given below :

Also state (i) median class (ii) the modal class.
Solution:

Question 102.
Draw an ogive for the following distribution :

Use the ogive drawn to determine :
(i) the median income,
(ii) the number of employees whose income exceeds Rs. 190.
Solution:

Question 103.
The result of an examination are tabulated below :

Draw the ogive for above data and from it determine :
(i) the number of candidates who got marks less than 45.
(ii) the number of candidates who got marks more than 75.
Solution:

Probability Chapterwise Revision Exercises

Question 104.
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
Solution:

Question 105.
A bag contains 6 red balls, 8 blue balls and 10 yellow balls, all the balls being of the same size. If a ball is drawn from the bag, without looking into it, find the probability that the ball drawn is
(i) yellow
(ii) red
(iii) blue
(iv) not yellow
(v) not blue
Solution:

Question 106.
Two dice are thrown at the same time. Write down all the possible outcomes. Find the probability of getting the sum of two numbers appearing on the top of the dice as :
(i) 13
(ii) less than 13
(iii) 10
(iv) less then 10
Solution:

Question 107.
Five cards : the ten, jack, queen, king and ace. of diamonds are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen ?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace ? (b) a queen ?
Solution:

Question 108.
(i) A lot of 20 bulbs contains 4 defective bulbs, one bulb is drawn at random, from the lot. What is the probability that this bulb is defective ?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?
Solution:

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapterwise Revision Exercises  are helpful to complete your math homework.

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles Ex 17C

Question 1.
In the given circle with diameter AB, find the value of x. (2003)

Solution:

∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a ∆)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180°
⇒ x + 120° = 180°
∴ x = 180°- 120° = 60°

Question 2.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.

Solution:

Radius of the circle whose centre is O = 5 cm
OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.
Join OA and OC, then OA = OC=5cm.
∴ OP ⊥ AB
∴ P is the mid-point of AB.
Similarly, Q is the mid-point of CD.
In right ∆OAP,
OA² = OP² + AP² (Pythagorous theoram)
⇒ (5)² =OP² +(4)² ( ∵ AP = AB = \(\frac { 1 }{ 2 }\) x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP³ = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC² = OQ² + CQ²
⇒ (5)² =OQ²+(3)² (∵ CQ = CD = \(\frac { 1 }{ 2 }\) x 6 = 3cm)
⇒ 25 = OQ² + 9
⇒ OQ³ = 25 – 9 = 16 = (4)²
∴ OQ = 4 cm
Hence, PQ = OP + OQ = 3-4 = 7 cm.

Question 3.
The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

Solution:

Join AP and produce AB to meet the bigger circle at C.
AB = AC – BC = 5 cm – 3 cm = 2 cm.
But, M is the mid-point of AB
∴ AM = \(\frac { 2 }{ 2 }\) = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)² = MP² -1- (1 )²
⇒ 25 = MP: + 1
⇒ MP: = 25 – 1 = 24

Question 4.
In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.

Solution:

Given: In the figure ABC is a triangle in winch ∠A = 30°
To Prove: BC is the radius of circumcircle of ∆ABC whose O is the centre.
Const: Join OB and OC.
Proof: ∠BOC is at the centre and ∠BAC is at the remaining part of the circle
∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°
Now in ∆OBC,
OB = OC (Radii of the same circle)
∴ ∠OBC = ∠OCB
But ∠OBC + ∠OCB + ∠BOC – 180°
∠OBC + ∠OBC + 60°- 180°
⇒ 2 ∠OBC = 180°- 60° = 120°
⇒ ∠OBC = \(\frac { { 120 }^{ circ } }{ 2 }\) = 60°
∴ ∆OBC is an equilateral triangle.
∴ BC = OB = OC
But OB and OC are the radii of the circumcircle
∴ BC is also the radius of the circumcircle.

Question 5.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:

Given: In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.
To Prove: D is the mid point of BC.
Const: Join AD.
Proof: ∠ 1 = 90° (Angle in a semi-circle)
But ∠ 1 + ∠ 2 – 180° (Linear pair)
∴ ∠ 2 = 90°
Now, in right ∆ ABD and ∆ ACD,
Hyp. AB – Hyp. AC (Given)
Side AD – AD (Common)
∴ ∆ABD = ∆ACD (RHS criterion of congruency)
∴ BD = DC (c.p.c.t.)
Hence D is he mid point of BC. Q.E.D.

Question 6.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Solution:
Join OE,
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°
Now, in ∆OEC, OE = OC (radii of the same circle)
∴ ∠OEC = ∠OCE
But ∠ OEC + ∠ OCE + ∠ EOC = 180°
⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°

Question 7.
Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.

Solution:
Given: Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.

To Prove: AB = CD.
Proof: ∵ Two chords AB and CD intersect each other inside the circle at P.
∴ AP x PB = CP x PD ⇒ \(\frac { AP}{ CP }\) = \(\frac { AD }{ PB }\)
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(ii)
Adding (i) and (ii),
AP + PB = CP + PD
⇒ AB = CD Q.E.D.

Question 8.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.
To Prove: PRQS is a cyclic quadrilateral.
Proof: In ∆APD,

∠1 +∠2 + ∠P= 180° ,…(i)
Similarly, in ∆BQC.
∠3 + ∠4 + ∠Q= 180° ….(ii)
Adding (i) and (ii), we get:
∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°
⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)
But ∠1+∠2+∠3+∠4 = \(\frac { 1 }{ 2 }\) (∠A+∠B+∠C+∠D)
= \(\frac { 1 }{ 2 }\) x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS
∴ Quad. PRQS is a cyclic quadrilateral. Q.E.D.

Question 9.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)

Solution:
∠DBC = 58°
BD is diameter
∴ ∠DCB=90° (Angle in semi circle)

(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°
(ii) ∠BEC =180°-32°
(opp. angle of cyclic quadrilateral)
= 148°
(iii) ∠BAC = ∠BDC = 32°
(Angles in same segment)

Question 10.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.
Solution:

Given: In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.
To Prove: B,C,E,D. are concyclic.
Proof: In ∆ABC, AB = AC
∴ ∠B = ∠C (Angles opposite to equal sides)
Similarly in ∆ADE, AD = AE (given)
∴ ∠ADE = ∠AED
In ∆ABC,
∵ \(\frac { AP }{ AB }\) = \(\frac { AE }{ AC }\)
∴ DE || BC.
∴ ∠ADE = ∠B (Corresponding angles)
But ∠B = ∠C (Proved)
∴ Ext. ∠ADE = its interior opposite ∠C.
∴ BCED is a cyclic quadrilateral.
Hence B,C, E and D are concyclic.

Question 11.
In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.

Solution:

In cyclic quad. ABCD,
AF || CB and DA is produced to E such that
∠ADC = 92° and ∠FAE – 20°
Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,
∠B – ∠D = 180° ⇒ ∠B + 92° = 180″
⇒∠B = 180°-92° = 88°
∵ AF || CB.
∴∠FAB = ∠B = 88°
But ∠FAE – 20° (Given)
Ext. ∠BAE – ∠BAF + ∠FAE
= 88° + 20° = 108°
But Ext. ∠BAE – ∠BCD
∴ ∠BCD = 108°

Question 12.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate
(i) ∠DBC,
(ii) ∠IBC,
(iii) ∠BIC.

Solution:

Join DB and DC, IB and IC.
∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.
(i) ∵ ∠DBC and ∠DAC are in the same segment
∴ ∠DBC – ∠DAC.
But ∠DAC = \(\frac { 1 }{ 2 }\)∠BAC = \(\frac { 1 }{ 2 }\) x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = \(\frac { 1 }{ 2 }\) ∠ABC = \(\frac { 1 }{ 2 }\) x 80° = 40°.
(iii) ∴∠BAC = 66° ∠ABC = 80°
∴ In ∆ABC,
∠ACB = 180° – (∠ABC + ∠CAB)
= 180°-(80°+ 66°)= 180°- 156° = 34°
∵ IC bisects the ∠C
∴ ∠ ICB = \(\frac { 1 }{ 2 }\) ∠C = \(\frac { 1 }{ 2 }\) x 34° = 17°.
Now in ∆IBC,
∠ IBC + ∠ ICB + ∠ BIC = 180°
⇒ 40° + 17° + ∠BIC = 180°
⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°
= 123°

Question 13.
In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :
(i) ∠ABD
(ii) ∠APB

Hence or otherwise prove that AP is parallel to DB.
Solution:
Given: In figure, AB = AD = DC = PB.
∠DBC = x. Join AC and BD.
To Find : the measure of ∠ABD and ∠APB.
Proof: ∠DAC=∠DBC= x(angles in the same segment)
But ∠DCA = ∠DAC (∵ AD = DC)
= x
But ∠ABD = ∠DAC (Angles in the same segment)
In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB
But ∠ BAP = ∠APB (∵ AB = BP)
2 x x = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2x
⇒ ∠APB = x
∵ ∠APB = ∠DBC = x
But these are corresponding angles
∴ AP || DB. Q.E.D.

Question 14.
In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.

Solution:
Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
Proof: In cyclic quad. ABEP,
∠APE+ ∠ABE= 180° ….(i)
Similarly in cyclic quad. BCQE

∠CQE +∠CBE = 180° ….(ii)
Adding (i) and (ii),
∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°
⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°
But ∠ABE + ∠CBE = 180° (Linear pair)
∴ ∠APE + ∠CQE + 180° = 360°
⇒ ∠APE + ∠CQE = 360° – 180° = 180°
Hence ∠APE and ∠CQE are supplementary. Q.E.D.

Question 15.
In the given figure, AB is the diameter of the circle with centre O.

If ∠ADC = 32°, find angle BOC.
Solution:
Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle

∴ ∠AOC = 2 ∠ADC
= 2 x 32° = 64°
∵ ∠AOC + ∠ BOC = 180° (Linear pair)
⇒ 64° + ∠ BOC = 180°
⇒ ∠ BOC=180° – 64° =116° .

Question 16.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.
If ∠ A : ∠ B = 2 : 1 ; find angles A and B.
Solution:
PQRS is a cyclic-quadrilateral in which ∠ PQR =135°
Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.

Question 17.
In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.

Solution:
Given : In the figure, AB is the diameter of the circle with centre O.
CD is the chord with length equal to the radius OA.
AC and BD are produced to meet at P.
To prove : ∠ APB = 60°
Const : Join OC and OD
Proof : ∵ CD = OC = OD (Given)
∴ ∆OCD is an equilateral triangle
∴ ∠ OCD = ∠ ODC = ∠ COD = 60°
In ∆ AOC, OA = OC (Radii of the same circle)
∴ ∠ A = ∠ 1
Similarly, in ∆ BOD,
OB = OD
∴∠2= ∠B
Now in cyclic quadrilateral ACDB,
∠ A CD + ∠B = 180°

∠60°+ ∠ 1 + ∠B = 180°
= ∠ 1 + ∠B = 180° – 60°
⇒∠ 1 + ∠B = 120°
But ∠ 1 = ∠ A
∴ ∠ A + ∠B = 120° …(i)
Now, in ∆ APB,
∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)
⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°
Hence ∠ P = 60° or ∠ APB = 60° Hence proved.

Question 18.
In the following figure,

ABCD is a cyclic quadrilateral in which AD is parallel to BC.
If the bisector of angle A meets BC at point E and the given circle at point F, prove that :
(i) EF = FC
(ii) BF = DF
Solution:

Given : ABCD is a cyclic quadrilateral in which AD || BC.
Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.
To prove :
(i) EF = FC
(ii) BF = DF
Proof : ∵ ABCD is a cyclic -quadrilateral and AD || BC
∵ AF is the bisector of ∠ A
∴ ∠ BAF = ∠ DAF
∴ Arc BF = Arc DF (equal arcs subtends equal angles)
⇒ BF = DF(equal arcs have equal chords)
Hence proved

Question 19.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Given : In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.

Question 20.
The following figure shows a circle with PR as its diameter.

If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)
Solution:
In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter
PQ = 7 cm,
QR = 3 RS = 6cm
∴ 3 RS = 6cm
and RS = \(\frac { 6 }{ 3 }\) = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR² = PQ² + QR² (Pythagoras theorem)
= (7)² + (6)² = 49 + 36 = 85
Again, in right ∆ PSQ, PR² = PS² + RS²
⇒ 85 = PS² + (2)²
⇒ 85 = PS² + 4
⇒ PS² = 85 – 4 = 81 = (9)²
∴ PS = 9cm
Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm

Question 21.
In the following figure, AB is the diameter of a circle with centre O.

If chord AC = chord AD, prove that :
(i)arc BC = arc DB
(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :
(a) ∠ BAC
(b) ∠ ABC
Solution:
Given : In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.
To prove : (i) arc BC = arc DB
(ii) AB is the bisector of ∠CAD
(iii) If arc AC = 2 arc BC, then find
(a) ∠BAC (b) ∠ABC

Construction: Join BC and BD.
Proof : In right angled A ABC and A ABD
Side AC = AD (Given)
Hyp. AB = AB (Common)
∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)
(i) ∴ BC = BD (C.P.C.T)
∴ Arc BC = Arc BD (equal chords have equal arcs)
(ii) ∠ BAC = ∠ BAD (C.P.C.T)
∴ AB is the bisector of ∠CAD.
(iii) If arc AC = 2 arc B
Then ∠ABC = 2 ∠BAC
But ∠ABC – 2 ∠BAC = 90°
∴ 2 ∠BAC + ∠BAC = 90°
⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°
and ∠ABC = 2 ∠BAC = 2 * 30° = 60°

Question 22.
In cyclic-quadrilateral ABCD ; AD = BC,
∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ ADC
Solution:
ABCD is a cyclic-quadrilateral and AD = BC
∠ BAC = 30°, ∠ CBD = 70°
∠ DAC = ∠ CBD , (Angles in the same segment)

But ∠ CBD = 70°
∴ ∠ DAC = 70°
⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°
But ∠ BAD + ∠ BCD = 180°
(Sum of opposite angles of a cyclic quad.)
⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°
∴ ∠ BCD = 80°
∵ AD = BC (Given)
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)
∴ ∠ ADC = 80°
But in ∆ BCD,
∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)
⇒ 70° + 80° + ∠ BDC = ∠ 180°
⇒ 150°+ ∠ BDC = 180°
∴ ∠ BDC = 180° – 150° = 30°
⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)
∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°
∠ ADC + ∠ABC = 180°
(Sum of opp. angles of a cyclic quadrilateral)
⇒ 80°+ABC = 180°
⇒ ∠ ABC=180°-80° = 100°

Question 23.
In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.

Solution:
Now, ∠ ACE = 43° and ∠ CAF = 62° (given)
In ∆ AEC,

∠ ACE + ∠ CAE + ∠ AEC = 180°
∴ 43° + 62° + ∠ AEC = 180°
105° + ∠ AEC = 180°
⇒ ∠ AEC = 180°- 105° = 75°
Now, ∠ ABD + ∠AED=180°
(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)
⇒ 0 + 75°= 180° a = 180° – 75° = 105°
∠ EDF = ∠ BAE (Angles in the alternate segments)
∴ c = 62°
In ∆BAF, ∠a + 62° + ∠b = 180°
⇒ 105°+ 62°+ ∠b= 180°
⇒ 167° + ∠6 = 180°
⇒ ∠b= 180°-167°= 13°
Hence, a= 105°, 6=13° and c = 62°.

Question 24.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .

Find:
(i) ∠CAD
(ii)∠CBD
(iii) ∠ADC
Solution:

In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
and Ext. ∠BCE = ∠BAD
{ Ext. angle of a cyclic quad, is equal to interior opposite angle}
∴ ∠BAD = 80° (∵ ∠BCE = 80°)
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°
(ii) ∠CBD = ∠CAD (Angles in the same segment)
= 55°
(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)
= 180°- ∠BCE =180°- 80° = 100°

Question 25.
ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.
Solution:

Given : In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle
To prove: ∠APB = 60°
Construction : Join OC and OD
Proof: ∵ chord CD = CO = DO (radii of the circle)
∴ ∆DOC is an equilateral triangle
∠DOC = ∠ODC = ∠OCD – 60°
Let ∠A = x and ∠B = y
∵ OA = OD = OC = OB (radii of the same circle)
∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y
∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y
But AOB is a straight line
∴ ∠AOD + ∠BOC + ∠COD = 180°
180°- 2x + 180° -2y + 60° = 180°
⇒ 2x + 2y = 240°
⇒ x + y = 120°
But ∠A + ∠B + ∠P = 180° (Angles of a triangle)
⇒ 120° + ∠P = 180°
⇒ ∠P = 180° – 120° = 60°
Hence ∠APB = 60°

Question 26.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.

Solution:
Given : In the figure,
CP is the bisector of ∠ACB
To prove : DP is the bisector of ∠ADB
Proof: ∵ CP is the bisector of
∴ ∠ACB ∠ACP = ∠BCP
But ∠ACP = ∠ADP {Angles in the same segment of the circle}
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ ∠ADP = ∠BDP
∴ DP is the bisector of ∠ADB

Question 27.
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :
(i) ∠BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ADB

Solution:
In the figure,
ABCD is a cyclic quadrilateral
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
∠CAD = ∠CBD = 70°
(Angles in the same segment)
Similarly ∠BAC = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°
(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)
⇒ ∠BCD + 100°= 180°
⇒ ∠BCD = 180°-100° = 80°
(ii) ∵ AD = BC (given)
∴ ∠ABCD is an isosceles trapezium
and AB || DC
∴ ∠BAC = ∠DCA (alternate angles)
⇒ ∠DCA = 30°
∠ABD = ∠D AC = 30° (Angles in the same segment)
∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°
(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°
(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)

P.Q.
In the given below figure AB and CD are parallel chords and O is the centre.
If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

Solution:
Given : AB = 24 cm, CD = 18 cm
⇒AM = 12 cm, CN = 9 cm
Also, OA = OC = 15 cm
Let MO = y cm, and ON = x cm
In right angled ∆AMO

(OA)² = (AM)² + (OM)²
(15)² = (12)² + (y²
⇒ (15)²-(12)²
⇒ y² = 225-144
⇒ y² = 81 = 9 cm
In right angled ∆CON
(OC)² = (ON)² + (CN)²
⇒(15 )²= x2 + (9)²
⇒ x² = 225-81
⇒x²= 144
⇒ x = 12 cm
Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm

Question 28.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:

(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.
∴ ∠ODB = ∠CBD = 32° (Alternate angles)
In ∆OBD,
OD = OB (Radii of the same circle)
⇒ ∠ODB = ∠OBD = 32°
(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.
∴∠AOB = ∠OBC (Alternate angles)
∠OBC = ∠OBD + ∠DBC
⇒ ∠OBC = 32° + 32°
⇒ ∠OBC = 64°
∴ ∠AOB = 64°
(iii) In AOAB,
OA = OB(Radii of the same circle)
∴ ∠OAB = ∠OBA = x (say)
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 64° = 180°
⇒ 2x = 180° – 64°
⇒ 2x= 116°
⇒ x = 58°
∴ ∠OAB = 58°
That is ∠DAB = 58°
∴ ∠DAB = ∠BED = 58°
(Angles inscribed in the same arc are equal)

Question 29.
In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.

(i) Prove ∆TPS ~ ∆TRQ
(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.
(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)
Solution:
(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°
(Since sum of the opposite angles of a cyclic quadrilateral is 180°)
⇒ ∠RQP = 180° – ∠RSP …(i)
∠RQT + ∠RQP = 180°
(Since angles from a linear pair)
⇒ ∠RQP = 180° – ∠RQT …(ii)
From (i) and (ii),
180° – ∠RSP = 180° – ∠RQT
⇒ ∠RSP = ∠RQT …(iii)
In ∆TPS and ∆TRQ,
∠PTS = ∠RTQ (common angle)
∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)
(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that

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Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

Question 1.
Prove that, of any two chords of a circle, the greater chord is nearer to the centre.
Solution:
Given: In circle with centre O and radius r.
OM ⊥ AB and ON ⊥ CD and AB > CD
To Prove: OM < ON
Construction: Join OA, OC

Question 2.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(i) If the radius of the circle is 10 cm, find the area of the rhombus.
(ii) If the area of the rhombus is 32√3 cm² find the radius of the circle.
Solution:

Question 3.
Two circles with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:


Two circles with centres A and B touch each other at C internally.
PQ is the perpendicular bisector of AB meeting the bigger circles at P and Q. Join AP.
Radius AC = 5 cm.
and radius BC = 3 cm.
AB = AC – BC = 5-3 = 2 cm.

Question 4.
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of angle BAC.
Solution:
Given: A circle in which two chords AC and AB are equal in length. AL is the bisector of ∠ BAC.
To Prove: O lies on the bisector of ∠ BAC
Proof: In ∆ ADC and ∆ ADB,

AD = AD (Common)
AB = AC (Given)
∠ BAD = ∠ CAD (Given)
∴ ∆ ADC = ∆ ADB (SAS postulate)
∴ BD = DC (C.P.C.T.)
and ∠ ADB = ∠ ADC (C.P.C.T.)
But ∠ ADB + ∠ ADC = 180° (Linear pair)
∴ ∠ ADB = ∠ ADC = 90°
∴ AD is the perpendicular bisector of chord BC.
∵ The perpendicular bisector of a chord passes through the centre of the circle.
∴ AD is the bisector of ∠ BAC passes through the centre O of the circle. Q.E.D.

Question 5.
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle ?
Solution:
AB is the diameter and AC is the chord
∴ AB = 20 cm and AC = 12 cm
Draw OL ⊥ AC
∵ OL ⊥ AC and hence it bisects AC, O is the center of the circle.

Question 6.
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:

ABCD is a cyclic quad, in which AD || BC
∠ ADC = 110°, ∠ BAC = 50°
∠B + ∠D= 180° (Sum of opposite angles of a cyclic quad.)
⇒∠B + 110°= 180°
∴ ∠ B or ∠ ABC = 180° – 110° – 70″
Now. in ∆ ABC,
∠ BAC + ∠ ABC – ∠ ACB = 180°
⇒ 50° + 70° + ∠ ACB – 180°
⇒ 120° – ∠ ACB = 180°
∴ ∠ ACB = 180° – 120° = 60″
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
OL ⊥ AC and hence it bisects AC, O is the center of the circle.
∵ AD || BC (Given)
∴ ∠ DAC = ∠ ACB (Alternate angles)
= 60°
Now, in ∆ ADC,
∠ DAC + ∠ ADC + ∠ DCA -= 180°
60°+ 110° + ∠ DCA = 180°
170° + ∠ DCA – 180°
∴ ∠ DCA = 180″ – 170° = 10°

Question 7.
In the given figure, C and D arc points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°. calculate angle BDC.

Solution:
∴ ABCD is a cyclic quad.
∠ BAD ∠ BCD 180 (Sum of opposite angles)
⇒ 70° + ∠ BCD = 180°
⇒ ∠ BCD = 180°- 70° = 110°
Now in ∆ BCD,
∠ BCD + ∠ DBC + ∠ BDC = 180°
⇒ 30°+ 110° + ∠ BDC = 180°
⇒ 140°+ ∠ BDC = 180°
∴ ∠ BDC = 180°- 140° = 40°

Question 8.
In cyclic quadrilateral ABCD, ∠ A = 3 ∠ C and ∠ D = 5 ∠ B. Find the measure of each angle of the quadrilateral.
Solution:

Question 9.
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:

Given: ∆ABC in which AB=AC and AB as diameter, a circle is drawn which intersects BC at D.
To Prove: BD = DC
Construction: Join AD
Proof: AB is the diameter
∴ ∠ ADB = 90° (Angle in a semi-circle)
But ∠ ADB + ∠ ADC =180° (A linear pair)
∴ ∠ ADC = 90°
Now in right angled ∆s ABD and ACD,
Hypotenuse AB = AC (given)
Side AD = AD (Common)
∴ ∆ ABD ≅ ∆ ACD (RHS postulate)
BD = DC (C.P.C.T.)
Hence the circle bisects base BC at D. Q.ED.

Question 10.
Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that.
angle EDF = 90° – \(\frac { 1 }{ 2 }\)∠A.
Solution:
Given: A ∆ ABC whose bisectors of angles A, B and C intersect the circumcircle at D, E and F respectively. ED, EF and DF are joined.
∠ EDF = 90° – \(\frac { 1 }{ 2 }\)∠A
Construction: Join BF, FA, AE and EC.
Proof: ∠ EBF = ∠ ECF = ∠ EDF ……..(i)
(Angles in the same segment)

Question 11.
In the flgure; AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠ C = 20°, find angle AOD.

Solution:

Join OB
In ∆ OBC, BC = OD = OB (radii of the circle)
∴ ∠ BOC = ∠ BCO = 20°
and Ext. ∠ ABO = ∠ BCO + ∠ BOC
= 20°+ 20° = 40° ….(i)
In ∆ OAB, OA = OB (radii of the circle)
∴ ∠ OAB = ∠ OBA = 40° [from (i)]
∠ AOB = 180°- ∠ OAB – ∠ OBA
= 180°-40°-40°= 100°
∵ DOC is a line
∴ ∠ AOD + ∠ AOB + ∠ BOC = 180°
⇒ ∠ AOD + 100° + 20° = 180°
⇒ ∠ AOD + 120° = 180°
∴ ∠ AOD = 180° – 120° = 60°

Question 12.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution:

Given: In ∆ ABC, ∠ B = 90° which is inscribed in a circle and O is the incentre of the incircle of ∆ABC.
D and d are the diameters of circumcircle and incircle of ∆ ABC.
To Prove: AB +BC + CA = 2D + d.
Construction: Join OL, OM and ON.
Proof: In circumcircle of ∆ ABC,
∠ B = 90° (given)
∴ AB is the diameter of circumcircle i.e. AB = D.
Let radius of incircle = r
∴ OL = OM = ON = r
Now from B, BL, BM are the tangents to the incircle
∴ BL = OM = r
Similarly we can prove that:
AM = AN and CL = CN = R (radius)
(Tagents from the point outside the circle)
Now AB + BC + CA = AM + BM + BL + CL + CA
= AN + r + r + CN + CA
= AN + CN + 2r + CA
= AC + AC + 2r = 2 AC + 2r = 2D + d Q.E.D.

Question 13.
P is the mid point of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution:
Given: A circle with centre O, AB is an arc whose mid point is P and AB is chord. TPS is the tangent at P.
To Prove: TPS ||AB.
Construction: Join AP and BP.
Prove: TPS is tangent and PA is chord of the circle

∠ APT = ∠ PBA (angles in the alternate segment)
But ∠ PBA = ∠ PAB (∵ PA = BP)
∴ ∠ APT = ∠ PAB
But these are alternate angles
∴ TPS || AB Q.E.D

Question 14.
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent Prove that the line NM produced bisects AB at P.

Prove that the line NM produced bisects AB at P.
Solution:
Given: Two circles intersect each other at M and N. AB is their common tangent, chord MN intersect the tangent at P.
To Prove: P is mid point of AB.
Proof: From P, AP is the tangent and PMN is the secant of first circle.
∴ AP² = PM x PN ….(i)
Again from P, PB is the tangent and PMN is the secant of the second circle.
PB2 = PM x PN ….(ii)
from (i) and (ii)
AP² = PB² ⇒ AP = PB
∴ P is the mid point of AB. Q.E.D.

Question 15.
In the given figure, ABCD is a cyclic- quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠ DCQ = 40° and ∠ ABD = 60°, find :
(i) ∠ DBC
(ii) ∠ BCP
(iii) ∠ ADB

Solution:
(i) PQ is the tangent and CD is the chord
∴ ∠ DCQ = ∠ DBC (angles in the alternate segment)
∴ ∠ DBC = 40° (∵ ∠ DCQ = 40°)
(ii) ∠ DCQ + ∠ DCB + ∠ BCP = 180°
⇒ 40° + 900 + ∠ BCP = 180°(∵ ∠ DCB = 90°)
⇒ 130°+ ∠ BCP = 180°
∵ ∠ BCP =180° -130° = 50°
(iii) In Δ ABD, ∠ BAD = 90° (Angle in a semi circle) and ∠ ABD = 60°
∴ ∠ ADB = 180°- (60° + 90°)
⇒ 1800- 150° = 30°

Question 16.
The given figure shows a circle with centre O and BCD is tangent to it at C. Show that:
∠ ACD + ∠ BAC = 90°.

Solution:
Given: A circle with centie O and BCD is a tangent at C.

To Prove: ∠ ACD + ∠ BAC = 90°
Construction: Join OC.
Proof: BCD is the tangent and OC is the radius
∴ OC ⊥ BD
⇒ ∠ OCD = 90°
⇒ ∠ OCA + ∠ ACD = 90° ….(i)
But in ∆ OCA
OA = OC (radii of the same circle)
∴ ∠ OCA = ∠ OAC [from (i)]
∠OAC + ∠ACD = 90°
⇒ ∠ BAC + ∠ ACD = 90° Q.E.D.

Question 17.
ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that 
(i) AC x AD = AB²
(ii) BD² = AD x DC.

Solution:
Given: A circle with BC as diameter meets the hypotenuse of right ∆ ABC with ∠ B = 90° meets at D. BD is joined.
To Prove:
(i) AC x AD = AB²
(ii) BD² = AD x DC
Proof:
(i) In ∆ABC, ∠ B = 90° and BC is the diameter of the circle.

Question 18.
In the given figure. AC = AE.
Show that :
(i) CP = EP
(ii) BP = DP.

Solution:
Given: In the figure, AC = AE
To Prove: (i) CP = EP (ii) BP = DP
Proof: In ∆ ADC and ∆ ABE,
AC = AE (given)
∠ ACD = ∠ AEB (Angles in the same segment)
∠ A = ∠ A (Common)
∆ ADC ≅ ∆ ABE (ASA postulate)
AB = AD (C.P.C.T.)
But AC = AE (given)
∴ AC – AB = AE – AD
⇒ BC = DE
Now in ∆ BPC and ∆ DPE,
BC = DE (proved)
∠ C = ∠ E (Angles in the same segment)
∠ CBP = ∠ CDE (Angles in the same segment)
∴ ∆ BPC ≅ ∆ DPE (S.A.S. postulate)
∴ BP=DP (C.P.C.T.)
CP = PE (C.P.C.T.) Q.E.D.

Question 19.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.
Calculate :
(i) ∠ BEC
(ii) ∠ BED
Solution:
(i) In cyclic pentagon, O is the centre of circle.
Join OB, OC
AB = BC = CD (given)
and ∠ ABC = 120°.
∴ ∠ BCD = ∠ ABC = 120°
OB and OC are the bisectors of ∠ ABC and ∠ BCD respectively.
∴ ∠ OBC = ∠ BCO = 60°
In ∆ BOC,

Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO=30°, find
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB

Solution:
Given: In the fig. O is the centre of the circle CA and CB are the tangents to the circle from C. ∠ACO = 30°
P is any point on the circle. PA and PB are joined.
To find:
(i) ∠BCO
(ii) ∠AOB
(iii) ∠APB
Proof:
(i) In ∆ OAC and ∆OBC,
OC=OC (common)
OA = OB (radius of the circle)
CA = CB (tangents to the circle)
∴ ∆OAC ≅ ∆OBC (SSS axion)
∴ ∠ACO = ∠BCO = 30°
(ii) ∴ ∠ACB = 30° + 30° = 60°
∴ ∠AOB + ∠ACB = 180°
⇒ ∠AOB+ 60° =180°
∴ ∠AOB = 180° – 60° = 120°
(iii) Arc AB, subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle
∴ ∠APB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) x 120° = 60°

Question 21.
ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. (2011)

Solution:
Given: ABC is a triangle with AB = 10 cm, BC = 8 cm, AC = 6 cm. Three circle are drawn with centre A, B and C touch each other at P, Q and R respectively

Question 22.
In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L.
Show that :
(i) ∠ ONL + ∠ OML = 180°
(ii) ∠ BAM = ∠ BMA
(iii) ALOB is a cyclic quadrilateral
Solution:

Question 23.
The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Solution:

Join PB
In cyclic quad. PBCQ,
∠ BPQ + ∠ BCQ = 180°
⇒ ∠ BPQ + 140° = 180°
∴ ∠ BPQ = 180° – 140° = 40°
Now, in ∆ PBQ,
∠ PBQ + ∠ BPQ + ∠ BQP = 180°
⇒ 90° + 40° + ∠ BQP = 180°
(∠ PBQ = 90° angle in a semicircle)
⇒ 130° + ∠ BQP = 180°
∴ ∠ BQP = 180° – 130°= 50°
Nowin cyclic quad. PQBA,
∠ PQB + ∠ PAB = 180°
⇒ 50° + ∠ PAB = 180°
∴ ∠ PAB = 180° – 50° = 130°
(ii) Now, in ∆ PAB,
∠ PAB + ∠ APB + ∠ ABP = 180°
⇒ 130° + ∠ APB + ∠ ABP = 180°
⇒ ∠ APB + ∠ ABP = 180° – 130° = 50°
But ∠ APB = ∠ ABP = 25°
∴ (PA = AB)
∠ BAQ = ∠ BPQ = 40°
(Angles in the same segment)
Now, in ∆ ABQ,
∠ AQB + ∠ QAB + ∠ ABQ = 180°
⇒ ∠ AQB + 40° + 115° = 180°
⇒ ∠ AQB + 155° = 180°
⇒ ∠ AQB = 180° – 155° = 25°
(iii) Arc AQ subtends ∠ AOQ at the centre and ∠APQ at the remaining part of the circle,
∠ AOQ = 2 ∠ APQ = 2 x 65° = 130°
Now, in ∆ AOQ,
∠ OAQ = ∠ OQA
(∵ OA = OQ radii of the same circle)
But ∠ OAQ + ∠ OQA + ∠ AOQ = 180°
⇒ ∠ OAQ + ∠ OAQ + 130° = 180°
2 ∠ OAQ = 180° – 130° = 50°
∴ ∠ OAQ = 25°
∵ ∠ OAQ = ∠ AQB (each = 25°)
But these are alternate angles.
∴ AO || BQ.
Q.E.D.

Question 24.
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate
(i) angle QTR
(ii) angle QRP
(iii) angle QRS
(iv) angle STR

Solution:

Question 25.
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segemcnt, show that :
(i) ∠ BAP = ∠ ADQ
(ii) ∠ AOB = 2 ∠ ADQ
(iii) ∠ ADQ = ∠ ADB

Solution:
Given: PAT is the tangent to the circle with centre O, at A. Chord BC || PAT is drawn.
CDQ is a line segment which intersects the circle at C and D and meets the tangent PAT at Q.
To Prove:
(i) ∠ BAP = ∠ ADQ (ii)∠ AOB = 2 ∠ ADQ
(iii) ∠ ADQ = ∠ ADB
Proof:
(i) ∵ PAT || BC
∴ ∠ PAB = ∠ ABC (Alternate angles) ….(i)
In cyclic quad. ABCD,
Ext. ∠ADQ = ∠ABC ….(ii)
∴ ∠ PAB = ∠ ADQ (from (i) and (ii))
(ii) Arc AB subtends ∠ AOB at the centre and ∠ADB at the remaining part of the circle.
∴ ∠ AOB = 2 ∠ ADB
= 2 ∠ PAB (In the alt. segment)
= 2 ∠ ADQ [proved in (i)]
(iii) ∵ ∠ BAP = ∠ ADB
(Angles in the alt. segment)
Bui ∠ BAP = ∠ ADQ (Proved in (i))
∴ ∠ ADQ = ∠ ADB Q.E.D.

Question 26.
AB is a line segment and M is its mid-point. Three semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semi-circles. Show that: AB = 6 x r
Solution:

Given: A line segment AB whose mid-point is M. Three circles are drawn on AB, AM and MB as diameler A circle with radius r is drawn which touches (he three circles externally at L, R and N respectively. M, P, Q are the centres of the three circles.
To Prove: AB= 6r
Construction: Join OP and OQ.
Proof: OM = ON = r

Question 27.
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution:

Given: A circle with centre C. From a point T outside th circle, TA and TB are two tangent to the circle OT intersects the circle at P, AP and AB are joined.
To Prove: AP is the bisector of ∠ TAB
Construction: Join PB.

Question 28.
Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution:

Given: Two circles intersect each other at P and Q. From P, a secant intersects the circles at A and B respectively. From A and B tangents are drawn which intersect each other at T.
To Prove: A. Q, B and T on a circle.
Construction: Join PQ.
Proof: AT is the tangent and AP is chord
∴ ∠ TAP = ∠ AQP (Angles in all. segment) …(i)
Similarly ∠ TBP = ∠ BQP ….(ii)
Adding (i) and (ii),
∠ TAP + ∠ TBP = ∠ AQP + ∠ BQP = ∠ AQB …..(iii)
Now, in ∆ TAB,
∠ ATB + ∠ TAP + ∠ TBP = 180°
⇒ ∠ ATB + ∠ AQB = 180° (from (iii)
∴ AQBT is a cyclic quadrilateral.
Hence A, Q, B and T lie on the same circle. Q.E.D.

Question 29.
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle).
Solution:
ABCDE is a regular pentagon.


To Prove: Any four vertices lie on the same circle.
Construction: Join AC.
Proof: Each angle of a regular pentagon

Question 30.
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD. [2000]
Solution:

Let CD = x
∴ chords AB and CD intersect each other at outside the circle.
∴ AX.XB = CX.XD
⇒(4+6)x6 = (x + 5)x5
⇒ 10 x 6 = 5x + 25
⇒ 60 = 5x + 25
⇒ 5x = 60 – 25 = 35
∴ x = \(\frac { 35 }{ 5 }\) = 7
CD = 7 cm

Question 31.
in the given figure. find TP if AT = 16 cm AB = 12 cm.
Solution:
In the figure.
PT is the tangent and TBA is the secant of the circle.
∴ TP² = TA x TB = 16 x (16 – 12) = 16 x 4 = 64 = (8)²
Hence, TP = 8 cm.

Question 32.
In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm. QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle. (1990)

Solution:
A circle with centre is inscribed (see the fig.)
in a quadrilateral ABCD. BC = 38 cm, QB = 27cm,
DC = 25 cm and AD ⊥ BC.
Join OP and QS.
∵ OP and OS are the radii of the circle
∴ OP ⊥ AD and OS ⊥ CD
∴ OPDS is a square
∴ OP = OS – DP = DS.
Let length of radius of the circle = r
then DP = DS = r
∴ CS = 25 – r
∵ EQ = BR = 27 cm (tangents to the circle from B)
∴ CR = BC – BR = 38 – 27 = 11 cm
Similarly CR = CS
∴ 25 – r = 11 ⇒ r = 25 – 11 = 14
∴ Radius of the circle = 14 cm

Question 33.
In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.

If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.
Solution:
In the above figure,
XY is a diameter of the circle PQ is tangent to the circle at Y.
∠AXB = 50° and ∠ABX = 70°
(i) In ∆AXB,
∠XAB + ∠ABX + ∠AXB = 180° (Angles of a triangle)
⇒ ∠XAB + 70° + 50° = 180°
⇒ ∠XAB + 120° =180°
⇒ ∠XAB = 180° – 120° = 60°
But, ∠XAY = 90° (Angle in a semicircle)
∴ ∠BAY = ∠XAY-∠XAB=90°- 60° = 30°
(ii) Similarly ∠XBY=90°(Angle in a semicircle) and ∠CXB = 70°
∴ ∠PBY = ∠XBY-∠XBA =90° – 70° = 20°
∵ ∠BYA = 180° – ∠AXB ( ∵ ∠BYA + ∠AYB = 180°) = 180°- 50° = 130°
∠PYA =∠ABY (Angles in the alternate segment) = ∠PBY = 20°
and ∠PYB = ∠PYA + ∠AYB
= 20° + 130° = 150°
∴ ∠APY = 180°-(∠PYA + ∠ABY)
= 180° -(150° +20°) =180° – 170° = 10°

Question 34.
In the given figi QAP is the tangent point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°, find:
(i) ∠ BAP
(ii) ∠ABD
(iii) ∠ QAD
(iv) ∠ BCD

Solution:
In the given figure, QAP is the tangent to the circle at A and PBD is a B straight line.

Question 35.
In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If ∠ CAB = 34°,
find :
(i) ∠CBA
(ii) ∠CQB
Solution:

In ∆ ABC,
we have ∠ ACB = 90°
[Angle in a semicircle is 90°]
(i) Also ∠ CBA + ∠ CAB + ∠ ACB = 180° [Angle sum property of a ∆ ]
⇒ ∠ CBA =180°- ∠ CAB – ∠ ACB = 180°-34°-90° = 180°-124° = 56°
(ii) CQ is a tangent at C and CB is a chord of the circle.
⇒ ∴ ∠ QCB = ∠ BAC = 34° [Angles in the alternate segments]
∠ CBQ =180°- ∠ ABC [Linear pair]
⇒ ∠ CBQ = 180°- 56° = 124° [From (i)]
In ∆ BCQ, we have
⇒ ∠ CQB = 180° – (∠ QCB + ∠ CBQ) [Angle sum property of a ∆ ]
= 180° -(34° + 124°) = 180°- 158° = 22°
Hence, ∠CQB = 22°

Question 36.
In the given figure, A O is the centre of the circle. The tangents at B and D intersect each other at point P.

If AB is parallel to CD and ∠ ABC = 55
find :
(i) ∠ BOD
(ii) ∠ BPD
Solution:

In the given figure,
O is the centre of the circle AB || CD,
∠ ABC = 55° tangents at B and D are drawn which meet at P.
∵ AB || CD
∴ ∠ ABC = ∠ BCD (Alternate angles)
∴ ∠ ABC = 55° (Given)
(i) Now arc BD subtands ∠ BOD at the centre and ∠ BCD at the remaining part of the circle.
∴ ∠BOD = 2∠BCD = 2 x 55° = 110°
(ii) In quad. OBPD,
∠OBP = ∠ODP = 90° (∵ BP and DP are tangents)
∴ ∠BOD + ∠BPD = 180°
⇒ 110° +∠BPD =180°
⇒ ∠BPD =180°-110°= 70°
Hence, ∠BOD = 110° and ∠BPD = 70°

Question 37.
In the figure given below PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
(i) ∠QOP
(ii) ∠QCP
Solution:

Question 38.
In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.

Solution:
Given: Two concentric circles with centre O AB and CD are two cords of outer circle which touch the inner circle at P and Q respectively
To prove: AB = CD
Construction : Join OA, OC, OP and OQ
Proof: ∵ OP and OQ are the radii of the inner circle and AB and CD are tangents
∴ OP ⊥ AB and OQ ⊥ CD
and P and Q are the midpoints of AB and CD Now in right AOAP and OCQ,
Side OP = OQ (radii of the inner circle)
Hyp. OA = OC (radii of the outer circle)
∴ ∆OAP = ∆OCQ (R.H.S. axiom)
∴ AP = CQ (c.p.c.t.)
But AP = \(\frac { 1 }{ 2 }\) AB and CQ = \(\frac { 1 }{ 2 }\) CD
∴ AB = CD Hence proved.

Question 39.
In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.

Given that AB = 5 cm , Calculate PQ.
Solution:
In the figure, two circles with centres P and Q and radii 6 cm and 3 cm respectively
ABC is the common transverse tangent to the two circles. AB = 8 cm
Join AP and CQ
∵ AC is the tangents to the two circles and PA and QC are the radii

Question 40.
In the figure, given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

Solution:
In the figure, a circle with centre O, is the circumcircle of ∆XYZ.
∠XOZ =140° (given)
Tangents from X and Y to the circle meet at T such that ∠XTY = 80°
∵ ∠XTY = 80°
∴ ∠XOY= 180°-80°= 100°
But ∠XOY + ∠YOZ + ∠XOZ = 360° (Angles at a point)
⇒ 100°+∠YOZ+ 140o = 360o
⇒ 240o+∠YOZ =360°
⇒ ∠YOZ =360°- 240°
⇒∠YOZ =120°
Now arc YZ subtends ∠YOZ at the centre and ∠YXZ at the remaining part of the circle
∴ ∠YOZ = 2 ∠YXZ
⇒ ∠YXZ= \(\frac { 1 }{ 2 }\) ∠YOZ ⇒ ∠YXZ = \(\frac { 1 }{ 2 }\) x 120° = 60°

Question 41.
In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, findAE.

Solution:
In the given circle,
Chords AE and BC interesect each other at D at right angle i.e., ∠CDE = 90°, AB is joined AB = 5cm, BD = 4 cm, CD = 9 cm
Now we have to find AE.
Let DE=xm
Now in right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
⇒ (5)²=AD² + (4)²
⇒ 25 = AD² + 16
⇒ AD² = 25-16 = 9 = (3)²
∴AD = 3cm
∵ Chords AE and BC intersect eachothcr at D inside the circle
∴ AD x DE = BD x DC
⇒ 3 x x = 4 x 9
⇒ x= \(\frac { 4×9 }{ 3 }\) = 12cm;
∴ AE=AD + DE = 3 + 12 = 15 cm

Question 42.
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

Solution:
ABCD is a cyclic quadrilateral
∴ ∠ABC + ∠ADC = 180°
100° +∠ADC = 180°
∠ADC = 180°- 100° = 80°
In ∆ADC
∠ACD + ∠CDA + ∠D AC =180°
40° + 80° +∠D AC = 180°
∠D AC = 180° – 80° – 40° = 60°
Now ∠DAC = 60°
⇒ ∠DCT = 60° [angle in alt. segment]

Question 43.
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x,y and z. (2015)

Solution:
In the given figure,
O is the centre of the circle.
SP is tangent ∠SRT =65°.
To find the values of x, y and z
(i) In ∆STR,
∠S = 90° (∵ OS is radius and ST is tangent)
∴ ∠T + ∠R = 90°
⇒ x + 65° = 90°
⇒ x = 90° – 65° = 25°
(ii) Arc CQ subtends ∠SOQ at the centre and
∠STQ at the remaining part of the circle.
∠y = ∠QOS = 2∠T = 2∠x = 2 x 25° = 50°
(iii) In ∆OSP,
∠S = 90°
∴ ∠SOQ or ∠SOP + ∠P = 90°
⇒ y+z=90o
⇒ 50° + z = 90°
⇒ z = 90°-50° = 40°
Hence x = 25°, y = 50° and z = 40°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C  are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

 

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

Question 1.
The radius of a circle is 8cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10cm from its centre.
Solution:

OP = 10 cm,
raduis OT = 8 cm
∵ OT ⊥ PT
∴ In right ΔOTP,
OP² = OT²+PT²
⇒  (10)² =(8)²+PT²
⇒  100 = 64+PT²
⇒ PT² = 100-64 = 36
∴ PT = \( \sqrt{36} \) = 6 cm

Question 2.
In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.


Solution:

∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)
⇒  OB² = OA² – AB² ⇒ r² = (r + 7.5)² – 15²
⇒ r² = r² + 56.25 + 15r – 225 168.75
⇒   15r= 168.75
⇒  r =\(\frac { 168.75 }{ 75 }\) ⇒ r=11.25
Hence, radius of the circle = 11.25 cm

Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.

Solution:
Given: Two circles with centre O and O’ touches at P externally. Q is a point on the common tangent through P.
QA and QB are tangents from Q to the circles respectively.
To Prove: QA=QB.
Proof: From Q, QA and QP are the tangents to the circle with centre O
∴  QA=QP ….(i)
Similarly, QP and QB are the tangents to the circle with centre O’
∴ QP=QB   ….(ii)
From (i) and (ii)
QA=QB                           Q.ED.

Question 4.
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent, are equal in length.
Solution:
Given: Two circles with centre O and O’ touch each other internally at P. Q is a point on the common tangent through P. QP an QB are tangents from Q to the circles respectively.

Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.

Solution:
Given: Two concentric circles with radius 5 cm and 3 cm with centre O. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.
OL=3 cm, OP = 5 cm

Question 6.
Three circles touch each other externally. A triangle is formed when the centres of these circles are j oined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:
Three circles touches each other externally
Δ ABC is formed by joining the centres A, B and C of the circles.
AB = 6 cm, AC = 8 cm and BC = 9 cm
Let radii of the circles having centres A, B and C be r₁, r₂, r₃ respectively

Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that: AB + CD = BC + AD.

Solution:
Given: A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q,R and S respectively.

To Prove: AB + CD = BC + AD
Proof: Since AP and AS are the tangents to the circle from external point A.
∴ AP = AS     ….(i)
Similarly, we can prove that,
BP=BQ               ….(ii)
CR=CQ             …(iii)
DR=DS          ….(iv)
Adding, we get:
AP + BP + CR + DR = AS + BQ + CQ + DS
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence AB + CD = BC + AD.     Q.E.D.

Question 8.
If the sides of a parallelogram touch a circle (refer figure of Q/7) prove that the parallelogram is a rhombus.
Solution:
Given : The sides AB, BC, CD and DA of ||gm ABCD touches the circle at P, Q, R and S respectively.

To Prove : ABCD is a rhombus.
Proof : From A, AP and AS are the tangents to the circle.
∴ AP = AS    ….(i)

Question 9.
From the given figure, prove that:
AP + BQ + CR = BP + CQ + AR.

Solution:
Given: In the figure, sides of Δ ABC touch a circle at P, Q, R.
To Prove:
(i) AP + BQ + CR = BP + CQ + AR
(ii) AP + BQ + CR = \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.
Proof :
∵  From B, BQ and BP are the tangents to the circle.
∴ BQ = BP   ….(i)
Similarly we can prove that
AP= AR    ……… (ii)
and CR = CQ  …..(iii)
Adding we get
AP + BQ + CR = BP + CQ + AR ….(iv)
Adding AP + BQ + CR both sides,
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR+CR.
⇒  2 (AP + BQ + CR) = AB + BC + CA
∴ AP + BQ + CR = \(\frac { 1 }{ 2 }\) (AB + BC + CA)
= \(\frac { 1 }{ 2 }\) Perimeter of Δ ABC.            Q.E.D.

Question 10.
In the figure of Q.9 if AB = AC then prove that BQ = CQ.
Solution:
Given:  A circle touches the sides AB, BC, CA of Δ ABC at P, Q and R respectively, and AB = AC

Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centres if 
(i) they touch each other externally,
(ii) they touch each other internally.
Solution:
Radius of bigger circle = 6.3 cm

and raduis of smaller circle = 3.6 cm.
(i) Two circles touch each other at P externally. 0 and O’ are the centres of the circles.
Join OP and OP’
OP = 3.6 cm, O’P = 6.3 cm.
Adding we get
OO’ = OP + O’P = 3.6 + 6.3 = 9.9 cm
(ii) If the circles touch each other internally at P.
OP = 3.6 cm and O’P = 6.3 cm.

∴ OO’ = O’P – OP
= 6.3 – 3.6 = 2.7 cm

Question 12.
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
(i) ∠AOP = ∠BOP,
(ii) OP is the ⊥ bisector of chord AB.
Solution:
Given:  A circle with centre 0. A point P out side the circle. From P, PA and PB are the tangents to the circle, OP and AB are joined.

To prove:
(i) ∠AOP = ∠BOP
(ii) OP is the perpendicular bisector of chord AB.
Proof : In ∆ AOP and ∆ BOP,
AP = BP (Tangents from P to the circle.)
OP = OP (Common)
OA = OB (Radii of the same circle)
∴ ∆ AOP s ∆ BOP (SSS postulate)
∴∠AOP = ∠BOP (C.P.C.T.)
Now in ∆ OAM and ∆ OBM,
OA = OB (Radii of the same circle)
OM = OM (Common)
∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)
∴ ∆ OAM = ∆ OBM (S.A.S. Postulate)
∴ AM = MB (C.P.C.T.)
and ∠OMA = ∠OMB (C.P.C.T.)
But ∠OMA + ∠OMB = 180° (Linear pair)
∴ ∠OMA = ∠OMB = 90°
Hence OM or OP is the perpendicular bisector of AB. Q.E.D.

Question 13.
In the given figure, two circles touch each other externally at point P, AB is the direct common tangent of these circles. Prove that:
(i) tangent at point P bisects AB.
(ii) Angle APB = 90°

Solution:
Given : Two circles with centre O and O’ touch each other at P externally. AB is the direct common tangent touching the circles at A and B respectively.

AP, BP are joined. TPT’ is the common tangent to the circles.
To Prove : (i) TPT’ bisects AB (ii) ∠APB = 90°
Proof :
∵ TA and TP arc the tangents to the circle
∴ TA = TP …(i)
Similarly TP. = TB ….(ii)
From (i) and (ii)
TA = TB
∴ TPT’ is the bisector of AB.
Now in ∆ ATP
TA = TP
∴ ∠TAP = ∠TPA
Similarly in A BTP.
∠TBP = ∠TPB
Adding we get.
∠TAP + ∠TBP = ∠APB
But ∠TAP + ∠TBP + ∠APB = 180°
∴ ∠APB = ∠TAP + ∠TBP = 90°. Q.E.D.

Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:
Given: A circle with centre O. two tangents PA and QA are drawn from a point A out side the circle OP, OQ. OA and PQ are joined.

Question 15.
Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.
Solution:
Given: A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined

Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.

Calculate the value of x, the radius of the inscribed circle.
Solution:
In ∆ ABC, ∠B = 90°
OL ⊥ AB, OM ⊥ BC and
ON ⊥ AC.

Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :
(i) ∠ QOR
(ii) ∠ QPR given that ∠ A = 60°.
Solution:

Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60°, calculate:
(i) ∠ QOR
(ii) ∠ OQR
(iii) ∠ QSR.

Solution:

Question 19.
In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.

Solution:

Question 20.
In quadrilateral ABCD; angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle. |1990]
Solution:
BQ and BR arc the tangenls from B to the circle.

∴ BR = BQ = 27 cm.
∴ RC = 38-27 = 11 cm.
Since CR and CS are the tangents from C to the circle
∴ CS = CR= 11 cm.
∴ DS = 25 – 11 = 14 cm.
DS and DP are the tangents to the circle
∴ DS = DP
∴ ∠ PDS = 90° (given)
and OP ⊥ AD, OS ⊥ DC
∴ Radius = DS = 14 cm

Question 21.
In the given, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given ∠ SPR = x° and ∠ QRP = y°;
Prove that
(i) ∠ ORS = y°
(ii) Write an expression connecting x and y. [1992]

Solution:
∠ QRP = ∠ OSR = y (Angles in the alternate segment)

But OS = OR (radii of the same circle)
∴ ∠ ORS = ∠ OSR = y°
∴ OQ = OR (radii of the same circle)
∴ ∠ OQR = ∠ ORQ = 90° – y° ….(i) (OR ⊥ PT)
But in ∆ PQR,
Ext. ∠ OQR = x° + y° ….(ii)
from (i) and (ii)
x° + y° = 90° – y°
⇒ x° + 2y° = 90°

Question 22.
PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50°; calculate :

Solution:

Question 23.
In the given figure. O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC 80°; find the angle BAC. [1996]

Solution:

Join OC
∴ PA and PC are the tangents
∴ OA ⊥ PA and OC ⊥ PC
In quad APCO,
∴ ∠ APC + ∠ AOC = 180°
⇒ 80° + ∠ AOC = 180°
∴ ∠ AOC = 180° – 80° = 100°
∠ BOC = 360° – (∠ AOB + ∠ AOC)
= 360° – (140° + 100°)
= 360° – 240° = 120°
Now arc BC subtends ∠ BOC at the centre and ∠BAC at the remaining part of the circle.
∴ ∠ BAC = \(\frac { 1 }{ 2 }\)∠BOC
= \(\frac { 1 }{ 2 }\) x 120° = 60°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A  are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B 

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B.

Other Exercises

• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18A
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B
• Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18C

Question 1.
(i) In the given figure 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

(ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4 cm. Find CD.

(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

Solution:

Question 2.
In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. (2014)

Solution:
PT is tangent and PDC is secant out to the circle
∴ PT² = PC x PD
PT² = (5 + 7.8) x 5 = 12.8 x 5
PT² = 64 ⇒ PT = 8 cm
In ΔOTP
PT² + OT² = OP²
8²+x² = (x + 4)²
⇒ 64 +x² = x² + 16 + 8x
64- 16 = 8x
⇒ 48 = 8x
x = \(\frac { 48 }{ 8 }\) = 6 cm
∴ Radius = 6 cm
AB = 2 x 6 = 12 cm

Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60°, calculate
(i) ∠ QAB
(ii) ∠ PAD
(iii) ∠ CDB.

Solution:
(i) PAQ is a tangent and AB is the chord, ∠ QAB = ∠ ADB (Angles in the alternate segment)
= 30°
(ii) OA = OD (Radii of the same circle)
∴ ∠ OAD = ∠ ODA = 30°
But OA ⊥ PQ
∴ ∠ PAD = ∠ OAP – ∠OAD = 90° – 30° = 60°
(iii) BD is diameter
∴ ∠ BCD = 90° (Angle in semi circle)
Now in ∆ BCD,
∠ CDB + ∠ CBD + ∠ BCD = 180°
⇒ ∠ CDB + 60° + 90° = 180°
⇒ ∠ CDB = 180°- (60° + 90°) = 180° – 150° = 30°

Question 4.
If PQ is a tangent to the circle at R; calculate:
(i) ∠ PRS
(ii) ∠ ROT.
Given O is the centre of the circle and angle TRQ = 30°.
Solution:

PQ is tangent and OR is the radius
∴ OR ⊥ PQ
∴ ∠ORT = 90° =
∠TRQ = 90° – 30° = 60°
But in ∆ OTR, OT = OR (Radii of the circle)
∴ ∠ OTR = 60° or ∠STR = 60°
But ∠PRS = ∠STR (Angles in the alternate segment) = 60°
In ∆ ORT, ∠ OTR = 60°, ∠ TOR = 60°
∴ ∠ROT= 180°-(60°+ 60°)= 180°-120° = 60°

Question 5.
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersect AB produced in D. Show that BC = BD.
Solution:


Given: In a circle, O is the centre,
AB is the diameter, a chord AC such that ∠ BAC = 30°
and a tangent from C, meets AB in D on producing. BC is joined.
To Prove: BC = BD
Construction: Join OC
Proof : ∠ BCD = ∠ BAC =30°(Angle in alternate segment)
Arc BC subtends ∠ DOC at the centre of the circle and ∠ BAC at the remaining part of the circle.
∴ ∠ BOC = 2 ∠ BAC = 2 x 30° = 60°
Now in ∆ OCD,
∠ BOC or ∠ DOC = 60° (Proved)
∠ OCD = 90° (∵ OC ⊥ CD)
∴ ∠ DOC + ∠ ODC = 90°
⇒ 60° + ∠ ODC = 90°
∴ ∠ ODC = 90°- 60° = 30°
Now in ∆BCD,
∵∠ ODC or ∠ BDC = ∠ BCD (Each = 30°)
∴ BC = BD Q.E.D.

Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that ∆PQR is isosceles.
Solution:

In a circumcircle of ∆PQR, a tangent TPS is drawn through P which is parallel to QR
To prove : ∆PQR is an isosceles triangle.
Proof:
∵ TS \(\parallel\) QR
∠TPQ = ∠PQR (Alternate angles) ….(i)
∵ TS is tangent and PQ is the chord of the circle
∴ ∠TPQ = ∠RP (Angles in the alternate segment) ….(ii)
From (i) and (ii),
∠PQR = ∠QRP
∴ PQ = PR (Opposite sides of equal angles)
∴ ∆PQR is an isosceles triangle
Hence proved

Question 7.
Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.

Solution:

Given: Two circles with centre O and O’ intersect each other at A and B, O lies on the circumference, of the other circle. CD is a tangent at A to the second circle. AB, OA are joined.
To Prove: OA bisects ∠ BAC.
Construction: Join OB, O’A, O’B and OO’
Proof: CD is the tangent and AO is the chord
∠ OAC = ∠ OBA …(i)
(Angles in alt. segment)
In ∆ OAB, OA = OB (Radii of the same circle)
∴ ∠ OAB = ∠ OBA ….(ii)
From (i) and (ii),
∠ OAC = ∠ OAB
∴ OA is the bisector of ∠ BAC Q.E.D.

Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠ CPA = ∠DPB.

Solution:
Given:Two circles touch each other internally at P. A chord AB of the bigger circle intersects the smaller circle at C and D. AP, BP, CP and DP are joined.
To Prove: ∠ CPA = ∠ DPB
Construction: Draw a tangent TS at P to the circles given.
Proof:
∵ TPS is the tangent, PD is the chord.
∴ ∠ PAB = ∠ BPS …(i) ( Angles in alt. segment)
Similarly we can prove that
∠ PCD = ∠ DPS …(ii)
Subtracting (i) from (ii), we gel
∠ PCD – ∠ PAB = ∠ DPS – ∠ BPS
But in ∆ PAC,
Ext. ∠ PCD = ∠ PAB + ∠ CPA
∴ ∠ PAB + ∠ CPA – ∠ PAB = ∠ DPS – ∠ BPS
∠ CPA = ∠ DPB           Q.E.D.

Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:

Given: ABCD is a cyclic quadrilateral and diagonal AC bisects ∠ BCD. AT A. a tangent TAS is drawn. BD is joined.
To Prove: TS || BD.
Proof: ∠ ADB = ∠ ACB …….(i) (Angles in the same segment)
Similarly ∠ ABD = ∠ ACD ……..(ii)
But ∠ ACB = ∠ ACD (AC is the bisector of ∠ BCD)
∴ ∠ ADB = ∠ ABD |From (i) and (ii)]
TAS is a tangent and AB is chord
∴ ∠ BAS = ∠ ADB (Angles in all segment)
But ∠ ADB = ∠ ABD (Proved)
∴ ∠ BAS = ∠ ABD
But these are alternate angles.
∴ TS || BD.      Q.E.D.

Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠ BCG = 108° and O is the centre of the circle,
Find:
(i) angle BCT
(ii) angle DOC

Solution:

Join OC, OD and AC.
(i) ∠ BCG + ∠ BCD = 180° (Linear pair)
⇒ 108° + ∠ BCD = 180°
(∵∠ BCG = 108° given)
∴∠ BCD = 180° – 108° = 72°
BC = CD (given)
∴ ∠ DCP = ∠ BCT
But ∠ BCT + ∠ BCD + ∠ DCP = 180°
∴∠ BCT + ∠ BCT + 72° = 180°
(∵∠ DCP = ∠ BCT)
2 ∠ BCT = 180° – 72° = 108°
∴∠ BCT = \(\frac { { 108 }^{ \circ } }{ 2 }\) = 54°
(ii) PCT is the tangent and CA is chord
∴ ∠ CAD = ∠ BCT = 54°
But arc DC subtends ∠ DOC at the centre and
∠ CAD at the remaining part of the circle
∴ ∠ DOC = 2 ∠ CAD = 2 x 54° = 108°.

Question 11.
Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T arc concyclic.
Solution:
Given: Two circles intersect each other at point A and B. PAQ is a line which intersects circles at P, A and Q. At P and Q, tangents are drawn to the circles which meet at T.
To Prove: P, B, Q, T are concyclic.

Construction: Join AB, BP and BQ.
Proof: TP is the tangent and PA. a chord
∴ ∠ TPA = ∠ ABP
(angles in alt. segment)
Similarly we can prove that
∠ TQA = ∠ ABQ …,(ii)
Adding (i) and (ii), we get
∠ TPA + ∠ TQA = ∠ ABP + ∠ ABQ
But in ∆ PTQ,
∠ TPA + ∠ TQA + ∠ PTQ = 180°
⇒ ∠ TPA + ∠ TQA = 180° – ∠ PTQ
⇒ ∠ PBQ = 180°- ∠ PTQ
⇒ ∠ PBQ + ∠PTQ = 180°
But there are the opposite angles of the quadrilateral
∴ Quad. PBQT is a cyclic
Hence P, B. Q and T are concyclic     Q.E.D.

Question 12.
In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC.
Show that triangle PAD is an isosceles triangle. Also shaw that:

Solution:
Given: In a circle PA is the tangent, PBC is the secant and AD is the bisector of ∠BAC which meets the secant at D.
To Prove:
(i) ∆ PAD is an isosceles triangle.
(ii) ∠CAD = \(\frac { 1 }{ 2 }\) [(∠PBA – ∠PAB)]
Proof:
(i) PA is the tangent and AB is chord.
∠PAB = ∠C ….(i)
(Angles in the alt. segment)
AD is the bisector is ∠BAC
∴ ∠1 = ∠2 ….(ii)
In ∆ ADC,
Ext. ∠ADP = ∠C + ∠1
= ∠PAB + ∠2 = ∠PAD
∴ ∆ PAD is an isosceles triangle.
(ii) In A ABC,
Ext. ∠PBA = ∠C + ∠BAC
∴∠BAC = ∠PBA – ∠C
⇒ ∠1 + ∠2 = ∠PBA – ∠PAB [from (i)]
⇒ 2 ∠1 = ∠PBA – ∠PAB
⇒ ∠1 = – [∠PBA – ∠PAB]
⇒ ∠CAD = – [∠PBA – ∠PAB] Q.E.D.

Question 13.
Two circles intersect each other at points A and B. Their common tangent touches the circles v at points P and Q as shown in the figure . Show that the angles PAQ and PBQ arc supplementary. [2000]

Solution:
Given: Two circles intersect each other at A and
B. A common tangent touches the circles at P and
Q. PA. PB, QA and QB are joined.
To Prove: ∠ PAQ + ∠ PBQ = 180°
or ∠ PAQ and ∠ PBQ are supplementary.
Construction: Join AB.
Proof: PQ is the tangent and AB is the chord
∴ ∠ QPA = ∠ PBA (alternate segment) ,…(i)
Similarly we can prove that
∠ PQA = ∠ QBA ,…(ii)
Adding (i) and (ii), we get
∠ QPA + ∠ PQA = ∠ PBA + ∠ QBA
But ∠ QPA + ∠ PQA = 180° – ∠ PAQ ,…(iii) (In ∆ PAQ)
and ∠ PBA + ∠ QBA = ∠ PBQ ,…(iv)
from (iii) and (iv)
∠ PBQ = 180° – ∠ PAQ
⇒ ∠ PBQ + ∠ PAQ = 180°
= ∠ PAQ + ∠ PBQ = 180°
Hence ∠ PAQ and ∠ PBQ arc supplementary Q.E.D.

Question 14.
In the figure, chords AE and BC intersect each other at point D.
(i) If ∠ CDE = 90°.
AB = 5 cm, BD = 4 cm and CD 9 cm;
Find DE.
(ii) If AD = BD, show that AE = BC.

Solution:

Question 15.
Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles arc congruent; show that CE = BD.

Solution:

Given: Two circles with centre P and Q intersect each other at A and B. CBD is a line segment and EBM is tangent to the circle with centre Q, at B. Radii of the cirlces are equal.
To Prove: CE = BD
Construction: Join AB and AD.
Proof: EBM is the tangent and BD is the chord
∴ ∠ DBM = ∠ BAD (Anglesi in alt. segment)
But ∠ DBM = ∠ CBE (Vertically opposite angles)
∴ ∠ BAD = ∠ CBE
∵ In the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
∴ CE = BD Q.E.D.

Question 16.
In the following figure O is the centre of ti.e circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO

Solution:
Given: ∠BDC = 65° and AB is tangent to circle with centre O.
∵ OB is radius
⇒ OB ⊥ AB
In ∆BDC
∠DBC + ∠BDC + ∠B CD = 180°
90° + 65° + ∠BCD =180°
⇒ ∠BCD = 25°
∵ OE = OC = radius
⇒ ∠OEC = ∠OCE
⇒ ∠OEC = 25°
Also, ∠BOE = ∠OEC + ∠OCE
[Exterior angle = sum of opposite interior angles in a ∆]
⇒ ∠BOE = 25°+ 25°
⇒∠BOE = 50°
⇒ ∠BOA = 50°
In ∆AOB
∠AOB + ∠BAO + ∠OBA = 180°
50° + ∠BAO + 90° = 180°
⇒ ∠BAO = 40°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B  are helpful to complete your math homework.

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