ICSE Class 10

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
If O is the centre of the circle, find the value of x in each of the following figures (using the given information):

Solution:
From the figure
(i) ABCD is a cyclic quadrilateral

Question 2.
(a) In the figure (i) given below, O is the centre of the circle. If ∠AOC = 150°, find (i) ∠ABC (ii) ∠ADC.

(b) In the figure (i) given below, AC is a diameter of the given circle and ∠BCD = 75°. Calculate the size of (i) ∠ABC (ii) ∠EAF.

Solution:
(a) Given, ∠AOC = 150° and AD = CD
We know that an angle subtends by an arc of a circle
at the centre is twice the angle subtended by the same arc
at any point on the remaining part of the circle.

Question 3.
(a) In the figure, (i) given below, if ∠DBC = 58° and BD is a diameter of the circle, calculate:
(i) ∠BDC (ii) ∠BEC (iii) ∠BAC

(b) In the figure (if) given below, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find:
(i) ∠CAD (ii) ∠CBD (iii) ∠ADC (2008)

Solution:
(a) ∠DBC = 58°
BD is diameter
∠DCB = 90° (Angle in semi circle)
(i) In ∆BDC
∠BDC + ∠DCB + ∠CBD = 180°
∠BDC = 180°- 90° – 58° = 32°

Question 4.
(a) In the figure given below, ABCD is a cyclic quadrilateral. If ∠ADC = 80° and ∠ACD = 52°, find the values of ∠ABC and ∠CBD.

(b) In the figure given below, O is the centre of the circle. ∠AOE =150°, ∠DAO = 51°. Calculate the sizes of ∠BEC and ∠EBC.

Solution:
(a) In the given figure, ABCD is a cyclic quadrilateral
∠ADC = 80° and ∠ACD = 52°
To find the measure of ∠ABC and ∠CBD

Question 5.
(a) In the figure (i) given below, ABCD is a parallelogram. A circle passes through A and D and cuts AB at E and DC at F. Given that ∠BEF = 80°, find ∠ABC.
(b) In the figure (ii) given below, ABCD is a cyclic trapezium in which AD is parallel to BC and ∠B = 70°, find:
(i)∠BAD (ii) DBCD.

Solution:
(a) ADFE is a cyclic quadrilateral
Ext. ∠FEB = ∠ADF
⇒ ∠ADF = 80°
ABCD is a parallelogram
∠B = ∠D = ∠ADF = 80°
or ∠ABC = 80°
(b)In trapezium ABCD, AD || BC
(i) ∠B + ∠A = 180°
⇒ 70° + ∠A = 180°
⇒ ∠A = 180° – 70° = 110°
∠BAD = 110°
(ii) ABCD is a cyclic quadrilateral
∠A + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180° – 110° = 70°
∠BCD = 70°

Question 6.
(a) In the figure given below, O is the centre of the circle. If ∠BAD = 30°, find the values of p, q and r.

(b) In the figure given below, two circles intersect at points P and Q. If ∠A = 80° and ∠D = 84°, calculate
(i) ∠QBC (ii) ∠BCP

Solution:
(a) (i) ABCD is a cyclic quadrilateral

Question 7.
(a) In the figure given below, PQ is a diameter. Chord SR is parallel to PQ.Given ∠PQR = 58°, calculate (i) ∠RPQ (ii) ∠STP
(T is a point on the minor arc SP)

(b) In the figure given below, if ∠ACE = 43° and ∠CAF = 62°, find the values of a, b and c (2007)

Solution:
(a) In ∆PQR,
∠PRQ = 90° (Angle in a semi circle) and ∠PQR = 58°
∠RPQ = 90° – ∠PQR = 90° – 58° = 32°
SR || PQ (given)

Question 8.
(a) In the figure (i) given below, AB is a diameter of the circle. If ∠ADC = 120°, find ∠CAB.
(b) In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, the sides AD and BC are produced to meet at F. If x : y : z = 3 : 4 : 5, find the values of x, y and z.

Solution:
(a) Construction: Join BC, and AC then
ABCD is a cyclic quadrilateral.

Question 9.
(a) In the figure (i) given below, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E. If ∠ADE = 70° and ∠OBA = 45°, calculate
(i) ∠OCA (ii) ∠BAC
(b) In figure (ii) given below, ABF is a straight line and BE || DC. If ∠DAB = 92° and ∠EBF = 20°, find :
(i) ∠BCD (ii) ∠ADC.

Solution:
(a) ABCD is a cyclic quadrilateral

Question 10.
(a) In the figure (ii) given below, PQRS is a cyclic quadrilateral in which PQ = QR and RS is produced to T. If ∠QPR = 52°, calculate ∠PST.

(b) In the figure (ii) given below, O is the centre of the circle. If ∠OAD = 50°, find the values of x and y.

Solution:
(a) PQRS is a cyclic quadrilateral in which
PQ = QR

Question 11.
(a) In the figure (i) given below, O is the centre of the circle. If ∠COD = 40° and ∠CBE = 100°, then find :
(i) ∠ADC
(ii) ∠DAC
(iii) ∠ODA
(iv) ∠OCA.
(b) In the figure (ii) given below, O is the centre of the circle. If ∠BAD = 75° and BC = CD, find :
(i) ∠BOD
(ii) ∠BCD
(iii) ∠BOC
(iv) ∠OBD (2009)

Solution:
(a) (i) ∴ ABCD is a cyclic quadrilateral.
∴ Ext. ∠CBE = ∠ADC
⇒ ∠ADC = 100°
(ii) Arc CD subtends ∠COD at the centre
and ∠CAD at the remaining part of the circle
∴ ∠COD = 2 ∠CAD

Question 12.
In the given figure, O is the centre and AOE is the diameter of the semicircle ABCDE. If AB = BC and ∠AEC = 50°, find :
(i) ∠CBE
(ii) ∠CDE
(iii) ∠AOB.
Prove that OB is parallel to EC.

Solution:
In the given figure,
O is the centre of the semi-circle ABCDE
and AOE is the diameter. AB = BC, ∠AEC = 50°

Question 13.
(a) In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.

(b) In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.

Solution:
(a) Given: Chord BC || ED,
ABE and ACD are straight lines.
To Prove: ∆AED is an isosceles triangle.
Proof: BCDE is a cyclic quadrilateral.
Ext. ∠ABC = ∠D …(i)
But BC || ED (given)

Question 14.
In the given figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.

Solution:
In the given figure,
∆ABC is an isosceles triangle in which AB = AC.
A circle passing through B and C intersects
sides AB and AC at D and E.
To prove: DE || BC
Construction : Join DE.
∵ AB = AC
∠B = ∠C (angles opposite to equal sides)
But BCED is a cyclic quadrilateral
Ext. ∠ADE = ∠C
= ∠B (∵ ∠C = ∠B)
But these are corresponding angles
DE || BC
Hence proved.

Question 15.
(a) Prove that a cyclic parallelogram is a rectangle.
(b) Prove that a cyclic rhombus is a square.
Solution:
(a) ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle
Proof: ABCD is a parallelogram
∠A = ∠C and ∠B = ∠D

Question 16.
In the given figure, chords AB and CD of the circle are produced to meet at O. Prove that triangles ODB and OAC are similar. Given that CD = 2 cm, DO = 6 cm and BO = 3 cm, area of quad. CABD

Solution:
In the given figure, AB and CD are chords of a circle.
They are produced to meet at O.
To prove : (i) ∆ODB ~ ∆OAC
If CD = 2 cm, DO = 6 cm, and BO = 3 cm
To find : AB and also area of the
\(\frac { Quad.ABCD }{ area\quad of\quad \Delta OAC } \)
Construction : Join AC and BD

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Chapter Test

Question 1.
Using the given information, find the value of x in each of the following figures :


Solution:
(i) ∠ADB and ∠ACB are in the same segment.
∠ADB = ∠ACB = 50°
Now in ∆ADB,
∠DAB + X + ∠ADB = 180°

Question 2.
If O is the centre of the circle, find the value of x in each of the following figures (using the given information):


Solution:
(i) ∠ACB = ∠ADB
(Angles in the same segment of a circle)
But ∠ADB = x°

Question 3.
(a) In the figure (i) given below, AD || BC. If ∠ACB = 35°. Find the measurement of ∠DBC.
(b) In the figure (ii) given below, it is given that O is the centre of the circle and ∠AOC = 130°. Find∠ABC

Solution:
(a) Construction: Join AB
∠A = ∠C = 35° [∵ Alt angles]

Question 4.
(a) In the figure (i) given below, calculate the values of x and y.
(b) In the figure (ii) given below, O is the centre of the circle. Calculate the values of x and y.

Solution:
(a) ABCD is a cyclic quadrilateral

Question 5.
(a) In the figure (i) given below, M, A, B, N are points on a circle having centre O. AN and MB cut at Y. If ∠NYB = 50° and ∠YNB = 20°, find ∠MAN and the reflex angle MON.
(b) In the figrue (ii) given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find
(i) ∠ACB
(ii) ∠OBC
(iii) ∠OAB
(iv) ∠CBA

Solution:
(a) ∠NYB = 50°, ∠YNB = 20°.

Question 6.
(a) In the figure (i) given below, O is the centre of the circle and ∠PBA = 42°. Calculate the value of ∠PQB
(b) In the figure (ii) given below, AB is a diameter of the circle whose centre is O. Given that ∠ECD = ∠EDC = 32°, calculate
(i) ∠CEF
(ii) ∠COF.

Solution:
(a) In ∆APB,
∠APB = 90° (Angle in a semi-circle)
But ∠A + ∠APB + ∠ABP = 180° (Angles of a triangle)
∠A + 90° + 42°= 180°
∠A + 132° = 180°
⇒ ∠A = 180° – 132° = 48°
But ∠A = ∠PQB

Question 7.
(a) In the figure (i) given below, AB is a diameter of the circle APBR. APQ and RBQ are straight lines, ∠A = 35°, ∠Q = 25°. Find (i) ∠PRB (ii) ∠PBR (iii) ∠BPR.
(b) In the figure (ii) given below, it is given that ∠ABC = 40° and AD is a diameter of the circle. Calculate ∠DAC.

Solution:
(a) (i) ∠PRB = ∠BAP
(Angles in the same segment of the circle)
∴ ∠PRB = 35° (∵ ∠BAP = 35° given)
(ii) In ∆PRQ,

Question 8.
(a) In the figure given below, P and Q are centres of two circles intersecting at B and C. ACD is a st. line. Calculate the numerical value of x.

(b) In the figure given below, O is the circumcentre of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate
(i)∠CAB
(ii)∠OAC

Solution:
Given that
(a) Arc AB subtends ∠APB at the centre
and ∠ACB at the remaining part of the circle

Question 9.
(a) In the figure (i) given below, chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

(b) In the figure (ii) given below, C is a point on the minor arc AB of the circle with centre O. Given ∠ACB = p°, ∠AOB = q°, express q in terms of p. Calculate p if OACB is a parallelogram.
Solution:
(a) ∠CBE = ∠CAE
(Angle in the same segment of a circle)
⇒ ∠CAE = 65°
∠AEC = 90° (Angle in a semi circle)
Now in ∆AEC
∠AEC + ∠CAE + ∠ACE = 180° (Angle of a triangle)
⇒ 90°+ 65° +∠ACE = 180°
⇒ 155° + ∠ACE = 180°
⇒ ∠ACE = 180° – 155° – 25°
∵AC || ED (given)
∴∠ACE = ∠DEC (alternate angles)
∴∠DEC = 25°

Question 10.
(a) In the figure (i) given below, straight lines AB and CD pass through the centre O of a circle. If ∠OCE = 40° and ∠AOD = 75°, find the number of degrees in :
(i) ∠CDE
(ii) ∠OBE.
(b) In the figure (ii) given below, I is the incentre of ∆ABC. AI produced meets the circumcircle of ∆ABC at D. Given that ∠ABC = 55° and ∠ACB = 65°, calculate
(i) ∠BCD
(ii) ∠CBD
(iii) ∠DCI
(iv) ∠BIC.

Solution:
(a) (i) ∠CED = 90° (Angle in semi-circle)
In ∆CED
∠CED + ∠CDE + ∠DCE = 180°
⇒ 90° +∠CDE + 40° = 180°
⇒ 130° + ∠CDE = 180°
⇒ ∠CDE = 180° – 130° = 50°

NCVT MIS 2019

Question 11.
O is the circumcentre of the triangle ABC and D is mid-point of the base BC. Prove that ∠BOD = ∠A.
Solution:
In the given figure, O is the centre of circumcentre of ∆ABC.
D is mid-point of BC. BO, CO and OD are joined.

Question 12.
In the given figure, AB and CD are equal chords. AD and BC intersect at E. Prove that AE = CE and BE = DE.

Solution:
In the given figure, AB and CD are two equal chords
AD and BC intersect each other at E.
To prove : AE = CE and BE = DE
Proof:
In ∆AEB and ∆CED
AB = CD (given)
∠A = ∠C (angles in the same segment)
∠B = ∠D (angles in the same segment)
∴ ∆AEB ≅ ∆CED (ASA axiom)
∴ AE = CE and BE = DE (c.p.c.t.)

Question 13.
(a) In the figure (i) given below, AB is a diameter of a circle with centre O. AC and BD are perpendiculars on a line PQ. BD meets the circle at E. Prove that AC = ED.
(b) In the figure (ii) given below, O is the centre of a circle. Chord CD is parallel to the diameter AB. If ∠ABC = 25°, calculate ∠CED.

Solution:
(a) Given: AB is the diameter of a circle with centre O.
AC and BD are perpendiculars on a line PQ,
such that BD meets the circle at E.

Question 14.
In the adjoining figure, O is the centre of the given circle and OABC is a parallelogram. BC is produced to meet the circle at D.
Prove that ∠ABC = 2 ∠OAD.

Solution:
Given: In the figure,
OABC is a || gm and O is the centre of the circle.
BC is produced to meet the circle at D.
To Prove : ∠ABC = 2∠OAD.
Construction: Join AD.

Question 15.
(a) In the figure (i) given below, P is the point of intersection of the chords BC and AQ such that AB = AP. Prove that CP = CQ.

(b) In the figure (i) given below, AB = AC = CD, ∠ADC = 38°. Calculate :
(i) ∠ABC (ii) ∠BEC.

Solution:
(a) Given: Two chords AQ and BC intersect each other at P
inside the circle. AB and CQ are joined and AB = AP.
To Prove : CP = CQ
Construction : Join AC.
Proof: In ∆ABP and ∆CQP
∴ ∠B = ∠Q

Question 16.
(a) In the figure (i) given below, CP bisects ∠ACB. Prove that DP bisects ∠ADB.
(b) In the figure (ii) given below, BDbisects ∠ABC. Prove that \(\frac { AB }{ BD } =\frac { BE }{ BC } \)

Solution:
(a)Given: In the figure, CP is the bisector of
∠ACB meeting the circle at P.
PD is joined

Question 17.
(a) In the figure (ii) given below, chords AB and CD of a circle intersect at E.
(i) Prove that triangles ADE and CBE are similar.
(ii) Given DC =12 cm, DE = 4 cm and AE = 16 cm, calculate the length of BE.

(b) In the figure (ii) given below, AB and CD are two intersecting chords of a circle. Name two triangles which are similar. Hence, calculate CP given that AP = 6cm, PB = 4 cm, and CD = 14 cm (PC > PD).

Solution:
(a) Given: Two chords AB and CD intersect each other
at E inside the circle.

Question 18.
In the adjoining figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE. (2008)

Solution:
In the figure, AE and BC intersect each other at D.
AB is joined.

Question 19.
(a) In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.
(b) In the figure (ii) given below, the diagonals of a cyclic quadrilateral ABCD intersect in P and the area of the triangle APB is 24 cm². If AB = 8 cm and CD = 5 cm, calculate the area of ∆DPC.

Solution:
(a) PR is the diameter of the circle
PQ = 7 cm, QR = 6 cm, RS = 2 cm.

Question 20.
(a) In the figure (i) given below, QPX is the bisector of ∠YXZ of the triangle XYZ. Prove that XY : XQ = XP : XZ,
(b) In the figure (ii) given below, chords BA and DC of a circle meet at P. Prove that:
(i) ∠PAD = ∠PCB
(ii) PA. PB = PC . PD.

Solution:
(a) Given: ∆XYZ is inscribed in a circle.
Bisector of ∠YXZ meets the circle at Q.
QY is joined.
To Prove : XY : XQ = XP : XZ

We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1 help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 15 Circles Ex 15.1, drop a comment below and we will get back to you at the earliest.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test

Question 1.
Draw a straight line AB of length 8 cm. Draw the locus of all points which are equidistant from A and B. Prove your statement.
Solution:
(i) Draw a line segment AB = 8 cm.

(ii) Draw the perpendicular bisector of AB intersecting AB at D.
∴ Every point P on it will be equidistant from A and B.
(iii) Take a point P on the perpendicular bisector.
(iv) Join PA and PB.
Proof: In ∆PAD and ∆PBD
PD = PD (common)
AD = BD (D is mid-point of AB)
∠PDA = ∠PDB (each 90°)
∴ ∆ PAD ≅ ∆ PBD (SAS axiom of congruency)
∴PA = PB (c.p.c.t.)
Similarly, we can prove any other point on the
perpendicular bisector of AB is equidistant from A and B.
Hence Proved.

Question 2.
A point P is allowed to travel in space. State the locus of P so that it always remains at a constant distance from a fixed point C.
Solution:
The point P is moving in the space and
it is at a constant distance from a fixed point C.
∴ Its locus is a sphere.

Question 3.
Draw a line segment AB of length 7 cm. Construct the locus of a point P such that area of triangle PAB is 14 cm².
Solution:
Base of ∆PAB = 7 cm
and its area = 14 cm²

Now draw a line XY parallel to AB at a distance of 4 cm.
Now take any point P on XY
Join PA and PB
area of ∆PAB = 14 cm.
Hence locus of P is the line XY
which is parallel to AB at a distance of 4 cm.

Question 4.
Draw a line segment AB of length 12 cm. Mark M, the mid-point of AB. Draw and describe the locus of a point which is
(i) at a distance of 3 cm from AB.
(ii) at a distance of 5 cm from the point M. Mark the points P, Q, R, S which satisfy both the above conditions. What kind of quadrilateral is PQRS? Compute the area of the quadrilateral PQRS.
Solution:
Steps of Construction :

(i) Take a line AB = 12 cm
(ii) Take M, the midpoint of AB.
(iii) Draw straight lines CD and EF parallel to AB at a distance of 3 cm.
(iv) With centre M and radius 5 cm,
draw areas which intersect CD at P and Q and EF at R and S.
(v) Join QR and PS.
PQRS is a rectangle where the length PQ = 8 cm.
Area of rectangle PQRS = PQ x RS = 8 x 6 = 48 cm²

Question 5.
AB and CD are two intersecting lines. Find the position of a point which is at a distance of 2 cm from AB and 1.6 cm from CD.
Solution:
(i) AB and CD are the intersecting lines which intersect each other at O.

(ii) Draw a line EF parallel to AB and GH parallel to CD intersecting each other at P
P is the required point.

Question 6.
Two straight lines PQ and PK cross each other at P at an angle of 75°. S is a stone on the road PQ, 800 m from P towards Q. By drawing a figure to scale 1 cm = 100 m, locate the position of a flagstaff X, which is equidistant from P and S, and is also equidistant from the road.
Solution:
1 cm = 100 cm
800 m = 8 cm.
Steps of Construction :
(i) Draw the lines PQ and PK intersecting each other
at P making an angle of 75°.

(ii) Take a point S on PQ such that PS = 8 cm.
(iii) Draw the perpendicular bisector of PS.
(iv) Draw the angle bisector of ∠KPS intersecting
the perpendicular bisector at X.
X is the required point which is equidistant from P and S
and also from PQ and PK.

Question 7.
Construct a rhombus PQRS whose diagonals PR, QS are 8 cm and 6 cm respectively. Find by construction a point X equidistant from PQ, PS and equidistant from R, S. Measure XR.
Solution:
Steps of Construction :

(i) Take PR = 8 cm and draw the perpendicular bisector
of PR intersecting it at O.
(ii) From O, out. off OS = OQ = 3 cm
(iii) Join PQ, QR, RS and SP.
PQRS is a rhombus. Whose diagonal are PR and QS.
(iv) PR is the bisector of ∠SPQ.
(v) Draw the perpendicular bisector of SR intersecting PR at X
∴ X is equidistant from PQ and PS and also from S and R.
On measuring length of XR = 3.2 cm (approx)

Question 8.
Without using set square or protractor, construct the parallelogram ABCD in which AB = 5.1 cm. the diagonal AC = 5.6 cm and the diagonal BD = 7 cm. Locate the point P on DC, which is equidistant from AB and BC.
Solution:
Steps of Construction :

(i) Take AB = 5.1 cm
(ii) At A, with readius \(\\ \frac { 5.6 }{ 2 } \) = 2.8 cm
and at B with radius \(\\ \frac { 7.0 }{ 2 } \) = 3.5 cm,
draw two arcs intersecting each other at O.
(iii) Join AO and produce it to C such that
OC = AD = 2.8 cm and
join BO and produce it to D such that
BO = OD = 3.5 cm
(iv) Join BC, CD, DA
ABCD is a parallelogram.
(v) Draw the angle bisector of ∠ABC intersecting CD at P.
P is the required point which is equidistant from AB and BC.

Question 9.
By using ruler and compass only, construct a quadrilateral ABCD in which AB = 6.5 cm, AD = 4cm and ∠DAB = 75°. C is equidistant from the sides AB and AD, also C is equidistant from the points A and B.
Solution:
Steps of Construction :

(i) Draw a line segment AB = 6.5 cm.
(ii) At A, draw a ray making an angle of 75° and cut off AD = 4 cm.
(iii) Draw the bisector of ∠DAB.
(iv) Draw perpendicular bisector of AB intersecting the angle bisector at C.
(v) Join CB and CD.
ABCD is the required quadrilateral.

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test 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.

ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Chapter Test

Question 1.
A point moves such that its distance from a fixed line AB is always the same. What is the relation between AB and the path traveled by P ?
Solution:
Let point P moves in such a way that
it is at a fixed distance from the fixed line AB.

∴ It is a set of two lines l and m parallel to AB
drawn on either side of it at equal distance from it.

Question 2.
A point P moves so that its perpendicular distance from two given lines AB and CD are equal. State the locus of the point P.
Solution:
(i) When two lines AB and CD are parallel,
then the locus of the point P which is equidistant
from AB and CD is a line (l)
in the midway of AB and CD and parallel to them

(ii) If AB and CD are intersecting lines,
then the locus of the point P will be a
pair of the straight lines l and m which bisect
the angles between the given lines AB and CD.

Question 3.
P is a fixed point and a point Q moves such that the distance PQ is constant, what is the locus of the path traced out by the point Q ?
Solution:
∴ P is a fixed point and Q is a moving point
such that it is always at an equidistant from P.

∴ P is the centre of the path of Q which is a circle.
The distance between P and Q is the radius of the circle.
Hence locus of point Q is a circle with P as centre.

Question 4.
(i) AB is a fixed line. State the locus of the point P so that ∠APB = 90°.
(ii) A, B are fixed points. State the locus of the point P so that ∠APB = 60°.
Solution:
(i) AB is a fixed line and P is a point
such that ∠APB = 90°.

The locus of P will be the circle whose diameter is AB.
We know that the angle in a semi-circle is always equal to 90°.
∠APB = 90°
(ii) AB is a fixed line and P is a point such that ∠APB = 60°.
The locus of P will be a major segment of a circle whose AB is a chord.

Question 5.
Draw and describe the locus in each of the following cases :
(i) The locus of points at a distance 2.5 cm from a fixed line.
(ii) The locus of vertices of all isosceles triangles having a common base.
(iii) The locus of points inside a circle and equidistant from two fixed points on the circle.
(iv) The locus of centres of all circles passing through two fixed points.
(v) The locus of a point in rhombus ABCD which is equidistant from AB and AD. (1998)
(vi) The locus of a point in the rhombus ABCD which is equidistant from points A and C.
Solution:
1. Draw a given line AB.
2. Draw lines of l and m parallel to AB at a distance of 2.5 cm.
Lines l and m are the locus of point P which is at a distance of 2.5 cm.

(ii) ∆ABC is an isosceles triangle in which AB = AC.
From A, draw AD perpendicular to BC.
AD is the locus of the point A the vertices of ∆ABC.
In rt. ∆ABD and ∆ACD
Side AD = AD (Common)
Hyp. AB = AC (given)
∴ ∆ABD = ∆ACD (R.H.S. Axiom)
∴ BD = DC (c.p.c.t.)
Hence locus of vertices of isosceles triangles
having common base is the perpendicular bisector of BC.
(iii) (i) Draw a circle with centre O.

(ii) Take points A and B on it and join them.
(iii) Draw a perpendicular bisector of AB
which passes from O and meets the circle at C.
CE the diameter, which is the locus of a point inside the circle
and equidistant from two points A and B at the circle.
(iv) Let C1, C2, C3 be the centres of the circle
which pass through the two fixed points A and B.

Draw a line XY passing through these centres C1, C2, C3.
Hence locus of centres of circles passing through two points A and B
is the perpendicular bisector of the line segment joining the two fixed points.
(v) In rhombus ABCD, Join AC.

AC is the diagonal of rhombus ABCD
∴ AC bisect ∠A.
∴ Any point on AC, is the locus which is equidistant from AB and AD.
(vi) ABCD is a rhombus. Join BD

BD is the locus of a point in the rhombus which is equidistant from A and C.
Diagonal BD bisects  ∠B and ∠D.
Any point on BD will be equidistant from A and C.

Question 6.
Describe completely the locus of points in each of the following cases :
(i) mid-point of radii of a circle.
(ii) centre of a ball, rolling along a straight line on a level floor.
(iii) point in a plane equidistant from a given line.
(iv) point in a plane, at a constant distance of 5 cm from a fixed point (in the plane).
(v) centre of a circle of varying radius and touching two arms of ∠ADC.
(vi) centre of a circle of varying radius and touching a fixed circle, centre O, at a fixed point A on it.
(vii) centre of a circle of radius 2 cm and touching a fixed circle of radius 3 cm with centre 0.
Solution:
(i) The locus of midpoints of the radii of a circle
is another concentric circle with radius is
half of the radius of the given circle.

(ii) AB is the straight line on the ground and the ball is rolling on it
∴ locus of the centre of the ball is a line parallel A lo the given line AB.

(iii) AB is the given line and P is a point in the plane.

From P, draw a line CD and another line EF from P’ parallel to AB.
Thus CD and EF are the lines which are the locus of the point equidistant from AB
(iv) Take a point O and another point P such that OP = 5 cm.
with centre O and radius equal to OP, draw a circle.
Thus this circle is the locus of point P
which is at a distance of 5 cm from O, the given point.

(v) Draw the bisector BX of ∠ABC.
This bisector of an angle is the locus of the centre of a circle with different radii.
Any point on BX, is equidistant from the arms BA and BC of the ∠ABC.

(vi) A circle with centre O is given and one point A on it.
The locus of the centre of a circle which touches the circle
at the fixed point A on it, is a line joining the points O and A.

(vii) (a) If the circle with 2 cm as radius touches the given circle
externally then the locus of the centre of the circle
will be a concentric circle with radius 3+2 = 5 cm.

If the circle with 2 cm as radius touches the given circle with 3 cm as radius internally,
then the locus of the centre of the circle will be a concentric circle with radius 3-2 = 1 cm.

Question 7.
Using ruler and compasses construct :
(i) a triangle ABC in which AB = 5.5 cm, BC = 3.4 cm and CA = 4.9 cm.
(ii) the locus of points equidistant from A and C.
Solution:
(i) Draw BC = 3.4 and mark the arcs of 5.5 add 4.9 cm from B and C.
Join A, B and C.
ABC is the required triangle.
(ii) Draw ⊥ bisector of AC.
(iii) Draw an angle of 90° at AB at A which intersects ⊥ bisector at O.
Draw circle taking 0 as centre and OA as the radius.

Question 8.
Construct triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C and
(ii) P is equidistant from AB and BC
(iii) Measure and record the length of PB. (2000)
Solution:

(i) Take BC = 8 cm a long line segment. At B,
draw a ray BX making an angle of 60° with BC.
Cut off BA = 7 cm. and join AC.
(i) Draw the perpendicular bisector of BC.
(ii) Draw the angle bisector of ∠B which intersect
the perpendicular bisector of BC at P. P is the required point.
(iii) On measuring the length of BP = 4.6 cm (approx.)

Question 9.
A straight line AB is 8 cm long. Locate by construction the locus of a point which is :
(i) Equidistant from A and B.
(ii) Always 4 cm from the line AB.
(iii) Mark two points X and Y, which are 4 cm from AB and equidistant from A and B. Name the figure AXBY.(2008)
Solution:
Steps of construction:

(i) Draw a line segment AB = 8 cm.
(ii) With the help of compasses and ruler,
draw the perpendicular bisector l of AB which intersects AB at O.
(iii) Then any point on l, is equidistant from A and B.
(iv) Cut off OX = OY = 4 cm. The X and Y are the required loci,
which is equidistant from AB and also from A and B.
(v) Join AX, XB, BY and YA.
The figures so formed AXBY is the shape of a square because
its diagonals are equal and bisect each other at right angles.

Question 10.
Use ruler and compasses only for this question.
(i) Construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure arid record the length of PB. (2010)
Solution:
In ∆ABC, AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°

Steps of construction :
(a) (i) Draw a line segment BC = 6 cm
(ii) At, B draw a ray BX making an angle of 60° and cut off BA = 3.5 cm.
(iii) Join AC.
The ∆ABC is the required triangle.
(b) Draw the bisector BY of ∠ABC.
(c) Draw the perpendicular bisector of BC which intersects BY at P.
P is the required point P which is equidistant from BC and BA
and also equidistant from B and C.
On measuring PB it is 3.4 cm (approx)

Question 11.
Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC = 105°. Hence:
(i) Construct the locus of points equidistant from BA and BC.
(ii) Construct the locus of points equidistant from B and C.
(iii) Mark the point which satisfies the above two loci as P. Measure and write the length of PC.
Solution:
Steps of construction :

Construct the triangle ABC with AB = 5.5 cm
∠BAC = 105° and AC = 6 cm
(i) Points which are equidistant from BA and BC lies on the bisector of ∠ABC.
(ii) Points equidistant from B and C lies on the perpendicular bisector of BC.
Draw perpendicular bisector of BC.
The required point P is the point of intersection of the bisector of
∠ABC and the perpendicular bisector of BC.
(iii) Required length of PC = 4.8 cm.

Question 12.
In the given diagram, A, B and C are fixed collinear points; D is a fixed point outside the line: Locate

(i) the point P on AB such that CP = DP.
(ii) the points Q such that CQ = DQ = 3 cm. How many such points are possible?
(iii) the points R on AB such that DR = 4 cm. How many such points are possible?
(iv) the points S such that CS = DS and S is 4 cm away from the line CD. How many such points are possible?
(v) Are the points P, Q, R collinear?
(vi) Are the points P, Q, S collinear?
Solution:
Points A, B and C are collinear and D is any point outside AB.

(i) Join CD.
(ii) Draw the perpendicular bisector of CD which meets AB in P.
(iii) P is the required point such that CP = DP
(iv) With centres C and D, draw two arcs with 3 cm radius
which intersect each other at Q and Q’.
Hence there are two points Q and Q’ which are equidistant from C and D.
(v) With centre D, and radius 4 cm draw an arc which intersects AB at R and R’
∴ R and R’ are the two point on AB.
(vi) With centre C and D, draw arcs with a radius equal to
4 cm which intersects each other in S and S’.
∴ There can be two such points which are equidistant from C and D.
(vii) No P, Q, R are not collinear.
(viii) Yes, P, Q, S are collinear.

Question 13.
Points A, B and C represent position of three towers such that AB = 60 m, BC = 73m and CA = 52 m. Taking a scale of 10 m to 1 cm, make an accurate drawing of ∆ABC. Find by drawing, the location of a point which is equidistant from A, B and C, and its actual distance from any of the towers.
Solution:
AB = 60 mm = 6.0 cm, BC = 73 mm = 7.3 cm
and CA = 52 mm = 5.2 cm.

(i) Draw a line segment BC = 7.3cm
(ii) With Centre B and radius 6cm and with centre C
and radius 5.2 cm, draw two arcs intersecting each other at A
(iii) Joining AB and AC.
(iv) Draw the perpendicular bisector of AB, BC and CA respectively,
which intersect each other at point P. Join PB.
P is equidistant from A, B and C on measuring PB = 3.7 cm.
Actual distance = 37 m.

Question 14.
Draw two intersecting lines to include an angle of 30°. Use ruler and compasses to locate points which are equidistant from these lines and also 2 cm away from their point of intersection. How many such points exist ? (1990)
Solution:
(i) Two lines AB and CD intersect each other at O.

(ii) Draw the bisector of ∠BOD and ∠AOD
(iii) With centre O and radius equal to 2 cm.
marks points on the bisector of angles at P, Q, R and S respectively.
Hence there are four points which are equidistant
from AB and CD and 2 cm from 0, the point of intersection of AB and CD.

Question 15.
Without using set square or protractor, construct the quadrilateral ABCD in which ∠BAD = 45°, AD = AB = 6 cm, BC = 3.6 cm and CD = 5 cm.
(i) Measure ∠BCD.
(ii) Locate the point P on BD which is equidistant from BC and CD. (1992)
Solution:
(i) Take AB = 6 cm long
(ii) AT A, draw the angle of 45° and cut off AD = 6 cm

(iii) With centre D and radius 5 cm and with centre B,
and radius 3.5 cm draw two arcs intersecting each other at C.
(iv) Join CD and CB and join BD
ABCD is the required quadrilateral.
(v) On measuring ∠BCD = 65°.
(vi) Draw the bisector of ∠BCD which intersects BD at P.
P is the required point which is equidistant from CD and CB.

Question 16.
Without using set square or protractor, construct rhombus ABCD with sides of length 4 cm and diagonal AC of length 5 cm. Measure ∠ABC. Find the point R on AD such that RB = RC. Measure the length of AR. (1990)

Solution:
(i) Take AB = 4 cm
(ii) With centre A, draw an arc of 5 cm radius
and with B draw another arc of radius 4 cm intersecting each other at C.
(iii) Join AC and BD.
(iv) Again with centre A and C,
draw two arcs of radius 4 cm intersecting each other on D.
(v) Join AD and CD.
ABCD is the required rhombus and on measuring the ∠ABC, it is 78°.
(vi) Draw perpendicular bisector of BC intersecting AD at R.
On measuring the length of AR, it is equal to 1.2 cm.

Question 17.
Without using set-squares or protractor construct :
(i) Triangle ABC, in which AB = 5.5 cm, BC = 3.2 cm and CA = 4.8 cm.
(ii) Draw the locus of a point which moves so that it is always 2.5 cm from B.
(iii) Draw the locus of a point which moves so that it is equidistant from the sides BC and CA.
(iv) Mark the point of intersection of the loci with the letter P and measure PC. (1994)
Solution:
Steps of Construction :
(i) Draw BC = 3.2 cm long.
(ii) With centre B and radius 5.5 cm
and with centre C and radius 4.8 cm
draw arcs intersecting each other at A.

(iiii) Join AB and AC.
(iv) Draw the bisector of ∠BCA.
(v) With centre B and radius 2.5 cm,
draw an arc intersecting the angle bisector of ∠BCA at P and P’.
P and P’ are two loci which satisfy the given condition.
On measuring CP and CP’
CP = 3.6 cm and CP’ =1.1 cm.

Question 18.
By using ruler and compasses only, construct an isosceles triangle ABC in which BC = 5 cm, AB = AC and ∠BAC = 90°. Locate the point P such that :
(i) P is equidistant from the sides BC and AC.
(ii) P is equidistant from the points B and C.
Solution:
Steps of Construction :
(i) Take BC = 5.0 cm and bisect it at D.

(ii) Taking BC as diameter, draw a semicircle.
(iii) At D, draw a perpendicular intersecting the circle at A
(iv) Join AB and AC.
(v) Draw the angle bisector of C intersecting
the perpendicular at P. P is the required point.

Question 19.
Using ruler and compasses only, construct a quadrilateral ABCD in which AB = 6 cm, BC = 5 cm, ∠B = 60°, AD = 5 cm and D is equidistant from AB and BC. Measure CD. (1983)
Solution:
Steps of Construction :
(i) Draw AB = 6 cm

(ii) At B, draw angle of 60° and cut off BC = 5 cm.
(iii) Draw the angle bisector of ∠B.
(iv) With centre A and radius 5 cm. draw an arc
which intersects the angle bisector of ∠B at D
(v) Join AD and DC.
ABCD in the required quadrilateral.
On measuring CD, it is 5.3 cm (approx).

Question 20.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB (2001)
Solution:
Steps of Construction :
(i) Draw a line AB = 6 cm

(ii) With centre A and B and radius 4 cm,
draw two arcs intersecting each other at C.
(iii) Join CA and CB
(iv) Draw the bisector of ∠C and cut off CP = 5 cm
(v) Draw a line XY parallel to AB at a distance of 5 cm.
(vi) From P, draw arcs of radius 5 cm each intersecting
the line XY at Q and R. Hence Q and R are the required points.

Question 21.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of length 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC, (1995)
Solution:
Steps of Construction :
(i) With centre 0 and radius 4 cm draw a circle.
(ii) Take point A on this circle.

(iii) With centre A and radius 6 cm draw an arc cutting the circle at B.
(iv) Again with radius 5 cm, draw another arc cutting the circle at C.
(v) Join AB and AC.
(vi) Draw the perpendicular bisector of AC.
Any point on it, will be equidistant from A and C.
(vii) Draw the angle bisector of ∠A intersecting
the perpendicular bisector of AC at P. P is the required locus.

Question 22.
Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm. and ∠ABC = 60°.
(ii) Construct the locus of all points, inside ∆ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to ∆ABC.
(iv) Mark the point Q, in your construction, which would make ∆QBC equal in area to ∆ABC, and isosceles.
(v) Measure and record the length of CQ. (1998)
Solution:
Steps of Construction :
(i) Draw AB = 9 cm
(ii) At B draw an angle of 60° and cut off BC = 6 cm.
(iii) Join AC

(iv) Draw perpendicular bisector of BC.
All points on it will be equidistant from B and C.
(v) From A, draw a line XY parallel to BC.
(vi) Produce the perpendicular bisector of BC to meet XY in Q.
(vii) Join QC and QB.
∆QBC will be the triangle equal in area to ∆ABC
because these are on the same base BC and between the same parallel lines.
On measuring, the length of CQ is 8.2 cm (approx.).

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 14 Locus Ex 14 are helpful to complete your math homework.

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ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

More Exercises

• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.2
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.3
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity MCQS
• ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test

Question 1.
In the given figure, ∠1 = ∠2 and ∠3 = ∠4. Show that PT x QR = PR x ST.
Solution:
Given: In the given figure,
∠1 = ∠1 and ∠3 = ∠4

To prove : PT × QR = PR × ST
Proof: ∠1 = ∠2
Adding ∠6 to both sides
∠1 + ∠6 = ∠2 + ∠6
∠SPT = ∠QPR
In ∆PQR and ∆PST

Question 2.
In the adjoining figure, AB = AC. If PM ⊥ AB and PN ⊥ AC, show that PM x PC = PN x PB.

Solution:
Given : In the given figure,
AB = AC, PM ⊥ AB and PN ⊥ AC
To prove : PM × PC = PN × PB
Proof: In ∆ABC, AB = AC
∠B = ∠C
Now in ∆CPN and ∆BPM,

Question 3.
(a) In the figure (1) given below. ∠AED = ∠ABC. Find the values of x and y.
(b) In the fig. (2) given below, CD = \(\\ \frac { 1 }{ 2 } \) AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD.

Solution:
(a) Given : In following figure, ∠AED = ∠ABC
Required: The values of x and y.
Now, in ∆ABC and ∆ADE
∠AED = ∠ABC (given)
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE
(By A.A. axiom of similarity)

Question 4.
In the given figure, 2 AD = BD, E is mid-point of BD and F is mid-point of AC and EC || BH. Prove that:
(i) DF || BH
(ii) AH = 3 AF.

Solution:
Given: E is the mid-point of BD
and F is mid-point of AC
also 2 AD = BD and EC || BH
To Prove : (i) DF || BH
(ii) AH = 3 AF

Question 5.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Solution:
Given : In ∆ABC, D and E are the points
on the sides AB and AC respectively
DE || BC
AD = 2.4 cm, AE = 3.2 cm,
DE = 2 cm, BC = 5 cm

Question 6.
In a ∆ABC, D and E are points on the sides AB and AC respectively such that AD = 5.7cm, BD = 9.5cm, AE = 3.3cm and AC = 8.8cm. Is DE || BC? Justify your answer.
Solution:
In ∆ABC, D and E are points on the sides AB and AC respectively
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and AC = 8.8 cm

Question 7.
In a ∆ABC, DE is parallel to the base BC, with D on AB and E on AC. If \(\frac { AD }{ DB } =\frac { 2 }{ 3 } ,\frac { BC }{ DE } \)
Solution:
In ∆ABC, DE || BC
D is on AB and E is on AC

Question 8.
If the area of two similar triangles are 360 cm² and 250 cm² and if one side of the first triangle is 8 cm, find the length of the corresponding side of the second triangle.
Solution:
Let ∆ABC and ∆DEF are similar and area of
∆ABC = 360 cm²

Question 9.
In the adjoining figure, D is a point on BC such that ∠ABD = ∠CAD. If AB = 5 cm, AC = 3 cm and AD = 4 cm, find
(i) BC
(ii) DC
(iii) area of ∆ACD : area of ∆BCA.

Solution:
In ∆ABC and ∆ACD
∠C = ∠C (Common)
∠ABC = ∠CAD (given)
∴ ∆ABC ~ ∆ACD

Question 10.
In the adjoining figure the diagonals of a parallelogram intersect at O. OE is drawn parallel to CB to meet AB at E, find area of DAOE : area of ||gm ABCD.

Solution:
(a) In the figure
Diagonals of parallelogram ABCD are
AC and BD which intersect each other at O.
OE is drawn parallel to CB to meet AB in E.

Question 11.
In the given figure, ABCD is a trapezium in which AB || DC. If 2AB = 3DC, find the ratio of the areas of ∆AOB and ∆COD.

Solution:
In the given figure, ABCD is trapezium in
which AB || DC, 2AB = 3DC

Question 12.
In the adjoining figure, ABCD is a parallelogram. E is mid-point of BC. DE meets the diagonal AC at O and meet AB (produced) at F. Prove that .

(i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4.
Solution:
Given : In || gm ABCD,
E is the midpoint of BC and DE meets the diagonal AC at O
and meet AB produced at F.
To prove : (i) DO : OE = 2 : 1
(ii) area of ∆OEC : area of ∆OAD = 1 : 4
Proof: In ∆AOD and ∆EDC
∠AOD = ∠EOC (vertically opposite angle)
∠OAD = ∠OCB (alt. angles)
∆AOD ~ ∆EOC (AA postulate)

Question 13.
A model of a ship is made to a scale of 1 : 250. Calculate :
(i) the length of the ship, if the length of model is 1.6 m.
(ii) the area of the deck of the ship, if the area of the deck of model is 2.4 m².
(iii) the volume of the model, if the volume of the ship is 1 km³.
Solution:
Scale factor (k) of the model of the ship = \(\\ \frac { 1 }{ 250 } \)
(i) Length of model = 1.6 m

Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Chapter Test 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.