CBSE Class 10

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-5/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm
Solution:
(i) 7 cm, 24 cm,-25 cm
(7)² + (24)² = 49 + 576 = 625 = (25)² = 25
∴ The given sides make a right angled triangle with hypotenuse 25 cm

(ii) 3 cm, 8 cm, 6 cm
(8)² = 64
(3)² + (6)² = 9 + 36 = 45
64 ≠ 45
The square of larger side is not equal to the sum of squares of other two sides.
∴ The given triangle is not a right angled.

(iii) 50 cm, 80 cm, 100 cm
(100)²= 10000
(80)² + (50)² = 6400 + 2500
= 8900
The square of larger side is not equal to the sum of squares of other two sides.
∴The given triangle is not a right angled.

(iv) 13 cm, 12 cm, 5 cm
(13)² = 169
(12)² + (5)²= 144 + 25 = 169
= (13)² = 13
Sides make a right angled triangle with hypotenuse 13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM • MR.

Solution:
In right angled ∆QPR,
∠P = 90°, PM ⊥ QR
∴ ∆PMQ ~ ∆RMP
[If ⊥ is drawn from the vertex of right angle to the hypotenuse then triangles on both sides of perpendicular are similar to each other, and to whole triangle]

⇒ [Corresponding sides of similar
⇒ PM x MP = RM x MQ ⇒ PM2 = QM.MR

Question 3.
In the given figure, ABD is a triangle right angled at A and AC i. BD. Show that
(i) AB² = BC.BD
(ii) AC² = BC.DC
(iii) AD² = BD.CD

Solution:

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Given: In ∆ABC, ∠C = 90° and AC = BC
To Prove: AB² = 2AC²
Proof: In ∆ABC,
AB²= BC² + AC²
AB² = AC² + AC² [Pythagoras theorem]
= 2AC²

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC² , Prove that ABC is a right triangle.
Solution:

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:

Given: In ∆ABC, AB = BC = AC = 2a
We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
and AD ⊥ BC
BD = \(\frac { 1 }{ 2 }\) x 2 a = a
In right angled triangle ADB,
AD² + BD² = AB²
⇒ AD² = AB² – BD²= (2a)² – (a)² = 4a²– a²= 3a²
AD = √3a

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:
Given: ABCD is a rhombus. Diagonals AC and BD intersect at O.
To Prove: AB²+ BC²+ CD²+ DA² = AC²+ BD²

Question 8.
In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF².

Solution:

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. ind the distance of the foot of the ladder from base of the wall.
Solution:

Let AC be the ladder of length 10 m and AB = 8 m
In ∆ABC, BC² + AB² = AC²
⇒ BC²= AC² – AB²= (10)² – (8)²
BC² = 100-64 – 36 BC = √36 = 6 m
Hence distance of foot of the ladder from base of the wall is 6 m.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Solution:

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1\(\frac { 1 }{ 2 }\) hours?
Solution:

Question 12.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Solution:

Length of poles is 6 m and 11m.
DE = DC – EC = 11m-6m = 5m
In ∆DAE,
AD² = AE² + DE² [ ∵AE = BC]
= (12)² + (5)² =144 + 25 = 169
AD = √l69 = 13

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB² + DE².
Solution:

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the figure). Prove that 2AB² = 2AC² + BC².

Solution:

Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = \(\frac { 1 }{ 3 }\)BC. Prove that 9AD² = 7AB².
Solution:

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Solution:

Question 17.
Tick the correct answer and justify : In ∆ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is:
(a) 120°
(b) 60°
(c) 90°
(d) 45
Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-4/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC.
Solution:
Since, ∆ABC ~ ∆DEF
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Solution:
ABCD is a trapezium with AB || DC and AB = 2 CD

Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that


Solution:

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.
Solution:

Question 5.
D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.
Solution:

Question 6.
Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:

Question 8.
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 :1
(b) 1:2
(c) 4 :1
(d) 1:4
Solution:

Question 9.
Sides of two similar triangles are in the ratio 4:9. Areas of these triangles are in the ratio
(a) 2:3
(b) 4:9
(c) 81:16
(d) 16:81
Solution:
Justification: Areas of two similar triangles are in the ratio of the squares of their corresponding sides.

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1

Question 1.
Fill in the blanks by using the correct word given in brackets.
(i) All circles are ……………. . (congruent/similar)
(ii) All squares are …………… . (similar/congruent)
(iii) All …………….. triangles are similar. (isosceles/equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are …………… and
(b) their corresponding sides are …………… (equal/proportional)
Solution:
Fill in the blanks.
(i) All circles are similar.
(ii) All squares are similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Question 2.
Give two different examples of pair of
(i) similar figures.
(ii) non-similar figures.
Solution:
(i) Examples of similar figures:

• Square
• Regular hexagons

(ii) Examples of non-similar figures:

• Two triangles of different angles.
• Two quadrilaterals of different angles.

Question 3.
State whether the following quadrilaterals are similar or not.

Solution:
No, the sides of quadrilateral PQRS and ABCD are proportional but their corresponding angles are not equal.

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-6-ex-6-2/

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

Question 1.
In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

Question 2.
E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:

Question 3.
In the given figure, if LM || CB and LN || CD.


Solution:

Question 4.
In the given figure, DE || AC and DF || AE.


Solution:

Question 5.
In the given figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

Question 6.
In the given figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

Question 7.
Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∆ABC in which D is the mid-point of AB and DE || BC

Question 8.
Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Given: A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the
Solution:

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) Show that ABCD is a trapezium.
Solution:

We hope the NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2, drop a comment below and we will get back to you at the earliest.