# NCERT Solutions For Class 10 Maths Chapter 6 Triangles Ex 6.5

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## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5

NCERT TEXTBOOK EXERCISES

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 = 1/2 x 2 a = a
In right angled triangle ADB,
AD2 + BD2 = AB2
⇒ AD2 = AB2 – BD2= (2a)2 – (a)2 = 4a2– a2= 3a2

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+ DA= 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 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 2AB= 2AC2 + BC2.

Solution:

Question 15. In an equilateral triangle ABC, D is a point on side BC, such that BD = 1/3BC. 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:

## NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.5 PDF

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