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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 1
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

NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 2
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 3
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 4

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 5

Question 6.
ABC is an equilateral triangle of side la. Find each of its altitudes.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 6
Given: In ∆ABC, AB = BC = AC = 2a
We have to find length of AD
In ∆ABC,
AB = BC = AC = 2a
and AD ⊥ BC
BD = 12 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+ DA= AC2+ BD2
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 7

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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 8
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 9

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 10
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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 11

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 112 hours?
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 12

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 13
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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 14

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.
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 15
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 16

Question 15.
In an equilateral triangle ABC, D is a point on side BC, such that BD = 13BC. Prove that 9AD2 = 7AB2.
Solution:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 17

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:
NCERT Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.5 18

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 Ex 6.5 19