Class 7 Ex. 6.5 Maths NCERT Solutions

Solutions For All Chapters Maths Class 7

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

Ex 6.5 Class 7 Maths Question 1.
PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:

In right angled triangle PQR, we have
QR2 = PQ2 + PR2 From Pythagoras property)
= (10)2 + (24)2
= 100 + 576 = 676

∴ QR = $\sqrt{676}$ = 26 cm

The, the required length of QR = 26 cm.

Ex 6.5 Class 7 Maths Question 2.
ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Solution:

BC2 + (7)2 = (25)2 (By Pythagoras property)
⇒ BC2 + 49 = 625
⇒ BC2 = 625 – 49
⇒ BC2 = 576

∴ BC = $\sqrt{576}$ = 24 cm

Thus, the required length of BC = 24 cm.

Ex 6.5 Class 7 Maths Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:

In right angled∆ABC, we have
Solution:
Here, the ladder forms a right angled triangle.

∴ a² + (12)² = (15)² (By Pythagoras property)

⇒ a²+ 144 = 225

⇒ a² = 225 – 144

⇒ a² = 81

∴ a = $\sqrt{81}$ = 9 m

Thus, the distance of the foot from the ladder = 9m

Ex 6.5 Class 7 Maths Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm.

(ii) 2 cm, 2 cm, 5 cm.

(iii) 1.5 cm, 2 cm, 2.5 cm

Solution:

(i) Given sides are 2.5 cm, 6.5 cm, 6 cm.

Square of the longer side = (6.5)² = 42.25 cm.

Sum of the square of other two sides

= (2.5)² + (6)² = 6.25 + 36

= 42.25 cm.
Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.
∴ The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .
Square of the longer side = (5)²= 25 cm Sum of the square of other two sides
= (2)² + (2)² =4 + 4 = 8 cm
Since 25 cm ≠ 8 cm
∴ The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)² = 6.25 cm Sum of the square of other two sides
= (1.5)² + (2)²= 2.25 + 4
Since 6.25 cm = 6.25 cm = 6.25 cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
∴ The given sides form a right triangle.

Ex 6.5 Class 7 Maths Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.

Solution:

Let AB be the original height of the tree and broken at C touching the ground at D such that

AC = 5 m
and AD = 12 m
In right triangle ∆CAD,

AD² + AC² = CD² (By Pythagoras property)
⇒ (12)² + (5)² = CD²
⇒ 144 + 25 = CD²
⇒ 169 = CD²

∴ CD = $\sqrt{169}$ = 13 m

But CD = BC
AC + CB = AB
5 m + 13 m = AB
∴ AB = 18 m .
Thus, the original height of the tree = 18 m.

Ex 6.5 Class 7 Maths Question 6.
Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.

(i) PQ² + QR² = RP²

(ii) PQ²+ RP² = QR²

(iii) RP² + QR² = PQ²

Solution:

We know that
∠P + ∠Q + ∠R = 180° (Angle sum property)
∠P + 25° + 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° – 90°
∆PQR is a right triangle, right angled at P
(i) Not True
∴ PQ² + QR² ≠ RP² (By Pythagoras property)
(ii) True
∴ PQ² + RP² = QP² (By Pythagoras property)
(iii) Not True
∴ RP² + QR² ≠ PQ² (By Pythagoras property)

Ex 6.5 Class 7 Maths Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:

Given: Length AB = 40 cm
Diagonal AC = 41 cm
In right triangle ABC, we have
AB² + BC²= AC² (By Pythagoras property)
⇒ (40)² + BC² = (41)²
⇒ 1600 + BC² = 1681
⇒ BC2 = 1681 – 1600
⇒ BC2 = 81

∴ BC = $\sqrt{81}$ = 9 cm

∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)
∴ The required perimeter
= AB + BC + CD + DA
= (40 + 9 + 40 + 9) cm
= 98 cm

Ex 6.5 Class 7 Maths Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:

Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
∴ OA = OC = 8 cm and OB = OD = 15 cm
In right ∆OAB,
AB² = OA² + OB² (By Pythagoras property)
= (8)²+ (15)² = 64 + 225
= 289

∴ AB = $\sqrt{289}$ = 17 cm

Since AB = BC = CD = DA (Property of rhombus)
∴ Required perimeter of rhombus
= 4 × side = 4 × 17 = 68 cm.