NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5

These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR. Ans: In the right APQR
Answer:
In the right ΔPQR

QR² = PQ² + PR²
(pythagoras property)
= 10² + 24²
= 100 + 576
= 676
QR² = 26²
∴ QR = 26 cm

Question 2.
ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.
Answer:
In the right ΔABC

AB² = AC² + BC²
(using pythagoras property)
25² = 7² + BC²
625 = 49 + BC²
625 – 49 = BC²
576 = BC²
24² = BC²
∴ BC = 24 cm

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.
Answer:
Let the distance of the foot of a ladder from the wall be ‘a’m

a2² + 12² = 15²
(using pythagoras property)
a² + 144 = 225
a² = 225 – 144
a² = 81
a² = 9²
a = 9 m
The distance of the foot of the ladder from the wall = 9 m.

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.
In the case of right-angled triangles, identify the right angles.
Answer:
(i) 2.5 cm, 6.5 cm, 6 cm².
The longest side is 6.5 cm
2.5² + 6² = 6.25 + 36
= 42.25
6.5² = 42.25
∴ 6.5² = 2.5² + 6²
(pythagoras property)
The given lengths can be the sides of a right triangle.
The right angle is the angle between the sides 2.5 cm and 6 cm

(ii) 2 cm, 2 cm, 5 cm
The longest side is 5 cm
2² + 2² = 4 + 4 = 8
5² = 25
5² ≠ 2² + 2²
∴ The given length cannot be the sides of a right triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm
The longest side is 2.5 cm
1.5² + 2² = 2.25 + 4
= 6.25
2.5² = 6.25
1.5² + 2² = 2.5²
(pythagoras property)
The given lengths can be sides of a right triangle.
The right angle is the angle between the sides 1.5 cm and 2 cm.

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.

Answer:
Let the tree BD be broken at the point C, such that CD = CA
In the right triangle ABC, using the pythagoras property; we get
AC² = AB² + BC²
AC² = 12² + 5²
= 144 + 25
AC² = 169
AC² = 13²
∴ AC = 13
Now, height of the tree
BD = BC + CD (CD = AC)
= 5 m + 13 m
= 18 m
The height of the tree is 18 m.

Question 6.
Angles Q and R of PQR are 25° and 65°. Write which of the following is true:
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
Answer:
In ΔPQR
∠P + ∠Q +∠R = 180°
∠P + 25° + 65° = 180°

∠P + 90° = 180°
∠P = 180° – 90°
= 90°
So, ΔPQR is a triangle right angled at P.
∴ QR is the hypotenuse.
∴ using the pythagoras property
QR² = PQ² + PR² So,

(ii) PQ² + RP² = QR² is true

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.
Answer:
Let the breadth AD be x cm.
In the right triangle ABD,

BD² = AD² + AB²
41²= x² + 40²
x² = 41² – 40²
= 1681 – 1600
x² = 81
x² = 9²
x = 9 cm
Perimeter of the rectangle = 2 (l + b)
= 2 (40 + 9)
= 2 × 49
= 98 cm
Thus, the perimeter of the rectangle
= 98 cm

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.
Answer:
Since the diagonals of a rhombus bisect each other at right angles.

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
AO = \(\frac { 1 }{ 2 }\) AC; AO = \(\frac { 1 }{ 2 }\) × 30 = 15
∵ AO = OC = 15 cm and BO = OD = 8 cm
In the right ΔAOB,
AB² = AO² + BO²
AB² = 15² + 8²
AB² = 225 + 64
AB² = 289
AB² = 17²
AB = 17 cm
∵ Side of the rhombus is 17 cm.
∴ Perimeter of the rhombus ABCD = 4 × 17
(all the four sides are equal) = 68 cm