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
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.
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.
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.
(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.

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
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.
In the right triangle ABD,

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.
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

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