NCERT Solutions Class 7 Math Chapter – 6 The Triangle and its Properties Exercise – 6.5

1. PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution: Let us draw a rough sketch of a right-angled triangle

By the rule of Pythagoras’ Theorem,
Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the square on the legs.
In the above figure RQ is the hypotenuse,
QR² = PQ² + PR²
QR² = 10² + 24²
QR² = 100 + 576
QR² = 676
QR = √676
QR = 26 cm
Hence, the length of the hypotenuse QR = 26 cm.

2. ABC is a triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Solution: Let us draw a rough sketch of a right-angled triangle

By the rule of Pythagoras’ Theorem,
Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of a square on the legs.
In the above figure RQ is the hypotenuse,
AB² = AC² + BC²
25² = 7² + BC²
625 = 49 + BC²
By transposing 49 from RHS to LHS it becomes – 49
BC² = 625 – 49
BC² = 576
BC = √576
BC = 24 cm
Hence, the length of the BC = 24 cm.

3. A 15 m long ladder reached a window 12 m high from the ground by placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution: By the rule of Pythagoras’ Theorem,
Pythagoras’ theorem states that for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the square on the legs.
In the above figure RQ is the hypotenuse,
15² = 12² + a²
225 = 144 + a²
By transposing 144 from RHS to LHS it becomes – 144
a² = 225 – 144
a² = 81
a = √81
a = 9 m
Hence, the length of a = 9 m.

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, 2cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm
Solution: Let us assume the largest value is the hypotenuse side i.e. b = 6.5 cm.
Then, by Pythagoras theorem,
b² = a² + c²
6.5² = 2.5² + 6²
42.25 = 6.25 + 36
42.25 = 42.25
The sum of the square of two sides of the triangle is equal to the square of the third side,
∴The given triangle is a right-angled triangle.
The right angle lies on the opposite of the greater side 6.5 cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm
Solution: Let us assume the largest value is the hypotenuse side i.e. c = 5 cm.
Then, by Pythagoras theorem,
c² = a² + b²
5² = 2² + 2²
25 = 4 + 4
25 ≠ 8
The sum of the square of two sides of triangle is not equal to the square of the third side,
∴The given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm
Solution: Let us assume the largest value is the hypotenuse side i.e. b = 2.5 cm.
Then, by Pythagoras theorem,
b² = a² + c²
2.5² = 1.5² + 2²
6.25 = 2.25 + 4
6.25 = 6.25
The sum of the square of two sides of the triangle is equal to the square of the third side,
∴The given triangle is a right-angled triangle.
The right angle lies on the opposite of the greater side 2.5 cm.

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 ABC is the triangle and B is the point where the tree is broken at the height of 5 m from the ground.
Treetop touches the ground at a distance of AC = 12 m from the base of the tree,

By observing the figure we came to conclude that right angle triangle is formed at A.
From the rule of Pythagoras theorem,
BC² = AB² + AC²
BC² = 5² + 12²
BC² = 25 + 144
BC² = 169
BC = √169
BC = 13 m
Then, the original height of the tree = AB + BC
= 5 + 13
= 18 m

6. Angles Q and R of a ΔPQR are 25^o and 65^o.
Write which of the following is true:
(i) PQ² + QR²  = RP² 
(ii) PQ²  + RP² = QR² 
(iii) RP²  + QR²  = PQ²

Solution: Given that ∠Q = 25^o, ∠R = 65^o
Then, ∠P =?
We know that sum of the three interior angles of triangle is equal to 180^o.
∠PQR + ∠QRP + ∠RPQ = 180^o
25^o + 65^o + ∠RPQ = 180^o
90^o + ∠RPQ = 180^o
∠RPQ = 180^o – 90^o
∠RPQ = 90^o
Also, we know that the side opposite the right angle is the hypotenuse.
∴ QR² = PQ² + PR²
Hence, (ii) is true

7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Solution:

Let ABCD be the rectangular plot.
Then, AB = 40 cm and AC = 41 cm
BC =?
According to Pythagoras theorem,
From right angle triangle ABC, we have:
= AC² = AB2 + BC2
= 41² = 40² + BC²
= BC² = 41² – 40²
= BC² = 1681 – 1600
= BC² = 81
= BC = √81
= BC = 9 cm
Hence, the perimeter of the rectangle plot = 2 (length + breadth)
Where, length = 40 cm, breadth = 9 cm
Then,
= 2(40 + 9)
= 2 × 49
= 98 cm

8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Solution:

Let PQRS be a rhombus, all sides of the rhombus have equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in the rhombus bisect each other at 90o.
So, PO = (PR/2)
= 16/2
= 8 cm
And, SO = (SQ/2)
= 30/2
= 15 cm
Then, consider the triangle POS and apply the Pythagoras theorem,
PS² = PO² + SO²
PS² = 8² + 15²
PS² = 64 + 225
PS² = 289
PS = √289
PS = 17 cm
Hence, the length of side of rhombus is 17 cm
Now,
Perimeter of rhombus = 4 × side of the rhombus
= 4 × 17
= 68 cm
∴ Perimeter of rhombus is 68 cm.