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# Mathematics – Class 7 – Chapter 6 – The Triangle And Its Properties – Exercise 6.5 – NCERT Exercise Solution

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 diagram of right-angled triangle:

In the above figure, RQ is the hypotenuse, PR is the base and PQ is the height

QR2 = PQ2 + PR2  (From the rule of Pythagoras property)

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 diagram of right-angled triangle:

In the above figure, RQ is the hypotenuse,

AB2 = AC2 + BC2  (From the rule of Pythagoras Property)

252 = 72 + BC2

625 = 49 + BC2

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.

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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 the above figure RQ is the hypotenuse,

152 = 122 + a2 (From the rule of Pythagoras Property)

225 = 144 + a2

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.

Solution:

(i) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

Let the largest value is the hypotenuse of the right-angled triangle i.e. b = 6.5 cm.

According to the rule of the Pythagoras theorem, we know that the square of the longer side is equal to the sum of the square of the other two sides.

b2 = a2 + c2

(6.5)2 = (2.5)2 + 62

42.25 = 6.25 + 36

42.25 = 42.25

Here the sum of the square of two sides of the triangle is equal to the square of the third side,

Hence 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

Let the largest value is the hypotenuse side of the right-angled triangle i.e. c = 5 cm.

C2 = a2+ b2 (From the rule of Pythagoras Property)

52 = 22 + 22

25 = 4 + 4

25 ≠ 8

Here, the sum of the square of the two sides of the triangle is not equal to the square of the third side.

Hence, the given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

Let the largest value is the hypotenuse side of right-angled triangle i.e. b = 2.5 cm.

b2 = a2 + c2 (From the rule of Pythagoras Property)

(2.5)2 = (1.5)2 + 22

6.25 = 2.25 + 4

6.25 = 6.25

Here, the sum of the square of the two sides of the triangle is equal to the square of the third side.

Hence, 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 PQR is the triangle and Q is the point where tree is broken at the height 5 m from the ground.

Tree top touches the ground at a distance of PR = 12 m from the base of the tree.

By observing the figure we came to conclude that right angle triangle is formed at P.

QR2 = PQ2 + PR2 (From the rule of Pythagoras theorem)

QR2 = 52 + (12)2

QR2 = 25 + 144

QR2= 169

QR = √169

QR = 13 m

Then, the original height of the tree = PQ + QR

= 5 + 13

= 18 m

6. Angles Q and R of a ΔPQR are 25o and 65o.

Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

Solution:

Given,

∠Q = 25o, ∠R = 65o

Then we have to find,

∠P =?

We know that sum of the three interior angles of triangle is equal to 180o.

So,

∠PQR + ∠QRP + ∠RPQ = 180o

25o + 65o + ∠RPQ = 180o

90o + ∠RPQ = 180o

∠RPQ = 180 – 90

∠RPQ = 90o

We also know that the side opposite to the right angle is the hypotenuse of that right-angled triangle.

So, 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:

First, we draw a rough sketch according to the question:

Let ABCD be the rectangle.

Given, AB = 40 cm and AC = 41 cm

BC =?

According to the rule of Pythagoras theorem,

From the right-angled triangle ABC,

AC2 = AB2 + BC2

412 = 402 + BC2

BC2 = 412 – 402

BC2 = 1681 – 1600

BC2 = 81

BC = √81

BC = 9 cm

length = 40 cm, breadth = 9 cm

Now, the perimeter of the rectangle = 2 (length + breadth)

Then,

= 2(40 + 9)

= 2 × 49

= 98 cm

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

Solution:

First, we draw a rough sketch according to the question:

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 we 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

Hence, Perimeter of rhombus is 68 cm.

👍👍👍

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TIMING: – Before & After your school timing. We will discuss further on this topic.

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