**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

QR^{2} = PQ^{2} + PR^{2 } (From the rule of Pythagoras property)

QR^{2} = 10^{2} + 24^{2}

QR^{2} = 100 + 576

QR^{2} = 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,

AB^{2} = AC^{2} + BC^{2} (From the rule of Pythagoras Property)

252 = 72 + BC^{2}

625 = 49 + BC^{2}

Transposing 49 from RHS to LHS it becomes – 49

BC^{2} = 625 – 49

BC^{2} = 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,

15^{2} = 12^{2} + a^{2} (From the rule of Pythagoras Property)

225 = 144 + a^{2}

Transposing 144 from RHS to LHS it becomes – 144

a^{2} = 225 – 144

a^{2} = 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.

b^{2} = a^{2} + c^{2}

(6.5)^{2} = (2.5)^{2} + 6^{2}

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.

C^{2} = a^{2}+ b^{2} (From the rule of Pythagoras Property)

5^{2} = 2^{2} + 2^{2}

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.

b^{2} = a^{2} + c^{2} (From the rule of Pythagoras Property)

(2.5)^{2} = (1.5)^{2} + 2^{2}

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.

QR^{2} = PQ^{2} + PR^{2 }(From the rule of Pythagoras theorem)

QR^{2} = 5^{2} + (12)^{2}

QR^{2} = 25 + 144

QR^{2}= 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 25 ^{o} and 65^{o}.**

Write which of the following is true:

(i) PQ^{2} + QR^{2} = RP^{2}

(ii) PQ^{2} + RP^{2} = QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

**Solution:**

Given,

∠Q = 25^{o}, ∠R = 65^{o}

Then we have to find,

∠P =?

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

So,

∠PQR + ∠QRP + ∠RPQ = 180^{o}

25^{o} + 65^{o} + ∠RPQ = 180^{o}

90^{o} + ∠RPQ = 180^{o}

∠RPQ = 180 – 90

∠RPQ = 90^{o}

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

So, QR^{2} = PQ^{2} + PR^{2} 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,

AC^{2} = AB^{2} + BC^{2}

41^{2} = 40^{2} + BC^{2}

BC^{2} = 41^{2} – 40^{2}

BC^{2} = 1681 – 1600

BC^{2} = 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 90^{o}.

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,

PS^{2} = PO^{2} + SO^{2}

PS^{2} = 82 + 152

PS^{2} = 64 + 225

PS^{2} = 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.

👍👍👍

**Live online tuition for CBSE board (Sanskrit tuition):**

Live online tuition for CBSE board (Sanskrit tuition): On “Google Meet” or “Zoom” is running.

Charges: Only ₹500/ month where you can ask your question face to face directly live. Each batch will have at max 100 students. Our main focus is to get full/full marks in your exam, in this way we will study there.

TUITION DURATION: – 1 hour: Monday to Saturday. Sunday- Off.

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

FREE DEMO CLASS: 3

Seats are limited because of zoom classes. So, book your seat right now for free. Pay fee after demo classes directly on my upi id: maketoss@ybl

Our WhatsApp Contact: Please Don’t call. Do only WhatsApp Message: 9117748369

Free Registration link: Click here