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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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