**(i)** Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)^{2} + (24)^{2} = (25)^{2}

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

**(ii)** Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3^{2} + 6^{2} ≠ 8^{2}

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

**(iii)** Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50^{2} + 80^{2} ≠ 100^{2}

As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

**(iv)** Given, sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169

Or, 12^{2} + 5^{2} = 13^{2}

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

Answered by Abhisek | 1 year agoIn an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm^{2}. Find ar(∆COD)

Find the length of the altitude of an equilateral triangle of side 2a cm.

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.