The perimeters of two squares are 120 cm and 64 cm. Find the perimeter of the square whose area is equal to the sum of the areas of these two squares.

Given:

The perimeters of two squares are 120 cm and 64 cm.

Let a be the side of the square.

As we know that perimeter of the square = 4 x side

⇒ 4 x a = 120

⇒ a = $\dfrac{120}{4}$

⇒ a = 30 cm

The perimeter of square is 64 cm.

Let b be the side of the square.

As we know that perimeter of the square = 4 x side

⇒ 4 x b = 64

⇒ a = $\dfrac{64}{4}$

⇒ a = 16 cm

It is given that area of third square = the sum of the areas of these two squares

Let R be the side of the square.

As we know that area of the square = side²

⇒ R² = 30² + 16²

⇒ R² = 900 + 256

⇒ R² = 1156

⇒ R = $\sqrt{1156}$

⇒ R = 34 cm

Perimeter of third square = 4 x 34 cm

= 136 cm

Hence, the perimeter of the square is 136 cm.