Question

The length (in cm) of the hypotenuse of a right-angled triangle exceeds the length of one side by 2 cm and exceeds twice the length of another side by 1 cm. Find the length of each side. Also, find the perimeter and the area of the triangle.

Answer 1

Let the length of one side = x cm

And other side = y cm.

Then hypotenues = x + 2, and 2y + 1

∴ x + 2 = 2y + 1 

⇒ x - 2y 1 - 2

⇒ x - 2y = -1

⇒ x = 2y - 1     ...(i)

and using Pythagorous theorem,

x² + y² = (2y + 1)²

⇒ x² + y² = 4y² + 4y + 1

⇒ (2y - 1)² + y² = 4y² + 4y + 1   ...[From (i)]

⇒ 4y² - 4y + 1 + y² = 4y² + 4y + 1

⇒ 4y² - 4y + 1 + y² - 4y² - 4y - 1 = 0

⇒ y² - 8y = 0

⇒ y(y - 8) = 0

Either y = 0,

but it is not possible

or

y - 8 = 0,

then y = 8

Substituting the value of y in (i)

x = 2(8) - 1 

= 16 - 1

= 15

∴ Length of one side = 15 cm

and length of other side = 8 cm

and hypotenuse

= x + 2

= 15 + 2

= 17

∴ Perimeter

= 15 + 8 + 17

= 40 cm

and Area

= `(1)/(2)` × one side × other side

= `(1)/(2) xx 15 xx 8`

= 60 cm²