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
= 40cm
and Area

= `(1)/(2)` x one side x other side

= `(1)/(2) xx 15 xx 8`
= 60cm².