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 other side by 1 cm. Find the length of each side. Also find the perimeter and the area of the triangle.

Let the lengths of the two sides other than hypotenuse be x cm and y cm

According to question,

Hypotenuse = x + 2 (in terms of 1st side)

Hypotenuse = 2y + 1 (in terms of 2nd side)

∴ x + 2 = 2y + 1
x = 2y + 1 - 2
x = 2y - 1

Hypotenuse = 2y + 1

As the given triangle is right-angled, by using Pythagoras theorem, we get:

$x^2 + y^2 = (2y + 1)^2$

Putting value of x = 2y - 1 in above equation

$\Rightarrow (2y - 1)^2 + y^2 = (2y + 1)^2 \\[1em] \Rightarrow 4y^2 + 1 - 4y + y^2 = 4y^2 + 1 + 4y \\[1em] \Rightarrow 4y^2 - 4y^2 + y^2 + 1 - 1 - 4y -4y = 0 \\[1em] \Rightarrow y^2 - 8y = 0 \\[1em] \Rightarrow y(y - 8) = 0 \\[1em] \Rightarrow y = 0 \text{ or } y - 8 = 0 \\[1em] y = 0 \text{ or } y = 8$

y ≠ 0 , as that will make value of x negative and length cannot be negative.

∴ y = 8 , x = 2y - 1 = 15 , hypotenuse = 2y + 1 = 17

Perimeter = 8 + 15 + 17 = 40 cm

$\text{Area} = \dfrac{1}{2} \times \text{first side } \times \text{ second side} \\[1em] = \dfrac{1}{2} \times 8 \times 15 \text{ cm}^2 \\[1em] = 60 \text{ cm}^2$

Hence, the value of first side = 8cm, second side = 15 cm , Hypotenuse = 17cm , Perimeter = 40cm , Area = 60 cm².