Question

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle. 

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of all sides. 

Answer 1

Let the base and altitude of the right-angled triangle be x and y cm, respectively Therefore, the hypotenuse will be `(x + 2) cm`. 

∴ `(x+2)^2=y^2+x^2`                 ...............(1) 

Again, the hypotenuse exceeds twice the length of the altitude by 1 cm. 

∴ `h=(2y+1)` 

⇒ `x+2=2y+1` 

⇒ ` x=2y-1` 

Putting the value of x in (1), we get: 

`(2y-1+2)^2=y^2+(2y-1)^2` 

⇒ ` (2y+1)^2=y^2+4y^2-4y+1` 

⇒ ` 4y^2+4y+1=5y^2-4y+1` 

⇒ `-y^2+8y=0` 

⇒ `y^2-8y=0` 

⇒ `y(y-8)=0` 

⇒ `y=8 cm` 

∴ `x=16-1=15 cm` 

∴ ` h=16+1=17 cm` 

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.