Question

The area of a right triangle is `600cm^2` . If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Answer 1

Let the altitude of the triangle be x cm 

Therefore, the base of the triangle will be `(x+10)cm` 

Area of triangle  = `1/2x(x+10)=600` 

⇒`(x+10)=1200` 

⇒ `x^2+10x-1200=0` 

⇒`x^2+(40-30)x-1200=0` 

⇒`x^2+40x-30x-1200=0` 

⇒`x(x+40)-30(x+40)=0` 

⇒`(x+40) (x-30)=0` 

⇒`x=-40  or  x=30` 

⇒`x=30`                      [∵ Altitude  cannot be negative] 

Thus, the altitude and base of the triangle are 30 cm and `(30 + 10 = 40)` cm, respectively. 

`"(Hypotenuse)^2=(Altitude)^2+(Base)^2"` 

⇒` "(Hypotenuse)^2=(30)^2(40)^2"` 

⇒` "(Hypotenuse)^2=900+1600=2500"` 

⇒ `"(Hypotenuse)^2=(50)^2"` 

⇒  `"(Hypotenuse)^2=50"` 

Thus, the dimensions of the triangle are:

Hypotenuse = 50 cm

Altitude = 30 cm

Base = 40 cm