Question

Each of the equal sides of an isosceles triangle measures 2 cm more than its height, and the base of the triangle measures 12 cm. Find the area of the triangle.

Answer 1

Let the height of the triangle be h cm.

Each of the equal sides measures  `a=(h+2)cm and b=12 cm(base)` 

Now, Area of the triangle = Area of the isosceles triangle 

=`1/2xxbasexxheight=1/4xxbsqrt(4a^2-b^2)` 

⇒ `1/2xx12xxh=1/4xx12xxsqrt(4(h+2)^2-144)` 

⇒`6h=3sqrt(4h^2+16h-144)` 

⇒`2h=sqrt(4h^2+16h+16-144)` 

On squaring both the sides, we get: 

⇒`4h^2=4h^2+16h+16-144` 

⇒`16h-128=0` 

⇒`h=8` 

Area of the triangle=`1/2xxbxxh` 

=`1/2xx12xx8` 

=`48cm^2`