Question

Each of equal sides of an isosceles triangle is 4 cm is greater than its height. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

• A. $64cm;108c{m}^2$
• B. $64cm;192c{m}^2$
• C. $64cm;156c{m}^2$
• D. $64cm;132c{m}^2$

Answer 1

The correct option is B $64cm;192c{m}^2$

Let the congruent side of the isosceles triangle be $a$ and the height be $h$.
Given $a=h+4$
Since the height of the triangle, divides it into two right angled triangles,
$a^2=h^2+{\left(\frac{b}{2}\right)}^2=>{(h+4)}^2=h^2+{12}^2$
$h^2+16+8h=h^2+144$
$8h=128$
$h=16cm$
$a=20cm$
Perimeter of the triangle $=$ Sum of all sides $=a+a+b=20+20+24=64cm$
Area of triangle $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times 24\times 16=192$ sq cm