Question

The perimeter of an isosceles triangle is 42 cm and its base is $1\frac{1}{2}$ times its congruent sides. the area of the triangle

Answer 1

Let the congruent sides of the
isosceles triangle $=a$ and the base $=b$.

Given, $b=1\frac{1}{2}\times a=\frac{3}{2}\times a$

Perimeter of the triangle $=$Sum of all sides $=$ a + a + b $=42$

$=>2a+\frac{3}{2}\times a=42$

On solving, a $=12cm$

$b=\frac{3}{2}\times a=18$ cm
Since, the height of the triangle, divides it into two right angled triangles,
$a^2=h^2+{\left(\frac{b}{2}\right)}^2=>{12}^2=h^2+9^2$

$h^2=144-81=63$

$h=\sqrt{63}=\sqrt{9\times 7}=3\sqrt{7}$ cm

Area of triangle $=\frac{1}{2}\times base\times height=\frac{1}{2}\times 18\times 3\sqrt{7}=27\sqrt{7}$ sq cm