Question

The perimeter of an isosceles triangle is $42$ $cm$ and its base is $\left(3/2\right)$ times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.

Answer 1

Let two equal sides $AB$ and $AC$ be $x$ & $x$

Base $BC=\frac{3}{2}x$

Perimeter$=42$

$x+x+\frac{3x}{2}=42\frac{3}{2}x+2x=42\frac{7x}{2}=42x=6\times 2=12$

$AB=AC=12BC=\frac{3}{2}\times 12=18$

In $△ADC$

${AC}^2={AD}^2+{DC}^2{12}^2={AD}^2+9^{2}144-81={AD}^2{AD}^2=63AD=\sqrt{63}=3\sqrt{7}$

Area of $△ABC=\frac{1}{2}\times AD\times BC$

$=\frac{1}{2}\times 3\sqrt{7}\times 18=27\sqrt{7}$

Height of triangle$=AD=3\sqrt{7}$