The perimeter of an isosceles triangle is 42 cm and its base is


The perimeter of an isosceles triangle is $42 \mathrm{~cm}$ and its base is $1 \frac{1}{2}$ times each of the equal sides. Find

(i) the length of each side of the triangle,

(ii) the area of the triangle, and

(iii) the height of the triangle.


Let the equal sides of the isosceles triangle be a cm each.

$\therefore$ Base of the triangle, $b=\frac{3}{2} \mathrm{a} \mathrm{cm}$

(i) Perimeter $=42 \mathrm{~cm}$

or, $a+a+\frac{3}{2} a=42$

or, $2 a+\frac{3}{2} a=42$

$\Rightarrow 2 a+\frac{3}{2} a=42$

$\Rightarrow \frac{7 a}{2}=42$

$\Rightarrow a=12$

So, equal sides of the triangle are 12 cm each.

Base $=\frac{3}{2} a=\frac{3}{2} \times 12=18 \mathrm{~cm}$


Area of isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$

$=\frac{18}{4} \times \sqrt{4(12)^{2}-18^{2}} \quad(a=12 \mathrm{~cm}$ and $b=18 \mathrm{~cm})$

$=4.5 \times \sqrt{576-324}$

$=4.5 \times \sqrt{252}$

$=4.5 \times 15.87$

$=71.42 \mathrm{~cm}^{2}$


Area of triangle $=71.42 \mathrm{~cm}^{2}$

$\Rightarrow \frac{1}{2} \times$ Base $\times$ Height $=71.42$

$\Rightarrow$ Height $=\frac{71.42 \times 2}{18}=7.94 \mathrm{~cm}$


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