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

Question:

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.

Solution:

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.
Also,

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

(ii)

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 $b=18 \mathrm{~cm})$

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

$=4.5 \times \sqrt{252}$

$=4.5 \times 15.87$

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

(iii)

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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