The perimeter of an isosceles triangle is $42\ cm$ and its base is ($\frac{3}{2}) 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.

Given:

The perimeter of an isosceles triangle is $42\ cm$ and its base is ($\frac{3}{2}) times each of the equal sides. 

To do:

We have to find the length of each side of the triangle, area of the triangle and the height of the triangle.

Solution:

The perimeter of the isosceles triangle $= 42\ cm$.

Let each equal side be $x$.

This implies,
Base $=\frac{3}{2}x$

Therefore,

$x+x+\frac{3}{2} x=42$

$\frac{7}{2} x=42$

$x=\frac{42 \times 2}{7}$

$x=12$

Each equal side $=12 \mathrm{~cm}$

The third side $=\frac{12 \times 3}{2}$

$=18 \mathrm{~cm}$

$s=\frac{a+b+c}{2}$

$=\frac{12+12+18}{2}$

$=\frac{42}{2}$

$=21$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{21 \times(21-12)(21-12)(21-18)}$

$=\sqrt{21 \times 9 \times 9 \times 3}$

$=9 \times \sqrt{63}$

$=9 \times \sqrt{3 \times 3 \times 7}$

$=9 \times 3 \sqrt{7}$

$=27 \sqrt{7}$

$=27 \times 2.645$

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

The height of the triangle $=\frac{\text { Area } \times 2}{\text { Base }}$

$=\frac{71.42 \times 2}{18}$

$=\frac{71.42}{9}$

$=7.94 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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