Question

An isosceles triangle has a base $118mm$ long and two equal sides are each $143mm$ long. calculate the are of the triangle.

Answer 1

As we know that, the area of an isosceles triangle is given as-

$A=\frac{1}{2}\times \text{base}\times \text{Altitude}$

Now,

Let $ABC$ be the given triangle.

Therefore,

$AB=AC=143mm$

$BC=118mm$

$BD=CD=\frac{1}{2}BC=59mm$

Now in $△ABD$

$\angle ADC=90°[\because AD\perp BC]$

By pythagoras theorem,

${AC}^2={AD}^2+{CD}^2$

${\left(143\right)}^2={AD}^2+{\left(59\right)}^2$

$\Rightarrow {AD}^2=20449-3481$

$\Rightarrow AD=\sqrt{16968}=130.26mm$

Therefore,

$\text{Area of}△ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times 118\times 130.26=7685.34{mm}^2$

Hence the correct answer is $7685.34{mm}^2$.