Question

In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides, find the areas of the remaining portion of the triangle.

Answer 1

Hint: Here to find the remaining area of the triangle, we consider the relation between area of triangle, semi- perimeter and radius of inner circle as the circle is inscribed in the triangle i.e.., $R={\displaystyle \frac{A}{s}}$

Complete step-by-step answer:
We know the relation between area of triangle, semi-perimeter, radius of inner circle is $R={\displaystyle \frac{A}{s}}$ , where A is area of triangle, s is semi perimeter, R radius of incircle.
Given, the side of an equilateral triangle is 24 cm.
Also, semi perimeter s can be given as $s={\displaystyle \frac{24+24+24}{2}}=36cm$.
The formula for the area A for the equilateral triangle is ${\displaystyle \frac{\sqrt{3}}{2}}(side{)}^2$ .
So, the area will be ${\displaystyle \frac{\sqrt{3}}{2}}(24{)}^2=144\sqrt{3}c{m}^2$ .
Now, we have got the value of A and s, so we can easily calculate the value of R as follows: $R={\displaystyle \frac{144\sqrt{3}}{36}}=4\sqrt{3}cm$ .
Formula for area for the circle is $\pi (radius{)}^2$ since we have got the radius, we can move further.
Area of the circle will be $\pi R^2=\pi \times 4^2\times 3=48\pi$ .
 Area of the remaining portion will be equal to the area of the circle subtracted from the total area of the triangle. Viz. $(144\sqrt{3}-48)\pi c{m}^2$

Note: We can also put the value of $\pi$ as ${\displaystyle \frac{22}{7}}$ or $3.14$ for further simplification. While solving these types of problems we need to find the relations between the triangles and circles such that it makes the simplification easier.