Question

The circle inscribed in an equilateral triangle of sides 24 cm just touches the sides of the triangle. Find the area of the rest part of the triangle (let $\sqrt{3}$=1.732)

• A. 93 sq.cm
• B. 98.54 sq.cm
• C. 96.67 sq.cm
• D. 92.70 sq.cm

Answer 1

The correct option is B

98.54 sq.cm

In the figure, $\Delta$ ABC is an equilateral triangle, each side of which is 24 cm. AD is the perpendicular drawn from A to BC. Since the triangle is equilateral, D bisects BC.
$\therefore$ BD = CD = $\frac{24}{2}$cm = 12 cm.
The incentre and centroid of the $\Delta$ ABC are the same point O.
$\therefore$ OD=$\frac{1}{3}$ AD.....(i)
Height of the triangle ABC = AD = $\frac{\sqrt{3}}{2}\times$ 24 cm = $12\sqrt{3}$ cm
From (i) we get, OD = $\frac{1}{3}\times$ AD = $\frac{1}{3}\times 12\sqrt{3}$ cm = 4$\sqrt{3}cm$.
Area of the circle = $\pi \times (OD{)}^2=\{\frac{22}{7}\times (4\sqrt{3}{)}^2\}=(\frac{22}{7}\times 48)=150.86$
Area of the $\Delta$ ABC = $\frac{\sqrt{3}}{4}\times (side{)}^2=\frac{\sqrt{3}}{4}\times (24{)}^2$ = 144 $\sqrt{3}$
= 144$\times$ 1.732 = 249.40
$\therefore$ Area of the rest part of the $\Delta$ ABC = (249.40-150.86) sq-cm = 98.54 sq-cm