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Area of Circle, Sector and Segment — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsArea of Circle, Sector and Segment4 Marks
Question
The area of an equilateral triangle is $49\sqrt{3}\text{cm}^2.$ taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. $\big[\text{Take }\sqrt{3}=1.73\big]$
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Answer

Area of equilateral triangle $\text{ABC}=49\sqrt{3}\text{cm}^2$

Let a be its side
$\therefore\frac{\sqrt{3}}{4}\text{a}^2=49\sqrt{3}$
Or $\text{a}^2=49\times4$
$\therefore\text{a}=7\times2$
$\Rightarrow\text{a}=14\text{cm}$
Area of sector $\text{BDF}=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$=\frac{22}{7}\times7\times7\times\frac{60}{360}\text{cm}$
$=\frac{11\times7}{3}\text{cm}^2=\frac{77}{3}\text{cm}^2$
Area pf sector $BDF =$ Area of sector $CDE =$ Area of sector $AEF$
Sum of area of all the sectors
$=\frac{77}{3}\times3\text{cm}^2=77\text{cm}^2$
$\therefore$ Shaded area = Area of $\triangle\text{ABC}$ - Sum of area of all sectors
$=49\sqrt{3}-77\text{cm}^2=(84.77-77.00)\text{cm}^2$
$=7.77\text{cm}^2$

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