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Perimeter and Area of Plane Figures — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsPerimeter and Area of Plane Figures5 Marks
Question
In the given figure, $\triangle\text{ABC}$ is an equilateral triangle the length of whose side is equal to $10\ cm,$ and $\triangle\text{DBC}$ is right-angled at $D$ and $BD = 8\ cm.$ Find the area of the shaded region. $\big[\text{Take}\sqrt{3}=1.41\big]$
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Answer

Given:
Side of equilateral $\triangle \text{ABC} = 10\text{cm}$
$BD = 8\ cm$
Area of equilateral $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{a}^2 ($where $a = 10\ cm)$
Area of equilateral $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{a}^2$
$=25\sqrt{3}$
In the right $\triangle\text{BDC},$ we have:
$BC^2= BD^2+ CD^2$
$\Rightarrow 10^2= 8^2+ CD^2$
$\Rightarrow CD^2= 10^2- 8^2$
$\Rightarrow CD^2= 36$
$\Rightarrow CD = 6$
Area of triangle $\triangle\text{BCD}=\frac12\times\text{b}\times\text{h}$
$=\frac12\times8\times6$
$=24\text{cm}^2$
Area of the shaded region = Area of $\triangle \text{ABC}$ - Area of $\triangle \text{BDC}$
$= 43.30 - 24$
$= 19.3\ cm^2.$

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