Question
The given figure shows a right$-$angled $\triangle ABC$ and an equilateral $\triangle BCD$. Find the area of the shaded portion.
✓
Answer
From $\triangle ABC,$
$AB =\sqrt{ AC ^2- BC ^2} $
$=\sqrt{16^2-8^2} $
$=\sqrt{192}$
Area of $ΔABC$
$ \triangle ABC =\frac{1}{2} \times 8 \times \sqrt{192} $
$=4 \sqrt{192}$
Area of $\triangle BCD$
$ \triangle BCD =\frac{\sqrt{3}}{4} \times 8^2$
$ =16 \sqrt{3}$
Now
$\text{ABD} = \text{ABC} - \text{BDC}$
$=4 \sqrt{192}-16 \sqrt{3} $
$ =32 \sqrt{3}-16 \sqrt{3} $
$ =16 \sqrt{3} sq \cdot cm $
$AB =\sqrt{ AC ^2- BC ^2} $
$=\sqrt{16^2-8^2} $
$=\sqrt{192}$
Area of $ΔABC$
$ \triangle ABC =\frac{1}{2} \times 8 \times \sqrt{192} $
$=4 \sqrt{192}$
Area of $\triangle BCD$
$ \triangle BCD =\frac{\sqrt{3}}{4} \times 8^2$
$ =16 \sqrt{3}$
Now
$\text{ABD} = \text{ABC} - \text{BDC}$
$=4 \sqrt{192}-16 \sqrt{3} $
$ =32 \sqrt{3}-16 \sqrt{3} $
$ =16 \sqrt{3} sq \cdot cm $
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