Question
In a $\triangle\text{ABC},$ AB = BC = CA = 2a and $\text{AD}\perp\text{BC}.$ Prove that
Area $(\triangle\text{ABC})=\sqrt{3}\text{ a}^2$
Area $(\triangle\text{ABC})=\sqrt{3}\text{ a}^2$
✓
Answer
$\triangle\text{ABC},$ AB = BC = AC = 2a
$\text{AD}\perp\text{BC}$
AD bisects BC at D
$\text{BD}=\text{DC}=\frac{1}{2}\text{BC}=\text{a}$
Area of $\triangle\text{ABC}$
$=\frac{1}{2}\text{BC}\times\text{AD}$
$=\frac{1}{2}\times2\text{a}\times\sqrt{3}\text{ a}=\sqrt{3}\text{ a}^2$
Hence proved.
$\text{AD}\perp\text{BC}$
AD bisects BC at D
$\text{BD}=\text{DC}=\frac{1}{2}\text{BC}=\text{a}$
Area of $\triangle\text{ABC}$
$=\frac{1}{2}\text{BC}\times\text{AD}$
$=\frac{1}{2}\times2\text{a}\times\sqrt{3}\text{ a}=\sqrt{3}\text{ a}^2$
Hence proved.
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