Find the length of altitude AD of an isosceles $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units.
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Given: $\triangle\text{ABC}$ in which AB = AC = 2a units and BC = a units. Const: Draw $\text{AD}\perp\text{BC}$ then D is the midpoint of BC.
In $\triangle\text{ABC}$ $\text{BC}=\text{a}$ and $\text{BD}=\frac{\text{BC}}{2}=\frac{\text{a}}{2}$ In $\triangle\text{ADB},$ $(\text{AB})^2=\text{AD}^2+\text{BD}^2$ $\text{AD}^2=\Big(\text{AB}^2-\text{BD}^2\Big)$ $\text{AD}^2=\bigg[(\text{2a})^2-\Big(\frac{\text{a}}{2}\Big)^2\bigg]$ $\text{AD}^2=\bigg[4\text{a}^2-\frac{\text{a}^2}{4}\bigg]=\frac{15\text{a}^2}{4}$ $\Rightarrow\text{AD}=\frac{\text{a}\sqrt{15}}{2}\text{units}$
In $\triangle\text{ABC}$ $\text{BC}=\text{a}$ and $\text{BD}=\frac{\text{BC}}{2}=\frac{\text{a}}{2}$ In $\triangle\text{ADB},$ $(\text{AB})^2=\text{AD}^2+\text{BD}^2$ $\text{AD}^2=\Big(\text{AB}^2-\text{BD}^2\Big)$ $\text{AD}^2=\bigg[(\text{2a})^2-\Big(\frac{\text{a}}{2}\Big)^2\bigg]$ $\text{AD}^2=\bigg[4\text{a}^2-\frac{\text{a}^2}{4}\bigg]=\frac{15\text{a}^2}{4}$ $\Rightarrow\text{AD}=\frac{\text{a}\sqrt{15}}{2}\text{units}$
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