Question : In $\triangle \mathrm{CAB}, \angle \mathrm{CAB}=90^{\circ}$ and $\mathrm{AD} \perp \mathrm{BC}$. If $\mathrm{AC}=24 \mathrm{~cm}, \mathrm{AB}=10 \mathrm{~cm}$. then find the value of $AD$ (in cm).
Option 1: 9.23
Option 2: 8.23
Option 3: 7.14
Option 4: 10.23
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Correct Answer: 9.23
Solution : Given, In $\triangle \mathrm{CAB}, \angle \mathrm{CAB}=90^{\circ}$ and $\mathrm{AD} \perp \mathrm{BC}$. $\mathrm{AC}=24 \mathrm{~cm}, \mathrm{AB}=10 \mathrm{~cm}$ Applying Pythagoras theorem, $BC^2=AC^2+AB^2$ ⇒ $BC^2 =24^2+10^2$ ⇒ $BC = 26$ cm Now, area of triangle = $\frac{1}{2}\times \text{base}\times \text{height}$ So, $\frac{1}{2}\times AC\times AB=\frac{1}{2}\times AD\times BC$ ⇒ $24\times 10 = AD \times 26$ ⇒ $AD = 9.23$ cm Hence, the correct answer is 9.23 cm.
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