In a ∆ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD ⊥ BC, then AD =


In a $\triangle \mathrm{ABC}, \angle \mathrm{A}=90^{\circ}, \mathrm{AB}=5 \mathrm{~cm}$ and $\mathrm{AC}=12 \mathrm{~cm}$. If $\mathrm{AD} \perp \mathrm{BC}$, then $\mathrm{AD}=$

(a) 132cm
(b) 6013cm
(c) 1360cm
(d) 21513cm


Given: In $\triangle \mathrm{ABC} \angle \mathrm{A}=90^{\circ}, \mathrm{AD} \perp \mathrm{BC}, \mathrm{AC}=12 \mathrm{~cm}$, and $\mathrm{AB}=5 \mathrm{~cm}$.

To find: AD

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

In ∆ACB and ∆ADC,

∠C=∠C        (Common)


∴ ∆ACB ~∆ADC     (AA Similarity)


$\mathrm{AD}=\frac{\mathrm{AB} \times \mathrm{AC}}{\mathrm{BC}}$

$\mathrm{AD}=\frac{12 \times 5}{13}$


We got the result as (b)

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