In given figure, $∠ C A B = 90^{0}$ and $A D ⊥ B C$ . If AC=75 cm. AB=1 m and BD=1.25 m, find AD

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Solution

We have,

AB=1 m=100 cm, AC=75 cm and BD=125 cm.

In $△ B A C$ and $△ B D A$ , we have

$∠ B A C = ∠ B D A$ [Each equal to $90^{\circ}$ ]

and, $∠ B = ∠ B$

So, by AA-criterion of similarity, we have

$△ B A C \sim △ B D A$

$\Rightarrow$ $\frac{B A}{B D} = \frac{A C}{A D}$

$\Rightarrow$ $\frac{100}{125} = \frac{75}{A D}$

$\Rightarrow$ $A D = \frac{125 \times 75}{100} c m = 93.75 c m$

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