In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm, AD = 4 cm, find CD.

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Solution

It is given that ABC is a right angled triangle and BD is the altitude drawn from the right angle to the hypotenuse.

$In △ D B A and △ D C B, we have : ∠ B D A = ∠ C D B ∠ D B A = ∠ D C B = 90 ° Therefore, by AA similarity theorem, we get : △ D B A ~ △ D C B \Rightarrow \frac{B D}{C D} = \frac{A D}{B D} \Rightarrow C D = \frac{B D^{2}}{A D}$

CD = $\frac{8 \times 8}{4} = 16$ cm

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Similar questions

In the given figure, $∠ A B C = 90^{o}$ and $B D ⊥ A C$ . If BD = 8 cm, AD = 4 cm, find CD.

In Fig. 7.139, $∠ A B C = 90^{\circ}$ and $B D ⊥ A C$ . If BD = 8 cm and AD = 4 cm, find CD.

∠ A B C = 90^{\circ}

B D ⊥ A C

A B = 5 \cdot 7 c m, B D = 3 \cdot 8 c m

C D = 5 \cdot 4 c m,

10 B C

△ A B C

∠ B = 90^{\circ}

B D ⊥ A C

B D = 8 c m, A D = 4 c m

C D

∠

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