In $△ A B C, ∠ A B C = 90^{o}, B D ⊥ A C$ If $A B = 5.7$ cm, $B D = 3.8$ cm, $C D = 5.4$ cm, find $B C$

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Solution

In $△ D A B$ and $△ D B C$ , it is given that $∠ A B C = 90^{0}$ , $B D ⊥ A C$ , $A B = 5.7$ cm, $B D = 3.8$ cm and $C D = 5.4$ cm, therefore,

$B D^{2} = A D \times D C \Rightarrow {3.8}^{2} = A D \times 5.4 \Rightarrow A D = \frac{3.8 \times 3.8}{5.4} = 2.67 = 2.7$

Now, $A C = A D + D C = 2.4 + 5.4 = 8.1$ cm

Thus,

$B C^{2} = A C \times D C \Rightarrow B C^{2} = 8.1 \times 5.4 \Rightarrow B C^{2} = 47.74 \Rightarrow B C = \sqrt{47.74}$

Hence $B C = \sqrt{47.74}$ cm.

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∠ A B C = 90^{o}

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In the given figure, $∠ A B C = 90^{o}$ . and $B D ⊥ A C$ . If AB = 5.7 cm, BD = 3.8 cm and CD = 5.4 cm, find BC.

∠

⊥

In Fig. 7.140, $∠ A B = 90^{\circ}$ and $B D ⊥ A C$ . If AB = 5.7 cm, BD = 3.8 cm and CD=5.4 cm, find BC.

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