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.

Open in App

Solution

According to the Pythagorean theorem,

$(h y p o t e n u s e)^{2} = (b a s e)^{2} + (a l t i t u d e)^{2}$

$B C^{2} = B D^{2} + D C^{2}$

$B C^{2} = {3.8}^{2} + {5.4}^{2}$

$B C^{2} = 14.44 + 29.16$

$B C^{2} = 43.6$

$B C = \sqrt{43.6}$

$B C = 6.6$

Suggest Corrections

Similar questions

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.

∠ A B C = 90^{o}

B D ⊥ A C

A B = 5.7 c m, B D = 3.8 c m

C D = 5.4 c m

B C

In the adjoining figure, $∠ A B C = 90^{\circ}$ and $B D ⊥ A C$ . If AB = 5.7 cm, BD = 3.8 cm, CD = 5.4 cm, then find the length of BC.

90^{0}

⊥

Join BYJU'S Learning Program