ABC is an isosceles triangle with AB = AC. Altitudes are drawn to the sides AB and AC from vertices B and C. One altitude CF is found to be 4 cm and BC = 5 cm. Find EC, where BE is altitude to side AC.

4 cm

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5 cm

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3 cm

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none of these

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Solution

The correct option is D 3 cm
ABC is an isosceles triangle with AB = AC. Altitudes are drawn to the sides AB and AC from vertices B and C. One altitude CF is found to be 4 cm and BC = 5 cm.
Then in triangle BFC the angle BFC =90
Then $B F^{2} = B C^{2} - C F^{2} = (5)^{2} - (4)^{2} = 9 \Rightarrow B F = 3$
In triangle BEF & triange BFC are similler
Then EC=BF=3

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Similar questions

In an isosceles triangle ABC, if AB = AC=13 cm and the altitude from A on BC is 5 cm find BC.

$Δ A B C$ is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.

A B C

B C = 5 c m, A C = 12 c m

A B = 13 c m

A B C, A B = A C = 25 c m, B C = 14 c m

A

B C

A B C, A B = A C = 25 c m

B C = 14

A

c m

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