In an isosceles triangle ABC, AB $=$ AC $=$ 25 cm, BC $=$ 14 cm. Calculate the altitude from A on BC.

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Given:

In the given isosceles triangle ABC, AB $=$ AC $=$ 25 cm, BC $=$ 14 cm.

 To do:

Here, we have to calculate the altitude from A on BC.

Solution:

In $∆ABD$ and $∆ACD$,

$\angle ADB = \angle ADC=90^o$

$AB = AC$   (given)

$AD = AD$  (Common)

Therefore,

$∆ABD ≅ ∆ACD$      (By RHS congruency)

This implies,

$BD = CD = 7\ cm$              (Corresponding parts of congruent triangles)

In $∆ADB$,

By Pythagoras theorem,

$AD^2 + BD^2 = AB^2$

$AD^2 + 7^2 = 25^2$

$AD^2 = 625 – 49$

$AD^2 = 576$

$AD = \sqrt576 = 24\ cm$

The altitude from $A$ on $BC$ is $24\ cm$.

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Updated on: 10-Oct-2022

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