Find the length of altitude AD of an isosceles $Δ A B C$ in which AB = AC = 2a units and BC = a units.

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Solution

In an isosceles triangle

AB = AC = 2a units
BC= a units

Given BC = a. Then, $B D = \frac{a}{2}$

In ΔABD,

$∠ A D B = 90^{o}$

$(A B)^{2} = (A D)^{2} + (B D)^{2}$ [by Pythagoras thorem]

$(A D)^{2} = (A B^{2} - B D^{2})$

$= [(2 a)^{2} - (\frac{a}{2})^{2}] s q . u n i t = (4 a^{2} - \frac{a^{2}}{4}) s q . u n i t$

$A D^{2} = \frac{16 a^{2} - a^{2}}{4}$

$A D = \sqrt{\frac{15 a^{2}}{4}} u n i t .$

$A D = \frac{a \sqrt{15}}{2} u n i t$

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Similar questions

$AD is an altitude of an isosceles Δ A B C in which AB = AC. Show that (i) AD bisects BC, (ii) AD bisects ∠ A .$

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

Δ A B C

A B = A C

A D

A

B C

A B

A D

A B C

A B = A C

A D

B C

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