An isosceles triangle has sides 4a, b and b. Find the length of the perpendicular dropped onto the unequal side from the vertex opposite to this side.

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Solution

The correct option is B

Since the perpendicular bisects the unequal side, in the $△$ ABC, BD = DC = 2a and BC $⊥$ AD, $△$ ABD and $△$ ADC are right angled triangles. They are also congruent.
So, by Pythagoras theorem, $A B^{2} = B D^{2} + A D^{2}$
$\Rightarrow A D = \sqrt{A B^{2} - B D^{2}}$
$\Rightarrow A D = \sqrt{b^{2} - (2 a)^{2}}$
$\Rightarrow A D = \sqrt{b^{2} - 4 a^{2}}$

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An isosceles triangle has sides 4a, b and b. Find the length of the perpendicular dropped onto the unequal side from the vertex opposite to this side.

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An isosceles triangle has sides 4a, b and b. Find the length of the perpendicular dropped onto the unequal side from the vertex opposite to this side.

The correct option is B Since the perpendicular bisects the unequal side, in the △ABC, BD = DC = 2a and BC⊥AD, △ABD and △ADC are right angled triangles. They are also congruent. So, by Pythagoras theorem, AB2=BD2+AD2 ⇒AD=√AB2−BD2 ⇒AD=√b2−(2a)2 ⇒AD=√b2−4a2