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ΔABC is an isosceles triangle and $$\overline {AB} $$  = $$\overline {AC} $$  = 2a unit, $$\overline {BC} $$  = a unit. Draw $$\overline {AD} $$ ⊥ $$\overline {BC} $$ , and find the length of $$\overline {AD} $$

ΔABC is an isosceles triangle and $$\overline {AB} $$  = $$\overline {AC} $$  = 2a unit, $$\overline {BC} $$  = a unit. Draw $$\overline {AD} $$ ⊥ $$\overline {BC} $$ , and find the length of $$\overline {AD} $$ Correct Answer $$\frac{{\sqrt {15} }}{2}$$ a unit

According to question,
Given :
Triangles mcq solution image
AB = AC = 2a
BC = a
AD ⊥ BC
In isosceles triangle perpendicular sides bisects the opposite side of the length
∴ BD = $$\frac{{BC}}{2}$$
   BD = $$\frac{a}{2}$$
In ΔADB using Pythagoras theorem
AB2 = BD2 + AD2
(2a)2 = $${\left( {\frac{a}{2}} \right)^2}$$ + AD2
4a2 = $$\frac{{{a^2}}}{4}$$ + AD2
AD2 = 4a2 - $$\frac{{{a^2}}}{4}$$
AD = $$\frac{{\sqrt {15} }}{2}$$ a units

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