Question 60

$$\triangle$$ ABC is an equilateral triangle and AD $$\perp$$ BC, where D lies on BC. If AD = $$4\sqrt{3}$$ cm, then what is the perimeter (in cm) of $$\triangle$$ABC?

Solution

AD = $$4\sqrt{3}$$ cm

In right angle $$\triangle$$ ABD,

($$\because \angle D = 90\degree$$)

ByPythagoras,

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

(AB = AC = BC)

(BD = BC/2 = AB/2)

$$(AB)^2 - (BD)^2 = (AD)^2 $$

$$(AB)^2 - (\frac{AB}{2})^2 = (4\sqrt{3})^2 $$

$$\frac{3}{4}(AB)^2= 48 $$

$$(AB)^2 = 64$$

AB = 8 cm

Perimeter of $$\triangle ABC = 3 \times side = 3 \times$$ 8 = 24 cm

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