Point Division in Equilateral Triangle

Question

The length of each side of an equilateral triangle $ABC$ is $3	ext{ cm}$. Let $D$ be a point on $BC$ such that the area of triangle $ADC$ is half the area of triangle $ABD$. Then the length of $AD$, in cm, is

Accepted Answer

D. The ratio of areas of $	riangle ADC$ to $	riangle ABD$ is $1 : 2$. Since they share the same altitude from $A$, $D$ divides $BC$ in the ratio $1 : 2$, so $DC : BD = 1 : 2$. Since $BC = 3	ext{ cm}$, $DC = 1	ext{ cm}$ and $BD = 2	ext{ cm}$. In $	riangle ABD$, using the Cosine Rule at $\angle B = 60^\circ$: $AD^2 = AB^2 + BD^2 - 2(AB)(BD)\cos 60^\circ$ $AD^2 = 3^2 + 2^2 - 2(3)(2)\left(\frac{1}{2}ight) = 9 + 4 - 6 = 7$. So $AD = \sqrt{7}	ext{ cm}$.

Answer

$\sqrt{6}$

Answer

$\sqrt{5}$

Answer

$\sqrt{8}$

Answer

$\sqrt{7}$

Point Division in Equilateral Triangle

Explanation

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