Question

AD and BE are the medians of a triangle ABC. If AD = 4, $\angle DAB=\frac{\pi}{6}$, $\angle ABE=\frac{\pi}{3}$, then area of triangle ABC equals

• A. $\frac{8}{3}$
• B. $\frac{16}{3}$
• C. $\frac{32}{3}$
• D. $\frac{32\sqrt{3}}{9}$

Answer 1

Answer: (D)

$\frac{32\sqrt{3}}{9}$

Answer 2

Let G be the centroid of triangle ABC, which is the intersection point of the medians AD and BE.

1. Properties of the Centroid: The centroid divides each median in the ratio 2:1. Given AD = 4, we have AG = (2/3)AD = (2/3) * 4 = 8/3.

2. Analyze Triangle ABG: We are given the angles: $\angle DAB = \frac{\pi}{6}$. This is $\angle GAB$. $\angle ABE = \frac{\pi}{3}$. This is $\angle GBA$. The sum of these two angles in triangle ABG is: $\angle GAB + \angle GBA = \frac{\pi}{6} + \frac{\pi}{3} = \frac{(\pi + 2\pi)}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$. Since the sum of two angles in $\Delta ABG$ is $\frac{\pi}{2}$, the third angle must be: $\angle AGB = \pi - (\angle GAB + \angle GBA) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$. Thus, triangle ABG is a right-angled triangle with the right angle at G.

3. Calculate side AB: In the right-angled triangle ABG, we know AG = 8/3 and $\angle GBA = \frac{\pi}{3}$. We can use the sine function: sin($\angle GBA$) = AG / AB sin($\frac{\pi}{3}$) = (8/3) / AB $\frac{\sqrt{3}}{2}$ = (8/3) / AB AB = (8/3) * ($\frac{2}{\sqrt{3}}$) = $\frac{16}{3\sqrt{3}}$.

4. Calculate Area of Triangle ABD: Since AD is a median, it divides triangle ABC into two triangles of equal area: $\Delta ABD$ and $\Delta ACD$. Therefore, Area($\Delta ABC$) = 2 * Area($\Delta ABD$). The area of $\Delta ABD$ can be calculated using the formula: Area = (1/2) * product of two sides * sine of the included angle. Area($\Delta ABD$) = (1/2) * AB * AD * sin($\angle DAB$) Substitute the known values: AB = $\frac{16}{3\sqrt{3}}$, AD = 4, and $\angle DAB = \frac{\pi}{6}$. Area($\Delta ABD$) = (1/2) * ($\frac{16}{3\sqrt{3}}$) * 4 * sin($\frac{\pi}{6}$) Area($\Delta ABD$) = (1/2) * ($\frac{16}{3\sqrt{3}}$) * 4 * (1/2) Area($\Delta ABD$) = (1/2) * ($\frac{16}{3\sqrt{3}}$) * 2 Area($\Delta ABD$) = $\frac{16}{3\sqrt{3}}$.

5. Calculate Area of Triangle ABC: Area($\Delta ABC$) = 2 * Area($\Delta ABD$) Area($\Delta ABC$) = 2 * ($\frac{16}{3\sqrt{3}}$) Area($\Delta ABC$) = $\frac{32}{3\sqrt{3}}$.

6. Rationalize the Denominator: To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: Area($\Delta ABC$) = ($\frac{32}{3\sqrt{3}}$) * ($\frac{\sqrt{3}}{\sqrt{3}}$) Area($\Delta ABC$) = $\frac{32\sqrt{3}}{3 * 3}$ Area($\Delta ABC$) = $\frac{32\sqrt{3}}{9}$.