Given that in triangle ABC, $\displaystyle \angle A =60^\circ, \angle B =45^\circ, AC =1cm. $ D and E are respectively the midpoints of AB and AC. Find the area of triangle ADE.

(SOLVED using sine rule and $\displaystyle \frac{1}{2}ab sinC$)

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- Aug 4th 2012, 03:09 AMshiny718Find the area of triangle
Given that in triangle ABC, $\displaystyle \angle A =60^\circ, \angle B =45^\circ, AC =1cm. $ D and E are respectively the midpoints of AB and AC. Find the area of triangle ADE.

(SOLVED using sine rule and $\displaystyle \frac{1}{2}ab sinC$) - Aug 4th 2012, 04:10 AMProve ItRe: Find the area of triangle
http://i22.photobucket.com/albums/b3...rtriangles.jpg

In triangle ABC, you have two angles, so finding the third angle won't be too hard. Then use the sine rule to evaluate another side length. Finally, you can evaluate the area using $\displaystyle \displaystyle \begin{align*} \frac{1}{2}ab\sin{C} \end{align*}$.

For triangle ADE, notice that you have reduced the side lengths by $\displaystyle \displaystyle \begin{align*} \frac{1}{2} \end{align*}$. This means that the area of the triangle has been reduced by $\displaystyle \displaystyle \begin{align*} \left(\frac{1}{2}\right)^2 = \frac{1}{4} \end{align*}$.