The area of $\triangle A B C$ is 60 square units. If $B D=8$ units and $D C=12$ units, the area (in square units) of $\triangle A B D$ is
(A) 24
(D) 36
(B) 40
(E) 6
(C) 48

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-143.jpg?height=276&width=479&top_left_y=1342&top_left_x=1273)
The area of $\triangle A B C$ is 60 square units. If $B D=8$ units and $D C=12$ units, the area (in square units) of $\triangle A B D$ is
(A) 24
(B) 40
(C) 48
(D) 36
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-290.jpg?height=325&width=556&top_left_y=214&top_left_x=1424)

## Solution

From $A$, draw a line perpendicular to $B C$ to meet $B C$ at $E$. Thus the line segment $A E$ which is labelled as $h$ is the height of $\triangle A B D$ and $\triangle A B C$. Since the heights of the two triangles are equal, their areas are then proportionate to their bases. If the area of $\triangle A B C$ is 60 , then the area of $\triangle A B D$ is $\frac{8}{20} \times 60=24$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-290.jpg?height=309&width=547&top_left_y=648&top_left_x=1426)

ANSWER: (A)