$\triangle A B C$ has side lengths 6,8 and 10 , as shown. Each of the side lengths of $\triangle A B C$ is increased by $50 \%$, forming a new triangle, $\triangle D E F$. The area of $\triangle D E F$ is
(A) 24
(B) 48
(C) 108
(D) 12
(E) 54
![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-105.jpg?height=412&width=588&top_left_y=558&top_left_x=1164)
Solution 1

Since the side lengths of $\triangle D E F$ are $50 \%$ larger than the side lengths of $\triangle A B C$, then these new side lengths are $\frac{3}{2}(6)=9, \frac{3}{2}(8)=12$, and $\frac{3}{2}(10)=15$.

We know that $\triangle D E F$ is right-angled, and this right angle must occur between the sides of length 9 and 12 (since it is opposite the longest side).

Therefore, the area of $\triangle D E F$ is $\frac{1}{2}(9)(12)=54$.

Solution 2

The area of $\triangle A B C$ is $\frac{1}{2}(6)(8)=24$.

Since the side lengths of $\triangle A B C$ are each multiplied by $\frac{3}{2}$ to get the side lengths in $\triangle D E F$, then the area of $\triangle D E F$ is $\left(\frac{3}{2}\right)^{2}$ times that of $\triangle A B C$.

Therefore, the area of $\triangle D E F$ is $\left(\frac{3}{2}\right)^{2}(24)=\frac{9}{4}(24)=54$.

ANSWER: (E)

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