Question : In $\triangle{ABC}, \angle A=90^{\circ}, {AB}=16\ cm $ and $AC =12 \ cm$. $D$ is the midpoint of $AC$ and $DE \perp CB$ at $E$. What is the area (in ${cm}^2$ ) of $\triangle{CDE}$?
Option 1: 8.64
Option 2: 7.68
Option 3: 5.76
Option 4: 6.25
Correct Answer: 8.64
Solution :
Given: In $\triangle {ABC}, \angle A=90^{\circ}, {AB}=16\ cm $ and $AC =12 \ cm$.
$D$ is the midpoint of $AC$ and $DE \perp CB$ at $E$.
The area of the triangle = $\frac{1}{2}\times \text{Base} \times \text{Height}$
Use the Pythagoras's theorem in $\triangle ABC$, $(BC)^2=(AB)^2+(AC)^2$
⇒ $(BC)^2=(16)^2+(12)^2$
⇒ $(BC)^2=256+144=400$
⇒ $BC=20\ cm$
The area of the $\triangle ABC= \frac{1}{2}\times 12 \times 16=96 \ cm^2$
Also, $\triangle ABC \sim \triangle EDC$.
⇒ $\frac{\text{The area of $\triangle EDC$}}{\text{The area of $\triangle ABC$}}=(\frac{\text{DC}}{\text{BC}})^2$
⇒ $\frac{\text{The area of $\triangle EDC$}}{96}=(\frac{6}{20})^2$
⇒ $\frac{\text{The area of $\triangle EDC$}}{96}=\frac{36}{400}$
⇒ $\text{The area of $\triangle EDC$}=\frac{36\times 96}{400}=8.64 \ cm^2$
Hence, the correct answer is 8.64.
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