3 Views

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


Team Careers360 5th Jan, 2024
Answer (1)
Team Careers360 15th Jan, 2024

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.

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books