5 Views

Question : In the given figure, the area of isosceles triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$ and $\mathrm{BE = AB}$ and $\mathrm{AB = 2 AD}$, $\mathrm{AE \parallel DC}$, then what is the area (in$\;\mathrm{cm^2}$) of the trapezium $\mathrm{ABCD}$?

Option 1: 108

Option 2: 124

Option 3: 136

Option 4: 144


Team Careers360 23rd Jan, 2024
Answer (1)
Team Careers360 25th Jan, 2024

Correct Answer: 144


Solution : Given that the area of triangle $\mathrm{ABE}$ is $72\;\mathrm{cm^2}$.
The triangle is an isosceles.
$\mathrm{AB} = \mathrm{BE} = \sqrt{2 \times 72} = 12\;\mathrm{cm}$
$\mathrm{AB = 2 AD}$
⇒ $\mathrm{AD} = \frac{\mathrm{AB}}{2} = 6\;\mathrm{cm}$
Since $\mathrm{AE} \parallel \mathrm{DC}$
$\mathrm{CE} = \mathrm{AD} = 6\;\mathrm{cm}$, and $\mathrm{BC} = \mathrm{BE} + \mathrm{CE} = 12 + 6 = 18\;\mathrm{cm}$
The area of trapezium $\mathrm{ABCD}$ $=\frac{1}{2} \times (\mathrm{sum\; of\; parallel\; sides}) \times \mathrm{height} = \frac{1}{2} \times (18 + 6) \times 12 = 144\;\mathrm{cm^2}$
Hence, the correct answer is 144.

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