2 Views

Question : If the perimeter of a rhombus is 80 cm and one of its diagonals is 24 cm, then what is the area (in cm2) of the rhombus?

Option 1: 218

Option 2: 192

Option 3: 384

Option 4: 768


Team Careers360 5th Jan, 2024
Answer (1)
Team Careers360 21st Jan, 2024

Correct Answer: 384


Solution :
Given: Perimeter of rhombus = 80 cm
Let AB, BD, DC, and CA be the sides, BC($d_1$) and AD($d_2$) be the diagonals of a rhombus meeting at O.
Side AB = $\frac{80}{4}$ = 20 cm
Diagonal ($d_1$) BC = 24 cm
OB = $\frac{1}{2}×24 = 12$ cm
We know that the diagonals of a rhombus bisect each other at 90$^\circ$.
Using the Pythagoras theorem,
$AB^2 = OA^2 + OB^2$
⇒ $20^2 = x^2 + 12^2$
⇒ $x^2 = 400–144$
⇒ $x^2 = 256$
⇒ $x = 16$
So, AD ($d_2$) = 16 × 2 = 32 cm
Now, area of the rhombus = $\frac{1}{2}×d_1×d_2$
= $\frac{1}{2} ×24×32$ = $384$ cm$^2$
Hence, the correct answer is 384.

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