11 Views

Question : If $a^2+b^2+c^2=62$ and $a + b + c = 12$, then what is the value of $4ab + 4bc + 4ca$?

Option 1: 164

Option 2: 148

Option 3: 152

Option 4: 160


Team Careers360 10th Jan, 2024
Answer (1)
Team Careers360 21st Jan, 2024

Correct Answer: 164


Solution : We can solve this problem using the identity,
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$
Given that $a^2 + b^2 + c^2 = 62$ and $a + b + c = 12$,
$⇒12^2 = 62 + 2(ab + bc + ca)$
$⇒144 = 62 + 2(ab + bc + ca)$
$⇒82 = 2(ab + bc + ca)$
$⇒41 = ab + bc + ca$
$⇒4(ab + bc + ca) = 4 \times 41$
$\therefore 4ab + 4bc + 4ca=164$
Hence, the correct answer is 164.

How to crack SSC CHSL

Candidates can download this e-book to give a boost to thier preparation.

Download Now

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