Question : If $2s = a + b + c$, then the value of $s(s - c) + (s - a)(s - b)$ is:
Option 1: $ab$
Option 2: $abc$
Option 3: $0$
Option 4: $\frac{a+b+c}{2}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $ab$
Solution :
Given: $2s = a + b + c$
⇒ $s = \frac{a\:+\:b\:+\:c}{2}$
Now, $s(s - c) = (\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:+\:c}{2}-c)$
⇒ $s(s - c) =(\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:-\:c}{2})$
Also, $(s - a)(s - b) = (\frac{a\:+\:b\:+\:c}{2}-a)(\frac{a\:+\:b\:+\:c}{2}-b)$
⇒ $(s - a)(s - b) = (\frac{b\:+\:c\:-\:a}{2})(\frac{a\:+\:c\:-\:b}{2})$
So, $s(s - c) + (s - a)(s - b)$
$= (\frac{a\:+\:b\:+\:c}{2})(\frac{a\:+\:b\:–\:c}{2})+(\frac{b\:+\:c\:–\:a}{2})(\frac{a\:+\:c\:–\:b}{2})$
$= \frac{1}{4}[(a+b)^2-c^2+c^2-(a-b)^2]$
$= \frac{1}{4}[(a+b)^2-(a-b)^2]$
$= \frac{1}{4}(4ab)$
$=ab$
Hence, the correct answer is $ab$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
