If $a + b + c = 9$ and $a^{2} + b^{2} + c^{2} = 21$ , then $a b + b c + c a$ is equal to

30

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

15

No worries! We‘ve got your back. Try BYJU‘S free classes today!

51

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $30$
Given, $a + b + c = 9$ and $a^{2} + b^{2} + c^{2} = 21$
We know ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$\Rightarrow 9^{2} = 21 + 2 (a b + b c + c a)$
$\Rightarrow a b + b c + c a = \frac{81 - 21}{2} = \frac{60}{2} = 30$

Suggest Corrections

Similar questions

a + b + c = 9

a b + b c + c a = 26

a^{2} + b^{2} + c^{2}

a^{2} + b^{2} + c^{2} = 31

a b + b c + c a = 25

a + b + c

a + b + c = 0

\frac{(a + b)^{2}}{a b} + \frac{(b + c)^{2}}{b c} + \frac{(c + a)^{2}}{c a}

Find $a^{2} + b^{2} + c^{2}; i f a + b + c = 9 a n d a b + b c + c a = 24$

Join BYJU'S Learning Program