Question

AB+BC+CA=96,ABC=144,A2+B2+C2=169 ,2 means sqare, so A,B,C=?

Answer 1

A=12,B=4,C=3

Step-by-step explanation:

It is given that,

AB+BC+CA=96,

ABC=144,

$\bold {A^2+B^2+C^2=169}$

We know that,

$(A+B+C)^2 = A^2+B^2+C^2+2(AB+BC+CA)$

$(A+B+C)^2=169+2(96)$

$(A+B+C)^2 = 169+192$

$(A+B+C)^2= 361$

Taking Square Root on both sides,

(A+B+C) = 19

$Since,ABC\ or\ A\times B\times C = 144 = 12\times 12 = 12\times 4\times 3$

On comparison,

A=12,B=4,C=3

Hence,A=12,B=4,C=3

Answer 2

A=12,B=4,C=3

Step-by-step explanation:

Given :

$AB+BC+CA=96$ .....(1)

$ABC=144$.......(2)

$A^2+B^2+C^2=169$ ......(3)

Using the algebraic identity,

$(A+B+C)^2 = A^2+B^2+C^2+2(AB+BC+CA)$

Substitute the values,

$(A+B+C)^2=169+2(96)$

$(A+B+C)^2 = 169+192$

$(A+B+C)^2= 361$

Square Root both sides,

$(A+B+C) = \sqrt{361}$

$A+B+C=19$

We have given that, $A\times B\times C = 144$

$A\times B\times C = 12\times 4\times 3$

As sum of $12+4+3=19$

On comparison,

A=12,B=4,C=3

#Learn more

If a + b + c = 14 and a2 + b2 +c2 = 96 then (ab + bc + ca) =?

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