Math, asked by waseem3, 1 year ago

find the third side of a triangle whose perimeter is 7x^2+7x-7 and the two sides are 5x+56 and -16x^2-3x-1

Answers

Answered by Anonymous
33
Hi !

Side A = 5x+56
Side B = 
-16x² -3x-1

Side C = ?

A + B + C = perimeter

A + B + C = 7x²+7x-7

 5x + 56 + -16x² -3x-1 + C =  7x²+7x-7

2x - 16x² + 55  + C = 7x² + 7x-7

-16x
² +2x + 55 + C = 7x²+ 7x-7

C = 7x² + 16x² + 7x - 2x -7 - 55
    = 23x² + 5x - 62

Third side = C = 23x² + 5x - 62
===============================

Verification

A + B + C = perimeter 


 5x + 56 + -16x² -3x-1 + 23x² + 5x - 62
  = 7x²+7x-7

Hope this helps you !
Similar questions