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
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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
-16x² +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 !
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
-16x² +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 !
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