The perimeter of triangle is 14xsquare +20x +15 and two of its sides are 3xsquare +5x +1 and X square +10x-6. Find its third side?
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Let the sides be a, b and c
a+b+c = 14x^2 +20x +15
a = 3x^2 +5x +1
b = x^2 +10x - 6
Third side = (a+b+c) -(a+b)
= 14x^2 +20x +15 - ( 3x^2 +5x +1 +x^2+10x - 6)
= 14x^2 +20x +15 - (4x^2 +15x - 5)
= 14^2 +20x +15 - 4x^2 - 15x +5
= 10x^2 +5x +20
Third side = 10x^2 +5x +20
a+b+c = 14x^2 +20x +15
a = 3x^2 +5x +1
b = x^2 +10x - 6
Third side = (a+b+c) -(a+b)
= 14x^2 +20x +15 - ( 3x^2 +5x +1 +x^2+10x - 6)
= 14x^2 +20x +15 - (4x^2 +15x - 5)
= 14^2 +20x +15 - 4x^2 - 15x +5
= 10x^2 +5x +20
Third side = 10x^2 +5x +20
Kalidasan:
thanks
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