Asked by Raj
Q. The sum of the two shortest sides of a right-angles triangle is 18cm.Calculate
a.the least possible length of the hypotenuse
b.the greatest possible area of the triangle
a.the least possible length of the hypotenuse
b.the greatest possible area of the triangle
Answers
Answered by
Reiny
shortest side ---- x
longer side = 18-x
h^2 = x^2 + (18-x)^2
= 2x^2 - 36x + 324
d(h^2)/dx = 4x - 36
= 0 for a max/min ( for h to be a minimum, h^2 would have to be a minimum) x = 9
when x = 9 , h^2 = 2(81) - 36(9) + 324 = 162
h = √162 or appr 12.728
area = (1/2)(x)(18-x)
= 9x - x^2/2
d(area)/dx = 9 - x/4 = 0 for maximum area
etc.
longer side = 18-x
h^2 = x^2 + (18-x)^2
= 2x^2 - 36x + 324
d(h^2)/dx = 4x - 36
= 0 for a max/min ( for h to be a minimum, h^2 would have to be a minimum) x = 9
when x = 9 , h^2 = 2(81) - 36(9) + 324 = 162
h = √162 or appr 12.728
area = (1/2)(x)(18-x)
= 9x - x^2/2
d(area)/dx = 9 - x/4 = 0 for maximum area
etc.
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