The largest possible area for the right triangle is 1250 in² in which the sum on the lengths of the two shorter sides is 100 in.
we know that in a right triangle the two shorter sides are the base and the perpendicular. Here we have to find the largest possible area of the triangle in which the sum of the shorter sides is 100 in.
Let the base and perpendicular be x and y respectively.
Therefore,
x+y = 100
y = 100 -x
Also, Area A
= 1/2 xy
= 1 x(100-x)/z
= 100x-x²/2
so the coordinate of vertex = -b/2a
= - 50/2(1-1/2)
= 50
Also,
y = 100-50 [Putting the value of x in equation 1]
= 50
Therefore Maximum, area:
= 1/2 (50)(50)
= 1250 in²
Hence we get the required maximum area.
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