Manhattan Geometry Advanced Problem Set

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Manhattan Geometry Advanced Problem Set

by theachiever » Thu Nov 29, 2012 1:32 am
The length of two shorter legs of a right triangle add upto 40 sq units.What is the maximum possible area of the triangle?

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by kartikshah » Thu Nov 29, 2012 2:27 am
Property: A right-angled triangle with maximum area has to be an isosceles right-angled triangle.
Thus,
Let x and y be the two equal arms of such a triangle:
x = y = 40/2 = 20

Now,
Area = (x*y)/2
= (20*20)/2
= 200 sq units

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by savvysatyam » Thu Nov 29, 2012 2:41 am
Area will be maximum when the two legs are of same length. this is because of the property that the product of any two numbers is maximum when both are equal, only if the addition of those two numbers is a constant. hence 40= 20+20. thus 20,20 are two sides and area is 200