The lengths of the two shorter legs of a right triangle add up to 40

Question

The lengths of the two shorter legs of a right triangle add up to 40 units. What is the maximum possible area of the triangle?

A. 40 square units

B. 80 square units

C. 120 square units

D. 200 square units

E. 400 square units

MGMAT Solution:

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200 Square Units. You can think of a triangle as half of a rectangle. Constructing this right triangle with legs adding up to 40 is equivalent to constructing the rectangle with a perimeter of 80. Since the area of the triangle is half that of the rectangle, you can use the previously mentioned technique for maximizing the area of the rectangle: of all rectangles with a given perimeter, the square has the greatest area. The desired rectangle is thus 20 by 20 square and the right triangle has the area (1/2)(20)(20)=200 units.

Answer is 200

Question:

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MGMT set it up as a square then went from there by calculating it as half the area of a maximized square.

My approach was different and it yielded a wrong result, of course

, but I am sure that I had the right idea in mind and something went wrong.

Since we want to maximize a triangle, the maximum area would be that of an Equalitral triangle which has 3 equal sides. Since s1+s2 = 40 (assuming we are dealing with an Equalitral) then S1 must be 20; therefore, s2 and s3 has to be each 20.

The area of an Equalitral triangle is s^2\sqrt{3}/4 which should give us
400 * \sqrt{3}/4 which is 100\sqrt{3}.

What did I do wrong there?

OA:

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200

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Official Answer

D

, but I am sure that I had the right idea in mind and something went wrong. Since we want to maximize a triangle, the maximum area would be that of an Equalitral triangle which has 3 equal sides. Since s1+s2 = 40 (assuming we are dealing with an Equalitral) then S1 must be 20; therefore, s2 and s3 has to be each 20. The area of an Equalitral triangle is s^2\sqrt{3}/4 which should give us 400 * \sqrt{3}/4 which is 100\sqrt{3}. What did I do wrong there? OA: 200

Bunuel · Accepted Answer

The lengths of the two shorter legs of a right triangle add up to 40 units. What is the maximum possible area of the triangle?

First of all:  a right triangle can never be equilateral, since the hypotenuse (the side opposite the right angle) is always longer than any the other two sides .

Next,  a right triangle with the largest area will be an isosceles right triangle , so the the lengths of the two shorter legs must be equal, so each must be 20. Area=1/2*20*20=200.

Hope it's clear.

Bunuel · Answer

Check geometry chapters of Math Book: TRIANGLES: https://gmatclub.com/forum/math-triangles-87197.html POLYGONS: https://gmatclub.com/forum/math-polygons-87336.html COORDINATE GEOMETRY: https://gmatclub.com/forum/math-coordina ... 87652.html CIRCLES: https://gmatclub.com/forum/math-circles-87957.html 3-D Geometries: https://gmatclub.com/forum/math-3-d-geom ... ml#p792331 Hope it helps.

AbeinOhio · Answer

I did this in a slightly different and clearly more difficult way than presented above but maybe it will help someone else..

I set the 2 shorter sides to be x & y

so:
x + y = 40

&

then (X+Y)^2 = 1600

The area of this triangle= xy/2

I expanded the (X+Y)^2 expression and moved XY to one side to get:
xy= (1600 -X^2 - y^2)/2

so XY/2 = (1600-X^2- y^2)4

Now to maximize this expression i tried extremes (39,1) (38,2) and realized that it was maxmized when X= Y

and was able to solve so that (1600-20^2-20^2)/4 = 200

Hope that helps you guys.

Bunuel · Answer

Consider this, we are given that a+b=40 and we want to maximize area=1/2*ab, so basically we want to maximize the value of ab. Now, for a given sum of two values the product is maximized when they are equal. Hence ab is maximized when a=b=20.

Basically the property saying that  a right triangle with the largest area will be an isosceles right triangle  is derived from that rule.

Hope it helps.

pbull78 · Answer

Nice expanatio bunuel , bunuel can u tell me link from where I can get all these formulaes of geometry section
Thanks

Posted from my mobile device

PareshGmat · Answer

Given that sum of two shorter sides of right triangle = 40

It means, they can be in any combination from 1 - 39, 2 - 38, 3-37.............. upto 39 - 1

The mean of this would be 20 - 20 (Means its a Isosceles right triangle)

When sides are 20 - 20, area would be maximum  = \frac{1}{2} * 20 * 20 = 200

Please note: If combinations like 21 - 19, 22 - 18, 23 - 17 etc. are tried,     there product would always be less than 400

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The lengths of the two shorter legs of a right triangle add up to 40

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The lengths of the two shorter legs of a right triangle add up to 40

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