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

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The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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Updated on: 26 Sep 2014, 00:01
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75% (00:00) correct 25% (01:14) wrong based on 18 sessions

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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?

MGMAT Solution:
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.

Question:
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:
200

Originally posted by Lstadt on 28 Feb 2012, 13:48.
Last edited by Bunuel on 26 Sep 2014, 00:01, edited 2 times in total.
Renamed the topic and edited the question.
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Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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28 Feb 2012, 14:02
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Question is from MGMT

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?

200

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?

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.
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Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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28 Feb 2012, 14:14
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.
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Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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28 Feb 2012, 15:05
1
AbeinOhio wrote:
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.

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.
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Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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29 Feb 2012, 09:03
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
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Posts: 50007
Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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29 Feb 2012, 10:31
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1
pbull78 wrote:
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

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

Hope it helps.
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The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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30 Sep 2014, 01:44
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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Re: The lengths of the two shorter legs of a right triangle add up to 40 u  [#permalink]

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07 Feb 2018, 11:23
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Re: The lengths of the two shorter legs of a right triangle add up to 40 u &nbs [#permalink] 07 Feb 2018, 11:23
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