Last visit was: 19 Nov 2025, 12:32 It is currently 19 Nov 2025, 12:32
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
505-555 Level|   Geometry|      
User avatar
Lstadt
User avatar
Joined: 08 Jun 2011
Last visit: 21 Oct 2013
Posts: 62
Own Kudos:
76
 [15]
Given Kudos: 65
Posts: 62
Kudos: 76
 [15]
The lengths of the two shorter legs of a right triangle add up to 40 Updated on: 10 Mar 2021, 13:39
Originally posted by Lstadt on 28 Feb 2012, 13:48.
Last edited by Bunuel on 10 Mar 2021, 13:39, edited 3 times in total.
Renamed the topic and edited the question.
Kudos
Add Kudos
15
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

15% (low)

Question Stats:

78% (01:13) correct 22% (01:18) wrong based on 261 sessions

History

Date
Time
Result
Not Attempted Yet
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:
Show Spoiler
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:
Show Spoiler
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:
Show Spoiler
200
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,341
 [11]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,341
 [11]
The lengths of the two shorter legs of a right triangle add up to 40 28 Feb 2012, 14:02
4
Kudos
Add Kudos
7
Bookmarks
Bookmark this Post
Lstadt
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?

Answer is
Show Spoiler
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?
Show more

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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,341
 [6]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,341
 [6]
The lengths of the two shorter legs of a right triangle add up to 40 29 Feb 2012, 10:31
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
pbull78
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
Show more

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.
General Discussion
User avatar
AbeinOhio
User avatar
Joined: 22 Feb 2012
Last visit: 02 Nov 2016
Posts: 74
Own Kudos:
75
Given Kudos: 25
Schools: HBS '16
GMAT 1: 670 Q42 V40
GMAT 2: 740 Q49 V42
GPA: 3.47
WE:Corporate Finance (Aerospace and Defense)
Schools: HBS '16
GMAT 2: 740 Q49 V42
Posts: 74
Kudos: 75
The lengths of the two shorter legs of a right triangle add up to 40 28 Feb 2012, 14:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,341
 [2]
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,341
 [2]
The lengths of the two shorter legs of a right triangle add up to 40 28 Feb 2012, 15:05
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
AbeinOhio
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.
Show more

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.
avatar
pbull78
avatar
Joined: 16 Dec 2011
Last visit: 13 Oct 2012
Posts: 28
Own Kudos:
24
Given Kudos: 12
GMAT Date: 04-23-2012
Posts: 28
Kudos: 24
The lengths of the two shorter legs of a right triangle add up to 40 29 Feb 2012, 09:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
avatar
PareshGmat
avatar
Joined: 27 Dec 2012
Last visit: 10 Jul 2016
Posts: 1,534
Own Kudos:
8,102
Given Kudos: 193
Status:The Best Or Nothing
Location: India
Concentration: General Management, Technology
WE:Information Technology (Computer Software)
Posts: 1,534
Kudos: 8,102
The lengths of the two shorter legs of a right triangle add up to 40 30 Sep 2014, 01:44
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,588
Own Kudos:
1,079
Posts: 38,588
Kudos: 1,079
The lengths of the two shorter legs of a right triangle add up to 40 19 Nov 2023, 07:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts