Last visit was: 29 Apr 2026, 05:45

It is currently 29 Apr 2026, 05:45

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

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

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

An isosceles right triangle with a leg of length "a" and per

1 2

PareshGmat

Joined: 27 Dec 2012

Last visit: 10 Jul 2016

Posts: 1,531

Own Kudos:

8,280

Given Kudos: 193

Status:The Best Or Nothing

Location: India

Concentration: General Management, Technology

WE:Information Technology (Computer Software)

Posts: 1,531

Kudos: 8,280

07 Jan 2015, 18:54

Kudos

Bookmarks

PareshGmat

DesiGmat

Just to add on Karishma.

As per my knowledge, an altitude doesn't divide a triangle into 2 equal triangles having the same area.
It's the median.
In this case as the triangle is right angled triangle, and Karishma have mentioned how to calculate the longest side i.e. the hypotenuse using the Pythogorus Theorem.

there's one property of the median drawn on the hypotenuse in a right angled triangle.
Median = 1/2 * Hypotenuse = a/sqrt2

and we can accordingly proceed with the question.
Correct me if I am wrong somewhere.

For an Isosceles right triangle, altitude (height) = median

Refer diagram below for further clarity. Actually Karishma has also drawn the same in her explanation (orientation is changed)

Attachment:

iso.png [ 2.13 KiB | Viewed 3158 times ]

PareshGmat

Joined: 27 Dec 2012

Last visit: 10 Jul 2016

Posts: 1,531

Own Kudos:

8,280

Given Kudos: 193

Status:The Best Or Nothing

Location: India

Concentration: General Management, Technology

WE:Information Technology (Computer Software)

Posts: 1,531

Kudos: 8,280

07 Jan 2015, 18:57

Kudos

Bookmarks

DesiGmat

Altitute can't be the median in isoscles right angle traingle.

In the figure attached, in isosceles right angled triangle ABC, AB is an altitude but not a median.
Similarly BD is perpendicular to AC, but it doesn't mean that AD = DC

and in isoscles right angled triangle PQR, PS can be the median(not altitude) and it divides the triangle PQR into two equal traingles of same area

Correct me if i am wrong

Triangle ABC does not seem to be on scale. It does not appear a 45-45-90 triangle, may be that's why the altitude BD does not seem to bisect

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 28 Apr 2026

Posts: 16,447

Own Kudos:

79,444

Given Kudos: 485

Location: Pune, India

Products:

Expert

Expert reply

Posts: 16,447

Kudos: 79,444

07 Jan 2015, 21:14

Kudos

Bookmarks

DesiGmat

Kindly refer to this post for information on median/altitude/angle bisectors in special triangles: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/05 ... -the-gmat/

In an isosceles triangle, median to the base (the base is the non-equal side) is the altitude too and vice versa.

KarishmaB

Tutor

Joined: 16 Oct 2010

Last visit: 28 Apr 2026

Posts: 16,447

Own Kudos:

79,444

Given Kudos: 485

Location: Pune, India

Products:

Expert

Expert reply

Posts: 16,447

Kudos: 79,444

08 Jan 2015, 21:07

Kudos

Bookmarks

DesiGmat

In another post, I have given you a link. Do check it out for a discussion of properties of altitudes and medians in an isosceles triangle.

msad

Joined: 18 Dec 2014

Last visit: 25 Mar 2016

Posts: 9

Own Kudos:

Given Kudos: 17

Posts: 9

Kudos: 5

13 Jan 2015, 18:25

Kudos

Bookmarks

PerfectScores

emmak

An isosceles right triangle with a leg of length "a" and perimeter "p" is further divided into two similar triangles of equal area. Which of the following represents the perimeter of one of these smaller triangles?

A. p/2
B. p−a
C. (p+a)/2
D. 2a
E. \(p-a\sqrt{2}\)

Let us plug in here:

I followed the exact same process of thinking, which lead to solve this question within 45 seconds or 1 minute tops.

Sash143

Joined: 04 Jun 2015

Last visit: 26 Jul 2022

Posts: 62

Own Kudos:

751

Given Kudos: 2

Posts: 62

Kudos: 751

Updated on: 01 Mar 2016, 04:41

Originally posted by Sash143 on 22 Jan 2016, 12:45.
Last edited by Sash143 on 01 Mar 2016, 04:41, edited 1 time in total.

Kudos

Bookmarks

emmak

Here's my take
Given \(a\) is leg of isosceles right triangle. Hence the other leg has to be \(a\).
Assume \(a=2\); hypotenuse= \(2\sqrt{2}\); Perimeter p= \(4+2\sqrt{2}\)
Now the area \((\frac{1}{2}*2*2= 2)\)is divided into two equal areas with legs \(\sqrt{2}\) and an area of \(\frac{1}{2}*\sqrt{2}*\sqrt{2}= 1\). So the new perimeter= \(2+2\sqrt{2}\)
Substitute the highlighted "p" and "a" in the ans choice to get \(2+2\sqrt{2}\). Only B satisfies

chetan2u

GMAT Expert

Joined: 02 Aug 2009

Last visit: 29 Apr 2026

Posts: 11,232

Own Kudos:

45,044

Given Kudos: 335

Status:Math and DI Expert

Location: India

Concentration: Human Resources, General Management

GMAT Focus 1: 735 Q90 V89 DI81 Verified score

Products:

Expert

Expert reply

GMAT Focus 1: 735 Q90 V89 DI81 Verified score

Posts: 11,232

Kudos: 45,044

01 Mar 2016, 01:15

Kudos

Bookmarks

mikemcgarry

emmak

An isosceles right triangle with a leg of length "a" and perimeter "p" is further divided into two similar triangles of equal area. Which of the following represents the perimeter of one of these smaller triangles?
A. p/2
B. p−a
C. (p+a)/2
D. 2a
E. \(p-a\sqrt{2}\)

This is a HARD question. A good challenging question :-)

--- probably at the upper limit of difficulty of what the GMAT would ask.

Here's a blog about the properties of 45-45-90 triangles:
https://magoosh.com/gmat/2012/the-gmats- ... triangles/

Attachment:

isosceles right triangle, split in two.JPG

We know the legs have length a, so the hypotenuse BC = \(a*sqrt(2)\). Therefore the perimeter, p, equals
p = \(2a + a*sqrt(2)\)

OK, hold onto that piece and put it aside a moment.

Now, we draw AD, dividing the ABC into two smaller congruent triangles. AC = a is now the hypotenuse, so each leg is

AD = CD = \(\frac{a}{sqrt(2)}\) = \(\frac{a*sqrt(2)}{2}\)

Incidentally, on that final step, "rationalizing the denominator", here's a blog article:
https://magoosh.com/gmat/2013/gmat-math- ... uare-root/

Therefore, AD + CD = \(a*sqrt(2)\), and

new perimeter = AC + AD + CD = \(a + a*sqrt(2)\)

The only difference between this "new perimeter" and p is the extra "a", so

p = \(2a + a*sqrt(2)\) = \(a + a + a*sqrt(2)\) = a + "new perimeter"

So,

new perimeter = p - a

Answer = (B)

Let me know if anyone reading this has any questions.

Mike :-)

hi,
another point to what mike has written above..
RULE - an isosceles triangle when divided in two equal areas by a line from the third vertex, the line will be a perpendicular altitude bisecting the third angle and the third side too..

We can find the altitude through AREA, apart from the use of congruency of triangle as shown above..
The area of this isosceles triangle will remain the same irrespective of what altitude and corresponding BASE we take..
in this area when base is one equal side= a^2/2...
the area when the hypotenuse is the base= \(\frac{1}{2} * a\sqrt{2}* altitude\)..
so \(\frac{1}{2} * a\sqrt{2}* altitude\) = a^2/2..
we can find altitude from this too and when we know all sides, we can easily find the perimeter..

1 2

Moderators:

Bunuel

Math Expert

109968 posts

Krunaal

Tuck School Moderator

852 posts