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

Question

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. \(p-a\sqrt{2}\)

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

mikemcgarry · Accepted Answer

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/ 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 :-)

PerfectScores · Answer

Let us plug in here:

pprange · Answer

Mike, you are amazing at explaining things! I did a Veritas Prep exam and had trouble understanding their explanation so I searched it up on the forum. Turns out you're explaining things here just as well as Magoosh. Thank you, I am so happy I purchased Magoosh!

sanket1991 · Answer

mikemcgarry If we draw a circumcircle around a 45-45-90 (this) triangle, we can solve this question within 1 minute :-D

By drawing circle we will get "a/ 2^1/2" as radii and we already know side as 'a' we get 45-45-90 triangles with side "a/ 2^1/2" and hypotenuse as 'a' perimeter _/2 * a + a = p - a Hope I am correct in this approach... :oops:

mikemcgarry · Answer

Dear sanket1991 , I'm happy to respond. :-)

Your approach is correct and insightful. I would say: if you are going to write math on GC, it would be good to learn the conventions for writing proper math symbols on this site. See: writing-mathematical-formulas-in-the-gmat-club-forums-this-72468.html Best of luck, my friend. Mike :-)

oss198 · Answer

Hi, 
I don't really get it. Could you tell me what do you mean by "AC = a is not the hypotenuse, so each leg is

AD = CD" ?

mikemcgarry · Answer

Dear oss198 , I'm sorry, my friend! That was a typo, and no one else pointed it out until now. When I said: AC = a is not the hypotenuse, so each leg is ... I meant AC = a is now the hypotenuse, so each leg is ... I just changed that in the post above. Thank you for pointing this out. Does all this make sense now? Mike :-)

oss198 · Answer

Thank you for your answer Mike, I'm sorry but can you now precise how do you conclude that AD = CD?

mikemcgarry · Answer

I'm happy to respond. :-)

My friend, you have to know the Isosceles Triangle Theorem, one of the cornerstones of Geometry: https://magoosh.com/gmat/2012/isosceles- ... -the-gmat/ We know that angle ACD = 45 degrees, and we know that segment AD bisects the right angle, so angle DAC is 45 degrees. Because those two angles are equal, it absolutely must be true that AD = CD. Does this make sense? Mike :-)

oss198 · Answer

ooh right i didn't see that :oops:

thank you so much!

usre123 · Answer

Hi,
Is AD=DC always (triangle on the right) in such a scenario. I thought it should be equal, but spent maybe a minute proving it to myself.

reddyMBA · Answer

Just wanted to share, Kind of an easy but tricky question...

Q.) 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}

Please give me Kudo's if you like it.

jayanth7290 · Answer

Pretty good question actually ,enjoyed solving it

Refer to the attachment ..Let me know if u have any query regarding my explanation

PareshGmat · Answer

Answer = B.) p−a rt.png Figure I Hypotenuse = \sqrt{a^2 + a^2} = a\sqrt{2} Given that a + a + a\sqrt{2} = p 2a + a\sqrt{2} = p ............... (1) Figure II Divide Fig I in two parts Two new equal size isosceles right triangles are formed, Height & base of small isosceles right triangle = \frac{1}{2} a\sqrt{2} = \frac{a}{sqrt{2}} Perimeter of small isosceles right triangle = a + \frac{a}{sqrt{2}} + \frac{a}{sqrt{2}} = a + a\sqrt{2} = 2a + a\sqrt{2} - a = p - a (From I)

KarishmaB · Answer

Ques3.jpg Given is the isosceles right triangle with leg of length a. The hypotenuse will be \sqrt{2}a using pythagorean theorem. We are given a + a + \sqrt{2}a = p p = 2a + \sqrt{2}a When you drop an altitude to make 2 smaller similar triangles, the length of the altitude is say L. Area of the big triangle = (1/2)*a*a = (1/2)*\sqrt{2}a * L L = a/\sqrt{2} Perimeter of small triangle = a + L + \sqrt{2}a/2 = a + a/\sqrt{2} + \sqrt{2}a/2 = a + \sqrt{2}a But we need the perimeter in terms of p too. p = 2a + \sqrt{2}a So \sqrt{2}a = p - 2a Perimeter of small triangle = a + p - 2a = p - a Answer (B)

DesiGmat · Answer

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.

PareshGmat · Answer

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

jayanth7290 · Answer

DesiGmat  altitude is median since its an isoceles triangle so it works in this case

DesiGmat · Answer

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