Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is the figure A square of maximum possible area is circumscribed by a right angle triangle ABC in such a way that one of its side just lies on the hypotenuse of the triangle.
What is the area of the square? actually the answer is given as $(abc/(a^2+b^2+ab))^2$ Please provide the approach to solve the problem.

share|improve this question
Presumably, $a$, $b$, and $c$ are the sides of the triangle, with $c$ being the hypotenuse? –  Gerry Myerson Mar 4 '13 at 12:11
What have you tried? –  dtldarek Mar 4 '13 at 12:11
Seems to me (draw a picture!) that there are several triangles similar to ABC, and you should be able to get some mileage from that. –  Gerry Myerson Mar 4 '13 at 12:13
@dtldarek i extended the triangle into a rectangle. and the square also. Then i used $1/2*d_1*d_2$ and the final answer as $ab/4$. –  cdummy Mar 4 '13 at 12:20
I don't understand how you extend the triangle and square, and I don't know what you mean by $d_1$ and $d_2$. Also, you haven't answered my question about the meanings of $a$, $b$, and $c$. –  Gerry Myerson Mar 4 '13 at 12:50

3 Answers 3

up vote 2 down vote accepted

Consider - a,b as right legs and c as the hypotenuse.

Let side of square = s AC = b, BC = a, AB = c.

Right Angled Triangle

FB = as/b and AE = bs/a as the colored triangles are similar to the bigger triangle.

Steps to calculate area (S^2) :

1)Calculate GB and AD using right angle triangle rule for triangles GBF and ADE.

2)Calculate GD using right angle triangle rule for triangle GCD.

3)GD^2 = s^2. You get a quadratic equation in s which can be factorized. You get s = (abc)/(a^2 + b^2 +a.b)

If still not clarified will post the answer then.

share|improve this answer
how you got FB=as/b –  cdummy Mar 4 '13 at 15:49
i got $FB=bs/a$ and $EA=sa/b$ and please post the clarification. –  cdummy Mar 4 '13 at 16:45
sry you are correct.. i just wrongly used the ratio. FB=as/b is correct.And thanks for the answer. I got it. –  cdummy Mar 4 '13 at 18:05



The red solid line is the height dropped onto the hypotenuse, i.e. $h = \frac{ab}{c}$ and the red dotted lines are of the same length. The green parallel lines are unnecessary, but might get you some intuitions.

Good luck! ;-)

share|improve this answer
how to get $h=ab/c$. which pair of similar triangles to consider for that. –  cdummy Mar 4 '13 at 15:39
@cdummy $P_{\triangle ABC} = \frac{1}{2}ab = \frac{1}{2}ch$. –  dtldarek Mar 4 '13 at 16:09
I appreciate your answer from my heart. But the problem is i need some simple solution. or a hard one that is explained. See your reputation is 6340, where as mine is 29. i m sry to say this i cant get your point of approach. Can u give a detailed solution only if u are interested... –  cdummy Mar 4 '13 at 16:30
@cdummy Well, I don't have time to write a fully detailed answer, but consider the following. First calculate the area of the triangle in two ways: $\frac{1}{2}ab = |\triangle ABC| = \frac{1}{2}ch$ to get $h = \frac{ab}{c}$. Let $x$ be the side of a square, then the picture implies (the black and gray triangles are similar) that $\frac{x}{c} = \frac{h}{c+h}$, so $$x = \frac{ch}{c+h} = \frac{c\frac{ab}{c}}{c+\frac{ab}{c}} = \frac{abc}{c^2+ab}.$$ –  dtldarek Mar 4 '13 at 17:03
thanks very much –  cdummy Mar 4 '13 at 17:09

Hint: The square formed some similar right-angled triangles, make use of the ratio of the sides.

share|improve this answer
Much as I noted earlier, in one of my comments. –  Gerry Myerson Mar 4 '13 at 12:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.