# square inscribed in a right triangle 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.

• Presumably, $a$, $b$, and $c$ are the sides of the triangle, with $c$ being the hypotenuse? – Gerry Myerson 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
• @Gerry Myerson is the answer $ab/4$ wrong? – cdummy Mar 4 '13 at 12:37
• 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

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

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

• 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

Hint:

$\hspace{60pt}$ 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! ;-)

• 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
• @cdummy $\triangle ABC$ is a right triangle, so $c^2 = a^2 + b^2$. To get the final answer, the area of a square with side $x$ is $x^2$. – dtldarek Mar 4 '13 at 17:14

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

• Much as I noted earlier, in one of my comments. – Gerry Myerson Mar 4 '13 at 12:51

The question can be easily solved by subtracting the two sides of a right triangle from the bigger triangle i.e. subtract 'as/b' and 'bs/a' from the hypotenuse of larger triangle 'c'.and it will be equal to the side of a square. S=c-(as/b+bs/a). And thus s=abc/a^2+b^2+ab

• Use MathJax and format your post for making your answer better. – SchrodingersCat Nov 12 '15 at 7:38