A square with side length x is inscribed in a right triangle with side

As discussed by most of the experts here, this is a question which tests your knowledge of similarity concepts.

When you have a situation where a smaller triangle is inside a larger triangle, such that one side of the smaller is parallel to the corresponding side of the larger, corresponding angles will be equal and you will be able to prove the triangles similar.

Because the triangles are similar, their sides will be in the same ratio and so will be all the other line segments of the two triangles, like the altitudes, medians and angle bisectors.

Drawing a diagram for the first situation will give us a figure similar to this:


Clearly, DE is parallel to BC, the sides AB and AC are the transversals. Angle A is common, angles ADE and ABC are corresponding angles and hence equal and so are angles AED and ACB. So, triangle ADE is similar to triangle ABC. Therefore,

\(\frac{DE}{BC}\) = \(\frac{AD}{AB}\)

\(\frac{x}{3}\) = \(\frac{(4-x)}{4}\)
which on solving gives us x = \(\frac{12}{7}\).

At this point, we do not have sufficient information to eliminate any of the options, so let us continue analyzing the second half of the question.

A diagram representing the second situation looks something like this:


Using similar methods, we can prove that triangle BPQ is similar to triangle BAC (or ABC).

Area of triangle ABC = ½ * 3 *4 = ½ * 5 * BD. Therefore, BD = \(\frac{12}{5}\).

Now, since triangle BPQ is similar to triangle BAC,

\(\frac{Altitude of triangle BPQ}{BD}\) = \(\frac{y}{5}\) (because ratio of altitudes = ratio of corresponding sides).

Therefore, Altitude of triangle BPQ = BD * \(\frac{y}{5}\) = \(\frac{12}{5}\) * \(\frac{y}{5}\) = \(\frac{12y}{25}\).

Altitude of triangle BPQ + y = BD.

\(\frac{12y}{25}\) + y = \(\frac{12}{5}\).

Simplifying, we get, y = \(\frac{60}{37}\).

Therefore, \(\frac{x}{y}\) = \(\frac{12}{7}\) * \(\frac{37}{60}\) which simplifies to \(\frac{37}{35}\).
So, the correct answer option is D.

Hope that helps!

nick1816 Please explain highlighted part.

AC=AZ+ZW+WC=5
\((\frac{3y}{4})+y+\frac{4y}{3}=5\)
\((\frac{9+12+16}{12})y=5\)
y=\(\frac{60}{37}\)

Can someone explain how are you proving the triangles similar?

Posted from GMAT ToolKit

To be honest, unless I had a ton of extra time on the Quant section, I'd take an informed guess on this one.

First, I noticed the 'evil twins' in the answer choices: A and E are a pair, and so are B and D. Therefore, I'm definitely going to eliminate C, which doesn't have a twin.

Next, draw it out:



The squares look almost the same size to me (so I'm leaning towards either B or D), but my intuition says that x is a tiny bit bigger - it seems like you can fit a sliiiightly bigger square into a right triangle by fitting it perfectly into the corner of the triangle, versus rotating it. D is a reasonable guess, although B would also be reasonable, since I don't have a great way to convince myself that x really is greater.

No problem. If you want to PROVE that triangles are similar, you need to show that their angles are the same. Here's how you'd do it for the first triangle, the one with the square whose sides have length x:



Since it's a square, all of its angles are 90 degrees. I've marked those.

Next, the angles that are supplementary to a 90 degree angle are also 90 degrees:



Next, let's label one of the unknown angles with a question mark (?). It doesn't really matter which one you pick; I chose the bottom right corner.



The sum of the angles in any triangle is 180. So, we can label some of the missing angles: 180 - 90 - ? = 90 - ?



Finally, the last angle we don't know is the third angle of the top, smallest triangle. Its other two angles are 90 and 90-? .So, the third angle is 180-90-(90-?) = ?.



Check out the three triangles in that image: the one on the top, the one on the right, and the larger one that encompasses the other two. All three of them have exactly the same set of angles: ?, 90, and 90-?. Therefore, all three are similar to each other.

However, I don't think that's what the folks posting in this thread actually did! After all, that would be time consuming. To save time next time, you can just memorize a rule of thumb: if you draw a straight line across a triangle (parallel to one of its bases), the resulting triangle is similar to the original. For example, this image shows two similar triangles:

Triangles are similar if they have similar angles /sides. In this cases triangle with a square inside it . We need to prove 3points to make triangles similar (SAS,SSS,AAA,ASA)
Here we have ASA
1)A =90degree similar
2)S=both triangles have hypotenuse (both are right angle triangles)
3)A=both triangles share same top corner angle .

ccooley Thanks a lot i understand it now