Triangle PQR is right angled at Q. QT is perpendicular to PR

Find the hypotenuse by using the Pythagorean theorem.

\(3^2 + 6 ^2 = 3\sqrt{5}\)

Area of the right angled triangle = \(\frac{1}{2}*base*height\)

\(\frac{1}{2}*QR*PQ\)

Above can be equated with \(\frac{1}{2}*PR*QT\)

and QT can be found easily

Hi Bunuel, I have a small doubt regarding similarity of triangles. Could you please explain the concept in little more detail or point me to a place where i can learn the concept.

eg lets say same diagram except now QT = 4 , PT = 3 .....we have to find TR = ?

Could you please explain this ?

If QT=4 and PT=3 --> corresponding legs in QTR and in PTQ are in the same ratio: TR/QT=QT/PT --> TR=16/3.

You can use area equation approach if you are not comfortable with similarity.

Bunuel ,on your comment , corresponding legs in QTR and in PTQ are in the same ratio: TR/QT=QT/PT --> TR=16/3. When I list out propertions, its coming TR=TP, Can you explain wat did I do wrong? Thanks!

You might want to refer to this, i found it very comprehensive, like a high school chapter https://www.nos.org/Secmathcour/eng/ch-17.pdf

1/2 * QT * PR = 1/2 * PQ * QR

QT = 18/root(45) = 6/root(5)

Answer - C

Hi Bunuel,

I'm a little confused here. let me try to explain:

If I equate the ratios:

I get: QT/6 = 5/3Root5
I'm equating the ratios as such: the height of the QTR over the height of the the main triangle which equals, the hypotenuse of QTR to the hypotenuse of the main, which i 3 Root 5.

Can you please clarify what the ratios actually are(height of one vs. base of another etc). It's a little confusing since the variables are repeated between the 3 triangles. My worry comes in because if I did this same ratio and I equated it to QTP, i would've gotten a completely different answer. I'm surely doing something wrong -- perhaps in how i'm setting up the ratios?

I see where I was going wrong. I was only accounting to align the hypotenuse but I would switch the other two angles. Thanks for clarifying.

One question to that extent: While aligning the ratios of the 3 triangles (2 inner and 1 outer), Side A/B of Inner #1 = Side A/B of Outer = Side A/B of Inner #2 right? What I mean by that is, I can equate any of the triangles to each other - not just inner/outer but also the two inner triangles because they ALL are similar triangles. Correct?

About to start work on the links above. Thanks again.

Yes, you can equate the ratios of corresponding sides in all three triangles.

Hey guys,

What I don't understand, however, is why can we assume that the perpendicular segment QT cuts through angle PQR, could it not also be anywhere else along QR? And would the theorem (that a perpendicular to the hypotenuse creates three similar triangles) still work even if the line segment does not cut through the right angle (is not the height of the big triangle)?

Bunuel theres a property that median from the right angle is half of the hypotenuse. would there be a way for that to be used here?

QT is perpendicular to the hypotenuse, not the median. A perpendicular to the hypotenuse coincides with the median to the hypotenuse only if a right triangle is also isosceles. This is not so in the question.

Corresponding sides are opposite the corresponding (equal) angles. Corresponding (equal) angles are marked in the diagram below:
Hi Bunuel,
How you arrived at the conclusion that these are the corresponding sides on the basis of corresponding angles (equal) ?
I am facing difficulty in understanding this. can you please elaborate ?