Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#p1154669

According to the above, triangles PQR, PSQ and QSR must be similar.

Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12.

The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150.

Answer: D.

Hope it's clear.
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In right triangle PQR right angled at Q, QS^2 = PS * SR --------> qs^2 = 16*9 ------> qs = 4*3 = 12

Area = \(\frac{1}{2} QS*PR\) -----> \(\frac{1}{2} 12*25\) ------> 150
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Dear Narenn,

Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR??

It's a standard relationship based on similar triangle theory.

We should note that triangle PQR, triangle PSQ, and triangle QSR are similar

In similar triangles, ratios of corresponding sides are equal.

We know that triangle PSQ is similar to triangle QSR


so, \(\frac{base of PSQ}{height of PSQ} = \frac{base of QSR}{height of QSR}\)


\(\frac{QS}{PS} = \frac{RS}{QS}\) --------> \(QS^2\) = PS*RS


Similarly you can also prove that \(PQ^2\)= PS*PR and \(QR^2\) = SR*PR
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Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?

Triangle QSR is the same as RSQ.
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Thanks Bunnel!
Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides
My question is, how can we find the right set of corresponding sides to match up in similar triangles?
Thank you in advance! I've always had this problem with similar triangles.

I think the following post might helps: in-the-diagram-above-if-arc-abc-is-a-semicircle-what-is-119652.html#p1374777

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i had the same issue as in finding corresponding sides in similiar triangle. i suggest you watch some videos in khan academy relating to similiar triangles. there are few videos pertaining to it .watch them they are really helpful.

Hi Bunnel,

I tried solving this problem by using the 45-45-90 triangle approach for triangle QSR !! Why is this approach incorrect ?

First of all, you don't show your work at all and asks what's wrong with it. Next, why do you assume that QSR is 45-45-90 triangle?
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Can we solve this by using the following approach??
the area of triangle PQR = area of triangle PQS+ area of QRS
But the side QS must be equal to what ?? IS that equal to PS or SR
Kindly help

But 16+9 is the hypotenuse of the triangle PQR

Why are we using it as the height?

We are considering PR = 25 as the base of triangle PQR and QS as the height:

The area of triangle PQR = 1/2*base*height = 1/2*PR*QS = 1/2*(16+9)*12 = 150.
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Here is the basic mathematical way of solving this question.



Let QR = x, PQ = y and QS = z
Given PS = 16 and SR = 9

Triangle PQR is right angle triangle
\(x^2 + y^2 = (16+9)^2 = 25^2\) -----------------------------------------Equation 1

Triangle QSR is right angle triangle

\(z^2 + 9^2 = x^2\) ------------------------------------------------Equation 2

Triangle QPS is right angle traingle

\(z^2 + 16^2 = y^2\) -----------------------------------------------Equation 3

Putting the values of x^2 and y^2 from equation 2 and 3 into equation 1, we get

\(z^2 + 9^2 + z^2 + 16^2 = 25^2\)

\(2z^2 + 9^2 + 16^2 = 25^2\) ---------------------------------Equation 4

Solving equation 4 for z, we get z=12

Now Area of the triangle is (\(\frac{1}{2}\))*base*hieght
So, area of triangle PQR is (\(\frac{1}{2}\))*PR*QS = (\(\frac{1}{2}\))*(16+9)*12 = (\(\frac{1}{2}\))*25*12 = 150

Hence D

Hi.

Can we solve this by using the property : The median from right angle in a triangle is half of the hypotenuse?

Implying QS = 1/2* PR and then using Pythagorean Theorem ? Is this method valid as well ?

QS is not the median, it's height.
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Standard type of right triangles are with sides 3 4 5 , 5 12 13, 8 15 17, and 7 24 25.
1) In this question we have hypotenuse as 25 (a multiple of 5) so we can choose 3 4 5 and 5 12 13.
2) Now hypotenuse is the largest side of a right triangle , this leaves only 3 4 5.
In this case multiply each side of 3 4 5 with 5 (since hypotenuse is a multiple of 5) we get 15 20 25.
Now calculate the area! (15*20)/2=150

hit kudos if you like or tell me where I am wrong

I think This picture will describe the formulae