![Image](https://s4.postimage.org/25xrukx0/TTT.jpg)
Inscribed triangle - semicircle
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- MAAJ
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From the graphic, I understand that triangle PQS is similar to PQR because they share xº and 90º, so the last angle must be equal. But I'm missing why is PQS similar to QSR, can someone help me out? (I know they both share 90º but not sure about the other angles) Thanks!
![Image](https://s4.postimage.org/25xrukx0/TTT.jpg)
![Image](https://s4.postimage.org/25xrukx0/TTT.jpg)
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PQS is similar to PQR for the below reason :-
It is only necessary to determine that two sets of angles are identical in order to conclude that two triangles are similar; the third set will be identical because all of the angles of a triangle always sum to 180º.
Hope this helps.
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Please see my explanation for the problem below.
![Image](https://www.beatthegmat.com/files/trianglequestion_517.jpg)
Triangle PQR is a right triangle. QS is a height drawn through the right angle to the opposite side.
In any right triangle, a height drawn through the right angle to the opposite side creates 3 similar triangles.
We can prove this by plugging in for the angles measurements. If ∠RPQ = 20, then ∠PQS = 70, since the sum of the 2 angles must be 90. This forces ∠RQS to be 20. This forces ∠QRS to be 70. The result is that all 3 triangles (PQS, RQS, and PQR) have the same combination of angles: 20-70-90. Thus, the 3 triangles are similar.
The corresponding sides of similar triangles must yield the same proportion.
In triangle PQS, the shorter leg = QS, the longer leg = PS = 16.
In triangle RQS, the shorter leg = RS = 9, the longer leg = QS.
Since shorter leg:longer leg must be the same for each triangle, we get:
QS/16 = 9/QS
(QS)² = 144
QS = 12.
Thus, in PQR, h = 12 and b = PS + SR = 16+9 = 25.
Area = 1/2*b*h = 1/2(25)(12) = 150.
The correct answer is D.
![Image](https://www.beatthegmat.com/files/trianglequestion_517.jpg)
Triangle PQR is a right triangle. QS is a height drawn through the right angle to the opposite side.
In any right triangle, a height drawn through the right angle to the opposite side creates 3 similar triangles.
We can prove this by plugging in for the angles measurements. If ∠RPQ = 20, then ∠PQS = 70, since the sum of the 2 angles must be 90. This forces ∠RQS to be 20. This forces ∠QRS to be 70. The result is that all 3 triangles (PQS, RQS, and PQR) have the same combination of angles: 20-70-90. Thus, the 3 triangles are similar.
The corresponding sides of similar triangles must yield the same proportion.
In triangle PQS, the shorter leg = QS, the longer leg = PS = 16.
In triangle RQS, the shorter leg = RS = 9, the longer leg = QS.
Since shorter leg:longer leg must be the same for each triangle, we get:
QS/16 = 9/QS
(QS)² = 144
QS = 12.
Thus, in PQR, h = 12 and b = PS + SR = 16+9 = 25.
Area = 1/2*b*h = 1/2(25)(12) = 150.
The correct answer is D.
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- MAAJ
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Wow, many thanks! now I get it... I was missing the part of picking numbers for angles, thank U!
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Note that in triangle PQR, (xº + yº) = 90º
Now, in triangle PQS, angle PQS = 90º - xº = yº
In triangle PQS and QSR,
- angle PSQ = angle QSR
angle PQS = angle QRS = yº
Just remember this fact for right-angled triangle that when you draw a perpendicular from the vertex with the right-angle on the hypotenuse, the new smaller two right-angled triangles are similar to each other and also similar to the original one.
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- tomada
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Mitch, in triangle RQS, we know that RS < QS based on the diagram? (i.e. QS appears to be longer)
GMATGuruNY wrote:Please see my explanation for the problem below.
Triangle PQR is a right triangle. QS is a height drawn through the right angle to the opposite side.
In any right triangle, a height drawn through the right angle to the opposite side creates 3 similar triangles.
We can prove this by plugging in for the angles measurements. If ∠RPQ = 20, then ∠PQS = 70, since the sum of the 2 angles must be 90. This forces ∠RQS to be 20. This forces ∠QRS to be 70. The result is that all 3 triangles (PQS, RQS, and PQR) have the same combination of angles: 20-70-90. Thus, the 3 triangles are similar.
The corresponding sides of similar triangles must yield the same proportion.
In triangle PQS, the shorter leg = QS, the longer leg = PS = 16.
In triangle RQS, the shorter leg = RS = 9, the longer leg = QS.
Since shorter leg:longer leg must be the same for each triangle, we get:
QS/16 = 9/QS
(QS)² = 144
QS = 12.
Thus, in PQR, h = 12 and b = PS + SR = 16+9 = 25.
Area = 1/2*b*h = 1/2(25)(12) = 150.
The correct answer is D.
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We know that RS < QS based upon the information given. Here again is the figure, with variables assigned to the angles:tomada wrote:Mitch, in triangle RQS, we know that RS < QS based on the diagram? (i.e. QS appears to be longer)
![Image](https://s2.postimage.org/35lzt4cro/3_similar_triangles.jpg)
Since all 3 triangles are similar, in each triangle we get the following proportion:
(leg opposite x):(leg opposite y) = (leg opposite x):(leg opposite y)
This gives us the proportion QS/16 = 9/QS.
Since QS=12 and RS=9, RS < QS.
Which leg is shorter doesn't really matter. Just be sure to set up the correct proportion, as shown above.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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For more information, please email me (Mitch Hunt) at [email protected].
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