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In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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26 Jun 2008, 20:39
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In ΔPQS above, if PQ =3 and PS = 4, then PR =? A. 9/4 B. 12/5 C. 16/5 D. 15/4 E. 20/3 Attachment:
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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05 Mar 2012, 10:52




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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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26 Jun 2008, 20:58
prasannar wrote: In ΔPQS attached, if PQ =3 and PS = 4, then PR=?
(A) 9/4 (B) 12/5 (C) 16/5 (D) 15/4 (E) 20/3 its just equating the ares 1/2(pq*ps)=1/2(qs*pr) 1/2(3*4)=1/2 (5*pr) pr=12/5




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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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26 Jun 2008, 20:51
Let QR=x and PR=h
x^2 +h^2 = 9
5x)^2 + h^2 = 16
Solving for x = 9/5 and h=12/5



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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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26 Jun 2008, 20:53
B
345 triangle
let qr = x then rs = 5x
let pr = h
2 equations:
3^2 = h^2 + x^2 and 4^2 = h^2 + (5x)^2
expand and substract them to get x = 9/5
substitute 9/5 into the first equation (or second, your choice) to get 9 = h^2 + 81/25
solve the equation to get h = 12/5



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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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27 Jun 2008, 06:39
This is quite a wellknown diagram in mathematics: it can be used to prove the Pythagorean Theorem (using a and b for the two legs, and not 3 and 4). There are (at least) three entirely different ways to solve this problem, two of which were described above. Call the length we're looking for 'd': The area of the triangle must be the same no matter which base you choose. Thus 3*4/2 = 5*d/2 > d = 12/5. Let QR = c; then QS = 5c. We have two right angled triangles, and can use Pythagoras to set up two equations, two unknowns (c and d; this is the most timeconsuming approach). The approach not mentioned above: notice that the two smaller triangles in the diagram are each similar to the 345 triangle. Because PQR is similar to QSP, we have d/3 = 4/5 > d = 12/5. It's the similarity of the triangles that can let you prove Pythagoras, incidentally.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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09 Nov 2010, 15:36
PQ =3 and PS = 4, So QS = 5
area of the triangle = 1/2 * PQ * PS = 1/2*QS*PR
OR, 1/2 * 3 * 4 = 1/2 * QS * PR OR PR = 12/5



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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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05 Mar 2012, 11:18
Be patient. I have not fully understood Why 5/2 * PQ...should be QS ?? Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ??? We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 306090 triangle. So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90). So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR). If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90. I know that it does not hold anywater, but is useful to understand Thanks
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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05 Mar 2012, 11:33
carcass wrote: Be patient. I have not fully understood Why 5/2 * PQ...should be QS ?? Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ??? We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 306090 triangle. So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90). So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR). If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90. I know that it does not hold anywater, but is useful to understand Thanks First of all 5/2 * PQ does not equal to QS. We equate the areas, which can be found in two ways: 1. 1/2*Leg1*Leg2 > \(area=\frac{1}{2}*PQ*PS=6\); 2. 1/2*Perpendicular to hypotenuse*Hypotenuse > \(area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR\) (since hypotenuse QS=5); Now, equate the areas: \(6=\frac{5}{2}*PR\) > \(PR=\frac{12}{5}\). Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\). In PQS sides PQ and PS are NOT in the ratio \(1 : \sqrt{3}\), so PQS is not a 30°, 60°, and 90° right triangle. I think you are mixing 345 Pythagorean Triples triangle with 30°60°90° triangle. Hope it's clear.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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07 Sep 2012, 00:39
As mentioned by Bunuel, there are other methods as well to solve this problem One of the method is SimilarityTriangle QPS is similar to PRS (because one common side & angle is 3 & 90 degree) So, QS/QP = PS/PR 5/4 = 3/x x = 12/5 Hope it helps.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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24 Feb 2016, 03:26
prasannar wrote: Attachment: Diag1.JPG In ΔPQS above, if PQ =3 and PS = 4, then PR =? A. 9/4 B. 12/5 C. 16/5 D. 15/4 E. 20/3 Two approaches. the area is 6. so area of pqr + area of prs = 6. solution will result in 12/5. B second: testing answer choices. if assume that line(pr) is an answer choice and try to rearrange to get line(rs), then you discover line rs is squareroot of some negative numbers in all the options except B



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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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09 Apr 2016, 09:19
prasannar wrote: Attachment: Diag1.JPG In ΔPQS above, if PQ =3 and PS = 4, then PR =? A. 9/4 B. 12/5 C. 16/5 D. 15/4 E. 20/3 QS will be 5 being the hypotenuse of 90 degree triangle. As we can calculate the area by two ways here: 1/2 * PS * PQ = 1/2 * PR * QS 4 * 3 = PR * 5 PR = 12/5
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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05 Oct 2016, 10:22
Very simple hypotenuse will be 5 as right angle triangle (3,4,5) . A perpendicular drawn to hypotenuse will bisect it , that means 5/2 or 2.5 ; 12/5 nearest option.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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04 Aug 2017, 00:32
Let QR=x and RS=y. Area of QPS= 1/2*4*3 = 6 We know, Area of QPS = Area of QRP + Area of PRQ > (1) Let PR=a. From Pythagoras Theorem, QS=5. ==> x+y=5 In (1), 6 = 1/2*y*a +1/2*x*a ==> 1/2(ay + ax) ==> a(x+y) =12 As x+y=5, a*5=12 ==> a=12/5 = PR Answer: B
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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