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Math Expert
Joined: 02 Sep 2009
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In the figure above, P, Q, R, and S are the midpoints of the sides of
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21 Dec 2017, 19:50
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Re: In the figure above, P, Q, R, and S are the midpoints of the sides of
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21 Dec 2017, 22:03
This is tough how to approach above question I thought of deleting the triangle bpq and srd area from triangle abcd and my result was polygon=3/4(area of abcd)..I don't know how to calculate abcd area. Sent from my Pixel 2 using GMAT Club Forum mobile app



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Re: In the figure above, P, Q, R, and S are the midpoints of the sides of
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22 Dec 2017, 00:09
From the figure above, ang(QPA) = ang(CRS) , the sum of all internal angles = 180, therefore 180x+180x+30+90=180, x=150, right angled triangle is 306090. From there, we can calculate BP = √3 and BQ=3
Area of rect(ABCD) = 12√3 Area of 2 triangles = 3√3
Area of polygon = 12√33√3 = 9√3
Please confirm if this is correct approach and answer



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Re: In the figure above, P, Q, R, and S are the midpoints of the sides of
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22 Dec 2017, 00:43
Area of PBQ is half of ABC, DRS half off ACD. Thus PBQ + SRD = ½ ABCD = PASRCQ. 2*sqrt(3) and x30 gives hint of a triangle with 906030. Thus, PB = sqrt(3), BQ =3. Area of ABCD is: 2*sqrt(3)*2*3=12*sqrt(3). Area of PASRCQ is 6*sqrt(3)
Please correct if it is wrong.



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Re: In the figure above, P, Q, R, and S are the midpoints of the sides of
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22 Dec 2017, 05:23
Bunuel wrote: In the figure above, P, Q, R, and S are the midpoints of the sides of rectangle ABCD. What is the area of the polygon APQCRS? A. \(12\sqrt{3}\) B. \(9\sqrt{3}\) C. \(6\sqrt{3}\) D. \(3\sqrt{3}\) E. \(24\) Attachment: The attachment Polygon.png is no longer available In addition to the solutions offered above, we'll also show a more visual, graphical answer. This is an Alternative approach and relies on the fact that there is only one way to draw the given figure (as the points P,Q,R,S are midpoints of a rectangle). Connecting the points P,Q,R and S to each other gives the below image. We can SEE that drawing the lines has divided the rectangle into 8 identical triangles. Therefore, our total area is the area of 6 of these triangles. To complete the calculation, notice that \(\angle QPR\) and \(\angle PRS\) are equal because PQ is parallel to RS. (once again  since there is only one way to draw this figure, you can SEE that they are parallel) This means that \(x  90 + x  120 = 90\) so \(2x = 300\) and \(x = 150\). Then \(\triangle PQO\) is a 306090 triangle with sides of \(\sqrt3\) and \(3\) so its area is \(1.5 \sqrt{3}\). Multiplying by 6 gives \(9 \sqrt{3}\). (B) is our answer.
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In the figure above, P, Q, R, and S are the midpoints of the sides of
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23 Dec 2017, 09:17
Bunuel wrote: In the figure above, P, Q, R, and S are the midpoints of the sides of rectangle ABCD. What is the area of the polygon APQCRS? A. \(12\sqrt{3}\) B. \(9\sqrt{3}\) C. \(6\sqrt{3}\) D. \(3\sqrt{3}\) E. \(24\) PQRS is a rectangle and the sum of the interior angles is: \(360 = x + (x30)+x +(x30)\) so \(x = 150\) \(BQP = BQC  x\); BQC is a straight line and \(x=150\), so: \(BQP = 180  150 = 30\) PBQ is a right angle and its angles are \(30:60:90\), which means that its sides must be: \(s:s√3:2s\) If PQ is the hypotenuse \(2√3=2s\), then the other sides \(BQ=(√3*√3)=3\) and \(BP=√3\) Area of PBQ = Area of SRD which is \(\frac{base*height}{2}=\frac{3√3}{2}\). Area of ABCD is \([(BP*2)*(BQ*2)]=[(2√3)*(6)]=12√3\) Area of APQCRS = Area of ABCD  Area of Triangles (PBQ and SRD): \(12√3  {(\frac{3√3}{2}+\frac{3√3}{2})} = 12√3  \frac{6√3}{2}=\frac{(24√36√3)}{2}=9√3\) (B) is the answer.




In the figure above, P, Q, R, and S are the midpoints of the sides of &nbs
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23 Dec 2017, 09:17






