Bunuel wrote:
In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?A) 3/2
B) 7/4
C) 15/8
D) 16/9
E) 2
Let \(PQ=x\), \(QR=y\) and \(PR=z\).
Given: \(x+y+z=60\) (i);
Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> \(xy=12z\) (ii);
Aslo \(x^2+y^2=z^2\) (iii);
So, we have:
(i) \(x+y+z=60\);
(ii) \(xy=12z\);
(iii) \(x^2+y^2=z^2\).
From (iii) \((x+y)^2-2xy=z^2\) --> as from (i) \(x+y=60-z\) and from (ii) \(xy=12z\) then (\(60-z)^2-2*12z=z^2\) --> \(3600-120z+z^2-24z=z^2\) --> \(3600=144z\) --> \(z=25\);
From (i) \(x+y=35\) and from (ii) \(xy=300\) --> solving for \(x\) and \(y\) --> \(x=20\) and \(y=15\) (as given that \(x>y\)).
Next,
perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQS and SQR are similar.
In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) --> \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\)
Answer: D.
Attachment:
Triangle PQR.GIF
Lets say PQ=a, QR =b, PS=c, SR =d, QS =e
a^2 - c^2 = 12^2 -------i
b^2 - d^2 = 12^2 -------ii
a^2 + b^2 = (c+d)^2 ------iii
Sum eq i+ii
a^2 + b^2 -c^2-d^2 = 2*12^2
replace a^2+b^2 by (c+d)^2 ref eq. iii
(c+d)^2 -c^2-d^2 = 2*12^2
c^2 + d^2 +2.c.d-c^2-d^2 = 2*12^2
solving it
cd = 12^2 =144
scan answer voice for cd = 144 only option 16*9 = 144 so the ratio = 16/9.
Bunuel pls suggest if this method is ok.