In the figure shown above, line segment has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
OA B.
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In the figure shown above, line segment QR has length 12,
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Official Guide
In the figure shown above, line segment has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
OA B.
In the figure shown above, line segment has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
OA B.
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Jay@ManhattanReview
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The question is posted correctly.AAPL wrote:Official Guide
In the figure shown above, line segment has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
OA B.
The correct version is this:
Area of MPRS = Area of MPQT + Area of TQRSIn the figure shown above, line segment QR has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
Say the length of the square MPQT = x, thus, Area of MPQT = x^2
And, Area of TQRS = 12.x
So, 540 = x^2 + 12x => x = 18.
Area of TQRS = 12*18 = 216
The correct answer: B
Hope this helps!
-Jay
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fskilnik@GMATH
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\[? = S\]AAPL wrote:Official Guide
In the figure shown above, line segment QR has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
\[540 = \frac{S}{{12}}\left( {12 + \frac{S}{{12}}} \right)\,\,\mathop = \limits^{x\,\,\left( {aux} \right)\,\, = \,\,\,S/12} \,\,\,x\left( {12 + x} \right)\]
\[540 = 54 \cdot 10 = 27 \cdot 2 \cdot 10 = {3^3} \cdot {2^2} \cdot 5\,\,\,\,\mathop \Rightarrow \limits^{12\,\,{\text{difference}}} \left\{ \begin{gathered}
\boxed{x = {3^2} \cdot 2 = 18} \hfill \\
12 + x = 3 \cdot 2 \cdot 5 = 30 \hfill \\
\end{gathered} \right.\]
\[\frac{S}{{12}} = x = 18\,\,\, \Rightarrow \,\,? = S = 12 \cdot 18 = 180 + 36 = 216\]
This solution follows the notations and rationale taught in the GMATH method.
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Scott@TargetTestPrep
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Since PQMT is a square, we can let PQ = PM = n. Since QR = 12, PR = n + 12. Finally, since the area of rectangle MPRS is 540, we have:AAPL wrote:Official Guide
In the figure shown above, line segment has length of 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?
A. 144
B. 216
C. 324
D. 360
E. 396
n(n + 12) = 540
n^2 + 12n = 540
n^2 + 12n - 540 = 0
(n + 30)(n - 18) = 0
n = -30 or n = 18
Since n can't be negative, n = 18, and the area of square MPQT is 18 x 18 = 324. Thus, the area of rectangle TQRS is 540 - 324 = 216. (Or the area of rectangle TQRS is 18 x 12 = 216.)
Answer: B
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