Source: GMAT Prep
The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
$$A.\ 9\sqrt{2}$$
$$B.\ \frac{3}{2}$$
$$C.\ \frac{9}{\sqrt{2}}$$
$$D.\ 15\left(1-\frac{1}{\sqrt{2}}\right)$$
$$E.\ \frac{9}{2}$$
The OA is A.
The shaded region in the figure above represents a
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From the question stem ("...the length and width of the picture have the same ratio as the length and width of the frame") we know:BTGmoderatorLU wrote:Source: GMAT Prep
The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
$$A.\ 9\sqrt{2} \,\,\,\,\,\, B.\ \frac{3}{2} \,\,\,\,\,\, C.\ \frac{9}{\sqrt{2}} \,\,\,\,\,\, D.\ 15\left(1-\frac{1}{\sqrt{2}}\right) \,\,\,\,\,\, E.\ \frac{9}{2}$$
The frame+picture (big rectangle) and the picture (small rectangle) are two SIMILAR rectangles.
(Important: from above we have proportionality on the corresponding sides. The necessary additional condition - equality in the corresponding internal angles - is guaranteed: they are all 90 degrees!)
Again from the question stem we know what the examiner defines as "length" and "width" (by the dimensions associated to these words), so that our FOCUS is:
\[? = x\,\,\,\,\left[ {{\text{inches}}} \right]\,\,\,\,\,\,\left( {{\text{See}}\,\,{\text{figure}}\,\,{\text{below}}} \right)\]
Still from the question stem ("The frame encloses a rectangular picture that has the same area as the frame itself.") we know that the "big" (rectangle) has TWICE the area of the "small" (rectangle).
To avoid using the second dimension of the picture (call it "y" and find another solution using "classical ratios"!), let´s remember an important geometric property:
In any two similar polygons, the ratio of their areas is equal to the square of the ratio of similarity of the polygons!
Therefore:
\[2 = \frac{{{S_{\,{\text{big}}}}}}{{{S_{\,{\text{small}}}}}} = {\left( {\frac{{18}}{x}} \right)^2}\,\,\,\,\mathop \Rightarrow \limits^{x\,\, > \,\,0} \,\,\,\,\,\sqrt 2 = \frac{{18}}{x}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x\sqrt 2 = 18\]
\[x\sqrt 2 = 18\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x\sqrt 2 \cdot \sqrt 2 = 18\sqrt 2 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = x = 9\sqrt 2 \]
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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IMPORTANT: the diagrams in GMAT problem solving questions are DRAWN TO SCALE unless stated otherwise.BTGmoderatorLU wrote:Source: GMAT Prep
The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
$$A.\ 9\sqrt{2}$$
$$B.\ \frac{3}{2}$$
$$C.\ \frac{9}{\sqrt{2}}$$
$$D.\ 15\left(1-\frac{1}{\sqrt{2}}\right)$$
$$E.\ \frac{9}{2}$$
The OA is A.
So, we can use this fact to solve the question by simply "eyeballing" the diagram.
See our video below on this topic as well as other assumptions we can make about diagrams on the GMAT
If you had to ESTIMATE the length of the picture, what would you say it is?
12? 13? 14? 15?
As long as you're in this range, you should be able to solve this one.
ASIDE: On test day, you should have memorized the following approximations:
√2 ≈ 1.4
√3 ≈ 1.7
√5 ≈ 2.2
Now let's check the answer choices....
A. 9√2 ≈ (9)(1.4) ≈ 13. This is within our estimated range. KEEP
B. 3/2 = 1.5. This is WAYYYY outside our estimated range. ELIMINATE
C. 9/√2 ≈ 9/1.4 ≈ 6. This is WAYYYY outside our estimated range. ELIMINATE
D. 15(1 - 1/√2) ≈ 15(1 - 0.7) ≈ (15)(0.3) ≈ 4.5. This is WAYYYY outside our estimated range. ELIMINATE
E. 9/2 = 4.5. This is WAYYYY outside our estimated range. ELIMINATE
Answer: A
RELATED VIDEO FROM OUR COURSE
[you-tube]https://www.youtube.com/watch?v=MaPuJh3m3Pk[/you-tube]
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We see that the total area of the frame and the picture is 18 x 15 = 270. Since we know that the length and width of the picture have the same ratio as the length and width of the frame, let's denote the length of the picture by 18k and the width of the picture by 15k, where k is some positive constant.BTGmoderatorLU wrote:Source: GMAT Prep
The shaded region in the figure above represents a rectangular frame with length 18 inches and width 15 inches. The frame encloses a rectangular picture that has the same area as the frame itself. If the length and width of the picture have the same ratio as the length and width of the frame, what is the length of the picture, in inches?
$$A.\ 9\sqrt{2}$$
$$B.\ \frac{3}{2}$$
$$C.\ \frac{9}{\sqrt{2}}$$
$$D.\ 15\left(1-\frac{1}{\sqrt{2}}\right)$$
$$E.\ \frac{9}{2}$$
Then, the area of the picture is (18k)(15k) = 270k^2
The area of the frame can be found by subtracting the area of the picture from the total area of the frame and the picture: 270 - 270k^2
Since the area of the frame is equal to the area of the picture, we have:
270 - 270k^2 = 270k^2
270(1 - k^2) = 270k^2
1 - k^2 = k^2
2k^2 = 1
k^2 = 1/2
k = 1/√2
Since the length of the picture was represented by 18k, the picture's length is 18(1/√2) = 18/√2 = (18/√2)*√2/√2= 18√2/2 = 9√2.
Answer: A
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