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The figure above shows four adjacent small squares, forming one large square. The vertices of square \(RSTU\) are midpoi

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by VJesus12 » Thu Nov 19, 2020 7:47 am

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phd03.png
The figure above shows four adjacent small squares, forming one large square. The vertices of square \(RSTU\) are midpoints of the sides of the small squares. What is the ratio of the area of \(RSTU\) to the area of the large outer square?


A. \(\dfrac12\)

B. \(\dfrac59\)

C. \(\dfrac7{12}\)

D. \(\dfrac35\)

E. \(\dfrac58\)

Answer: E

Source: Princeton Review
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Re: The figure above shows four adjacent small squares, forming one large square. The vertices of square \(RSTU\) are mi

by Scott@TargetTestPrep » Sat Jan 02, 2021 2:03 pm
VJesus12 wrote: ↑
Thu Nov 19, 2020 7:47 am
 phd03.png

The figure above shows four adjacent small squares, forming one large square. The vertices of square \(RSTU\) are midpoints of the sides of the small squares. What is the ratio of the area of \(RSTU\) to the area of the large outer square?


A. \(\dfrac12\)

B. \(\dfrac59\)

C. \(\dfrac7{12}\)

D. \(\dfrac35\)

E. \(\dfrac58\)

Answer: E

Solution:

If we let each side of the small squares be 2, then the side of the large outer square is 4, and thus it has an area of 4^2 = 16.
Let’s label the upper left vertex of the figure as Q, and let’s focus on right triangle RUQ. We know that the length of side QR is 3, and the length of side QU is 1. Using the Pythagorean theorem, we find that the length of the hypotenuse UR is √((3^2 + 1^2) = √10. Since UR is also the length of the side of square RSTU, we see that the area of RSTU = 10.
Thus, the ratio of the area of RSTU to that of the larger square is 10/16 = 5/8.
Answer: E

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