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# In the figure shown, PQRS is a square, T is the midpoint of

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In the figure shown, PQRS is a square, T is the midpoint of [#permalink]  21 Nov 2019, 11:35
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72% (01:16) correct 27% (01:56) wrong based on 22 sessions
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#greprepclub In the figure shown.jpg [ 27.25 KiB | Viewed 662 times ]

In the figure shown, $$PQRS$$ is a square, $$T$$ is the midpoint of side $$PS$$, and $$U$$ is the midpoint of side $$QR$$. The area of the shaded region is what fraction of the area of square $$PQRS$$ ?

A. $$\frac{1}{6 }$$

B. $$\frac{1}{8}$$

C. $$\frac{1}{5}$$

D. $$\frac{1}{4}$$

E. $$\frac{1}{3}$$

Kudos for the right answer and solution.
[Reveal] Spoiler: OA

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Re: In the figure shown, PQRS is a square, T is the midpoint of [#permalink]  20 Apr 2020, 15:55
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PTUR is a parallelogram so the area is x/2*x = x^2/2
the shaded portion is half of this
so it is x^2/4

Area of square x^2

(x^2/4)/x^2

=1/4
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Re: In the figure shown, PQRS is a square, T is the midpoint of [#permalink]  28 Jul 2020, 06:26
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Carcass wrote:
Attachment:
#greprepclub In the figure shown.jpg

In the figure shown, $$PQRS$$ is a square, $$T$$ is the midpoint of side $$PS$$, and $$U$$ is the midpoint of side $$QR$$. The area of the shaded region is what fraction of the area of square $$PQRS$$ ?

A. $$\frac{1}{6 }$$

B. $$\frac{1}{8}$$

C. $$\frac{1}{5}$$

D. $$\frac{1}{4}$$

E. $$\frac{1}{3}$$

Kudos for the right answer and solution.

Let's say that each side of the square has length 2
Which means its area = (2)(2) = 4

Since T and U are midpoints, we get the following measurements:

Since URPT is a parallelogram, its area = (base)(height) = (1)(2) = 2

Since the blue line divides parallelogram URPT into 2 EQUAL pieces, the area of the shaded region = 1/2 of 2, which is 1

The area of the shaded region is what fraction of the area of square PQRS?
Fraction = 1/4

Cheers,
Brent
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Re: In the figure shown, PQRS is a square, T is the midpoint of   [#permalink] 28 Jul 2020, 06:26
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