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

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In the figure shown, PQRS is a square, T is the midpoint of side PS  [#permalink]

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New post 22 Nov 2019, 03:19
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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. \(\frac{1}{3}\)


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Re: In the figure shown, PQRS is a square, T is the midpoint of side PS  [#permalink]

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New post 22 Nov 2019, 04:19
giving a try

let side of square = 4
area = 16
∆QRS area = 1/2 * 4*4 ; 8
line UP =2√5 so until point of intersection of line QU be point A i.e QA=QU=UA=2 ; equilateral ∆ similarly for side TS = TB=BS = 2
area of ∆ QAP = 1/2 * 2 * √5 = √5
area of ∆ BTS = √3/4 * 4 = √3
Area of shaded region = 16- ( √5+√3+ 8 ) ; 4
fraction area = 4/16 ;
1/4
IMO D


Bunuel wrote:
Image
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. \(\frac{1}{3}\)


Are You Up For the Challenge: 700 Level Questions

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Re: In the figure shown, PQRS is a square, T is the midpoint of side PS  [#permalink]

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New post 22 Nov 2019, 04:23
Each right-angle triangle has equal area=1/2*s/2*s= 1/4 *s^2

Area of grey+orange shaded area= 1/2 of area of square

Diagonal divides it into 2 equal trapezium; Hence, area of each trapezium is 1/4 of area of square.



Bunuel wrote:
Image
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. \(\frac{1}{3}\)


Are You Up For the Challenge: 700 Level Questions

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Re: In the figure shown, PQRS is a square, T is the midpoint of side PS  [#permalink]

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New post 22 Nov 2019, 08:34
Bunuel wrote:
Image
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. \(\frac{1}{3}\)


\(area:TSR = PQU = (x/2)x/2=x^2/4\)
\(area:Square-(TSR+PQU)=PTRU…x^2-2x^2/4=2x^2/4=x^2/2\)
\(area:PTRU/2=shaded…(x^2/2)/2=shaded…shaded=x^2/4\)
\(area:shaded/square=(x^2/4)/x^2=1/4\)

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Re: In the figure shown, PQRS is a square, T is the midpoint of side PS  [#permalink]

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New post 22 Nov 2019, 10:20
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 side PS   [#permalink] 22 Nov 2019, 10:20
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