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

Question

https://gmatclub.com/forum/download/file.php?id=51769 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 1.jpg

BrentGMATPrepNow · Accepted Answer

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, and since sides PQ and RS are  parallel , we get the following measurements and equal angles: 
 https://i.imgur.com/jW9vudt.png

Let's also add areas A, B, C, D, E, and F:
 https://i.imgur.com/HdZqMmd.png

Since diagonal QS divides the square into two equal areas, we can conclude that  A + B + C = D + E + F 
We can see that since the areas of regions A and F have the exact same shape and measurements, their areas must be equal. 
Applying similar logic we can conclude that the areas of regions D and C are equal

So it must also be true that the areas of regions E and B must be equal.
Since URPT is a parallelogram, its area = (base)(height) = (  1  )(  2  ) =  2

So the area of region E = the area of region B =   1

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

Answer: D

Cheers,
Brent

Archit3110 · Answer

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

nick1816 · Answer

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.

exc4libur · Answer

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

Ans (D)

ManjariMishra · Answer

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

fmcgee1 · Answer

Why are you assuming that QA=QU=UA=2 or that QUP is equilateral?

Fdambro294 · Answer

First, label in each full side as 4:

Which makes QU = UR = 2 = PT = TS

2 triangles can be shown to be congruent making the calculation much easier.

Mark the intersection point of UP and QS as ——-point X

The mark the intersection point of RT and QS ———point Y

Triangle QXU is congruent to Triangle SYT

The diagonal QS creates a 45 degree angle near the vertex of the square.

Further, since UP and RT are drawn from the midpoints of 2 parallel sides to the opposite vertex ——-> UP and RT are 2 parallel lines cut by the transversal QS

In triangle SYT, call angle at vertex Y ——-> A degrees

The vertically opposite angle will be equal.

This vertically opposite angle is a corresponding angle to the angle at vertex X in triangle QXU

Thus, by the Angle - Angle Postulate the 2 triangles QXU and SYT are similar. They have an equal corresponding side of 2, which is half of the side of the square (QU = 2 = TS).

Therefore, triangles QXU and SYT are also congruent.

To find the shaded area, from the entire area of the square, we can remove:

Triangle QPU

And

triangle QRS from the square.

We will be counting Triangle QXU TWICE, but since we are not including the congruent triangle SYT in the removal, we will have effectively removed all the outer non-shaded area from the square.

(In other words, removing triangle QXU twice is the same thing as removing it once along with triangle SYT...since they are congruent)

(4 * 4) - (1/2 * 2 * 4) - (1/2 * 4 * 4) =

16 - 4 - 8 =

4

And divide this by the entire area of the square of 16

4/16 = 1/4

Posted from my mobile device

andreagonzalez2k · Answer

Without any calculation:
Triangles QPU and RST are equal and if you joint them (in your mind move triangle RST to the left) clearly their combined area is a half of the square area. So area of paralelogram PURT must be the other half. And the shaded region is a half of the area of PURT, so a quarter of the square area.

chirag95 · Answer

How can we conclude that the blue line divides the parallelogram equally?

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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

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