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

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

Ans (D)

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

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.

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

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

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

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

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