Draw a line between point P to line R. We get a triangle and a parallelogram.
Triangle has a base of a/2 and height of a. Area of triangle is (a^2)/4
Parallelogram has a base of a and a height of a/2. Area of parallelogram is (a^2)/2
combine triangle and parallelogram we get (3a^2)/4

just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

How to get the area of the above triangle though? I don't know OP and therefore don't know the height.

To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

It's not square!

just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole rectangle) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

We can solve this by using parallel lines.

How do you calculate area of triangle above trapezoid?

After getting the length of OP (a/2), we know that the full length of the Y axis is a from (a,a) coordinate so the remaining length is a/2. This is one perpendicular side of the upper triangle. The other perpendicular side is a, from the (a,a) coordinate. So the area = a/2 * (a) = a^2/2.

All trapezoids have one pair of parallel sides, so we can deduce that the slope of the line from $O$ to $R$ has to have the same slope from $P$ to $Q$.

Since $O$ is the origin, the slope of the line $OR$ is:

$\frac{a-0}{2a-0}$

$\frac{a}{2a}$

$\frac{1}{2}$

This means that the slope of the line $PQ$ has a slope of $\frac{1}{2}$.

To find the area of the above triangle, we would need the base and its height. Looking at the figure, we know the height (the distance from the y-axis to $Q$) is $a$. Now we need to find the base.

We can use the slope formula to find it, since we know that the slope from the $P$ to point $Q$ has to be $\frac{1}{2}$.

Since the point $P$ is on the y axis, let $P$ = (0,$y$).

Using the slope formula we get:

$\frac{y-a}{0-a} = \frac{1}{2}$

$2y - 2a = -a$

$2y = a$

$y = \frac{a}{2}$

So Point $P$ = (0,$\frac{a}{2}$)

That must mean that the base of the triangle is $\frac{a}{2}$.

So now we solve for the area of the above triangle:

$\frac{1}{2}* a * \frac{a}{2}$ =

$\frac{a^{2}}{4}$

And there's the area of the above triangle.

Basically
Slope is = Rise/Run. For parallel lines it is same.

For line OR --> for the run 2a rise is a.
For line QP --> for a run of a i know the rise should be a/2 (half of run) but I have to reach to a rise of "a" therefore I must start from a/2 which is the point P.

Cheers

Good question ! fifan for suggesting a clever solution !!

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