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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