Quote:
How do you calculate area of triangle above trapezoid?
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.