It will be nice if we find the location of A in the figure.

Hahahha right

Edited the original post by sandy.

Thank you to pointed out

Okay, this is an isosceles triangle, and the largest area will be when the third side is also 5, and that area will be 1/2*5*5*sqrt3, which is greater than ten. But picture the triangle squashed down so that the height is .001. The base is less than 10, so area will be less than 1/2*.001*10. So answer D.

Fact: given two sides of a triangle, the area is maximized when the triangle is a right triangle with the given two sides as legs.


Therefore the max area would be

(1/2)(5)(5)=25/2= 12.5

12.5>10
Choice A


Given the two sides the area of the triangle can be made arbitrarily small by increasing the angle of OQP more and more.
The height will be approaching zero, therefore the area becomes smaller and smaller with 0 as a lower bound.

Therefore We can make the triangle have an area smaller than 10
Choice B

Since we cases for Choice A and Choice B

Final Answer: D

Let \(P\) be \((x,y)\).
We can assume \(-5 \le y \le 5\).
Then we have \(0 \le |y| \le 5\).
The area of the triangle \(OPQ\) is \(\frac{1}{2} \cdot 5|y|\) since its base is \(5\) and its height is \(|y|\).
We have \(0 \le \frac{1}{2} \cdot 5|y| \le \frac{1}{2} \cdot 5 \cdot 5 = \frac{25}{2}\), since \(0 \le |y| \le 5\).

If \(y = 5\), then its area is \(\frac{25}{2}\), which is greater than \(10\).
If \(y = 1\), then its area is \(\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}\), which is less than \(10\).

Therefore, the right answer is D.

We know that \(PQ = 5\) and \(OQ = 5\), so those sides are fixed. But let's pretend we can grab point \(P\) with our fingertips and rotate it around while conserving the lengths of 5. This action would change the length of \(PO\).

We can certainly drag \(P\) so that \(PQ\) is perpendicular to \(OQ\), making it a right triangle. That would maximize the area at:

\(\frac{5*5}{2} = 12.5\)

However, we can drag point \(P\) to the right of point \(Q\) towards the x-axis and almost flatten it to a line, or go the opposite way towards point \(O\). In that case we'd be shrinking the area towards zero.

So the range of our area is:

\(0 < Area <= 12.5\)

The answer is therefore D.

As an aside, playing with that point \(P\) you'll get to see why rotating it to become perpendicular maximizes the area. It's the greatest height any triangle can have. Remembering that property and being able to visually see the changing area as you rotate \(P\) is valuable come test day.

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