# Finding area of the triangle with vertices on a quarter circle

How to find area of Blue shaded region if it is given that AQ=2 and BP=4?

• Is $ARB$ assumed to be a quarter circle?
– kccu
Commented Dec 25, 2019 at 19:35
• What have you tried? Commented Dec 25, 2019 at 19:35
• @kccu Though is not given in the question that it is a quarter circle but YOu can assume it to be Commented Dec 25, 2019 at 19:36
• @B.Núñez I tried solving it using similarity. There are 3 triangles in the figure and all of them are similar Commented Dec 25, 2019 at 19:37

Let $$r$$ be the radius of the circle.

We can apply the Pythagorean theorem to triangle $$OQR$$. Leg $$OQ$$ has length $$r-2$$, leg $$QR$$ has length $$r-4$$, and hypotenuse $$OR$$ has length $$r$$, so: $$(r-2)^2+(r-4)^2=r^2$$ from which we obtain $$r=2$$ or $$r=10$$. We can rule out $$r=2$$ based on the diagram, hence $$r=10$$.

Now assuming $$O=(0,0)$$, you can find coordinates for all of the points and calculate the area of the shaded triangle.

As @kccu showed, it is easy to find that $$r=10$$.

Next, the lengths of the triangle's sides are:

$$a = \sqrt{10^2 + 10^2} = \sqrt{200}$$

$$b = \sqrt{4^2 + (10-2)^2} = \sqrt{80}$$

$$c = \sqrt{2^2 + (10-4)^2} = \sqrt{40}$$

To find the triangle's area, use Heron's formula for a SSS (side-side-side) triangle:

$$A = \sqrt{s (s-a)(s-b)(s-c)}$$

where $$s = \frac{a + b + c}{2}$$ to find $$A = 20$$.