How to find area of Blue shaded region if it is given that AQ=2 and BP=4?
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$\begingroup$ Is $ARB$ assumed to be a quarter circle? $\endgroup$– kccuCommented Dec 25, 2019 at 19:35
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$\begingroup$ What have you tried? $\endgroup$– B. NúñezCommented Dec 25, 2019 at 19:35
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$\begingroup$ @kccu Though is not given in the question that it is a quarter circle but YOu can assume it to be $\endgroup$– Kshitij SinghCommented Dec 25, 2019 at 19:36
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$\begingroup$ @B.Núñez I tried solving it using similarity. There are 3 triangles in the figure and all of them are similar $\endgroup$– Kshitij SinghCommented Dec 25, 2019 at 19:37
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2 Answers
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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.
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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 $A = 20$.
answered Dec 25, 2019 at 20:04