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Take a look at figure given below. Give it a try and when you are ready then watch the solution.
This is a pretty hard puzzle. But you can solve it by basic Formulas and extraordinary thinking capacity.
For solving this puzzle first I will draw some basic construction lines.
1) Join F and P. As you can observe FP will be radius of circle.
2) Construct CM perpendicular yo FP
3) Construct DN perpendicular yo FP
We will solve only by these constructions.
Now here is simple geometry part.
You will get a diagram as follows:
⇒ FC = FD = FP = r
Now we will focus on FP
1) FP = r
MP = 3
⇒ FM = r-3
1) FP = r
NP = 6
⇒ FN = r-6
Now observe angle CFD
CFD = 90
CFP = θ
PFD = 90-θ
In triangle CFM
\[cosθ=\frac{r-3}{r}\]In triangle DFN
\[cos(90-θ)=\frac{r-6}{r}\]
\[\Rightarrow sinθ=\frac{r-6}{r}\]
We know that:
\[sin^2θ+cos^2θ=1\]
\[\Rightarrow\left(\frac{r-3}{r}\right)^2+\left(\frac{r-6}{r}\right)^2=1\]
\[\Rightarrow\left(r-3\right)^2+\left(r-6\right)^2=r^2\]
\[\Rightarrow r^2-6r+9+r^2-12r+36=r^2\]
\[\Rightarrow r^2-18r+45=0\]
\[\Rightarrow (r-3)(r-15)=0\]
\[(r=3)or(r=15)\]
\[But(r\neq3)\]
\[\Rightarrow(r=15)\]
Here if r=3 then given figure can't be possible.
Now we can calculate area of Quarter Circle with r=15
\[Area=\frac{\pi r^2}{4}\]
\[Area=\frac{\pi 15^2}{4}\]
\[Area=\frac{225\pi}{4}\]
In this way we got:
Area of Quarter Circle = 225π/4
