The question is as follows:
Find the area of the shaded region in the terms of π. (No decimals)
To figure this problem out, would I figure out what the area of a whole circle is and then somehow figure out what is missing from the circle and subtract so I only have the shaded region?
Calling the marked segments we can call $x$, it looks like the radius of the circle will be $2x$, the problem is much harder if it is not the radius. Since there is a right angle it sections off a quarter of the circle so the area of the shaded region will be $\pi(2x)^2-\frac{\pi(2x)^2}{4}$. Solving for $x$ comes from the Pythagorean theorem
Let the radius of the circle is $R$
thus you can see that $$(R/2)^2+(R/2)^2=14^2 $$ $$\frac{R^2}{2}=14^2 $$
Now you want to find the area of the shaded region which is nothing but $\theta \frac{R^2}{2}$
where $\theta=\frac{3\pi}{2}$ Thus the area of the shaded region is $$\frac{3\pi} {2}\cdot14^2 \hspace{5pt}cm^2$$
Hope it helps.
We can see that there is an isosceles right triangle with sides 14,r/2 and r/2 where r is the radius of the circle. Apply Pythagoras Theorem, 2•(r/2)²=14² Thus r²=2•14². Now the shaded area is 3/4th of the complete circle so the required area is: 3/4(pi)r²=3/4*pi*2*14²=294*pi.
