MATH SOLVE

4 months ago

Q:
# In the diagram above, B is the center of the larger circle. AB is the radius of the larger circle, and it is the diameter of the smaller circle. The shaded area is what fraction of the larger circle?

Accepted Solution

A:

let's keep in mind that, the large circle has a radius of AB, whilst the small circle has a radius of AB/2.

[tex]\bf \textit{area of the large circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=AB \end{cases}\implies A=(AB)^2\pi \\\\ -------------------------------\\\\ \textit{area of the small circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=\frac{AB}{2} \end{cases}\implies A=\left( \frac{AB}{2} \right)^2\pi\implies A=\cfrac{(AB)^2}{2^2}\pi \\\\ -------------------------------[/tex]

[tex]\bf \textit{now subtracting the small area from the large one}\\\\ \textit{what's leftover is the \underline{shaded area}} \\\\\\ \stackrel{\textit{large's area}}{(AB)^2\pi }~~-~~\stackrel{\textit{small's area}}{\cfrac{(AB)^2\pi }{4}}~~\stackrel{LCD~of~4}{\implies }~~\cfrac{4(AB)^2\pi -(AB)^2\pi }{4} \\\\\\ \cfrac{3(AB)^2\pi }{4}\implies \cfrac{3}{4}(AB)^2\pi \impliedby \frac{3}{4}\textit{ of the large's area}[/tex]

[tex]\bf \textit{area of the large circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=AB \end{cases}\implies A=(AB)^2\pi \\\\ -------------------------------\\\\ \textit{area of the small circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=\frac{AB}{2} \end{cases}\implies A=\left( \frac{AB}{2} \right)^2\pi\implies A=\cfrac{(AB)^2}{2^2}\pi \\\\ -------------------------------[/tex]

[tex]\bf \textit{now subtracting the small area from the large one}\\\\ \textit{what's leftover is the \underline{shaded area}} \\\\\\ \stackrel{\textit{large's area}}{(AB)^2\pi }~~-~~\stackrel{\textit{small's area}}{\cfrac{(AB)^2\pi }{4}}~~\stackrel{LCD~of~4}{\implies }~~\cfrac{4(AB)^2\pi -(AB)^2\pi }{4} \\\\\\ \cfrac{3(AB)^2\pi }{4}\implies \cfrac{3}{4}(AB)^2\pi \impliedby \frac{3}{4}\textit{ of the large's area}[/tex]