Question

William ordered a 6 inch uncut pizza. Gina and Rina cut out slices of the pizza as shown in the figure:

What is the area of the pizza cut out by Gina, Rina and William?

• A. Area left for William $=(21\pi +18)inc{h}^2$
  Area cut out by Gina $=6\pi inc{h}^2$
  Area cut out by Rina $=(9\pi -18)inc{h}^2$
• B. Area left for William $=(21\pi +18)inc{h}^2$
  Area cut out by Gina $=6\pi -18inc{h}^2$
  Area cut out by Rina $=(9\pi -18)inc{h}^2$
• C. Area left for William $=(21\pi +18)inc{h}^2$
  Area cut out by Gina $=6\pi inc{h}^2$
  Area cut out by Rina $=(9\pi +18)inc{h}^2$
• D. Area left for William $=(21\pi -18)inc{h}^2$
  Area cut out by Gina $=6\pi inc{h}^2$
  Area cut out by Rina $=(9\pi -18)inc{h}^2$

Answer 1

The correct option is A Area left for William $=(21\pi +18)inc{h}^2$

Area cut out by Gina $=6\pi inc{h}^2$
Area cut out by Rina $=(9\pi -18)inc{h}^2$
Gina:
Gina cut out the pizza slice which contains a sector of ${60}^{\circ }$ angle.
Area of the sector $=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times r^2$

Area of the sector $=\frac{1}{6}\times \pi \times 6^2$
Area of the sector = Gina's sector $=6\pi inc{h}^2$

Rina:
Rina cut out the slice which contains a segment of a circle.
Area of the segment = Area of the sector - Area of the triangle
Area of the sector $=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi r^2$
Area of the sector $=\frac{1}{4}\times \pi \times (6{)}^2$
Area of the sector $=9\pi inc{h}^2$
Area of the triangle $=\frac{1}{2}\times b\times h$
Here, b = 6 inches and h = 6 inches
Area of the triangle$=\frac{1}{2}\times 6\times 6$
Area of the triangle $=18inc{h}^2$

Area of the segment $=9\pi -18inc{h}^2$
Rina's segment $=9\pi -18inc{h}^2$

William:
Area left for William = Area of the circle - area cut out by Gina - Area cut out by Rina
Area left for William $=\pi \left(6^2\right)-\left(6\pi \right)-(9\pi -18)$
Area left for William $=21\pi +18inc{h}^2$