CPM Homework Help : CCG Problem 9-81

A quadrilateral O, D, I, F where side O, D and side F, I are parallel. Side O, D is 20 meters. Side F, I is 14 meters. Side O, F is 18 meters. Angle F, is 120 degrees. The floor plan of Marina’s local drugstore is shown at right. While shopping one day, Marina tied her dog, Mutt, to the building at point $F$ . If Mutt’s leash is $4$ meters long, what is the area that Mutt can roam? Draw a diagram and show all work.

Step 1:

$\text{Area of the whole circle}=\pi r^2$ . Since $r=4$ , the $\text{area}=16\pi$ .

Added to quadrilateral, side, F I, is extended away from vertex, I, & part of a circle that starts at side, O F, ends at side, F I, with vertex F as its center, extended segment labeled, r = 4.

Step 2:

The area Mutt can roam = (fraction of circle Mutt can roam)(area of entire circle)

$\frac{360^{\circ}-120^{\circ}}{360^{\circ}}=\frac{240^{\circ}}{360^{\circ}}(16\pi)$

Answer:

Area Mutt can roam $\approx33.51\text{ m}^2$