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6) diagram: circle with a triangle (base as diameter) and radius 4 ft. …

Question

  1. diagram: circle with a triangle (base as diameter) and radius 4 ft. 7) diagram: rectangle with length 14 yd, and a semicircular (radius 5 yd) cutout on the right side, height of rectangle is 10 yd (diameter of semicircle)

Explanation:

Response
Problem 6:

Step1: Find area of circle

Radius \( r = 4 \) ft, area of circle \( A_{circle} = \pi r^2 = \pi(4)^2 = 16\pi \) sq ft.

Step2: Find area of triangle

Base of triangle is diameter \( = 8 \) ft, height \( = 4 \) ft. Area \( A_{triangle} = \frac{1}{2} \times 8 \times 4 = 16 \) sq ft.

Step3: Find shaded area

Shaded area = Area of circle - Area of triangle \( = 16\pi - 16 \approx 16(3.14) - 16 = 50.24 - 16 = 34.24 \) sq ft. (Wait, maybe misinterpretation. Wait, the figure: maybe the shaded is circle minus triangle? Or maybe the lower half is semicircle? Wait, original figure: circle with a triangle (base diameter, height radius). Wait, maybe the shaded area is the circle minus the triangle? Wait, no, maybe the lower half is semicircle and upper two segments. Wait, maybe re - examine. Wait, radius is 4, so diameter 8. Triangle area: \( \frac{1}{2} \times 8 \times 4 = 16 \). Circle area: \( \pi\times4^2 = 16\pi\approx50.24 \). If shaded is circle minus triangle, then \( 50.24 - 16 = 34.24 \) sq ft. But maybe the lower part is a semicircle? Wait, no, the radius is 4, so diameter 8. Wait, maybe the figure has a circle with a triangle (isosceles, base diameter, height radius) and the shaded area is the circle minus the triangle. So:

Step1: Find area of rectangle

Length \( l = 14 \) yd, width \( w = 10 \) yd (since the radius of the semicircle is 5 yd, so diameter is 10 yd, which is the width of the rectangle). Area \( A_{rectangle}=14\times10 = 140 \) sq yd.

Step2: Find area of semicircle

Radius \( r = 5 \) yd, area of semicircle \( A_{semicircle}=\frac{1}{2}\pi r^2=\frac{1}{2}\pi(5)^2=\frac{25\pi}{2}\approx39.25 \) sq yd.

Step3: Find shaded area

Shaded area = Area of rectangle - Area of semicircle \( = 140-\frac{25\pi}{2}\approx140 - 39.25 = 100.75 \) sq yd.

Answer:

\( 16\pi - 16\approx34.24 \) square feet (or exact form \( 16(\pi - 1) \))

Problem 7: