Answer:
8)2.86 cm^2
9) 100.48 m^2
10) 23.08 in^2
Step-by-step explanation:
8) In the figure , there's a right angled triangle in which a circle of radius 1 cm is inscribed in that triangle.
Area of the right angled triangle =
[tex] \frac{1}{2} \times 4 \times 3 = 6 \: {cm}^{2} [/tex]
Area of the circle =
[tex]\pi {r}^{2} = \pi {(1)}^{2} = \frac{22}{7} = 3.14 \: {cm}^{2}
[/tex]
Area of the shaded portion ( although itz not shaded..... i mean remaining portion ) = 6 - 3.14 =2.86cm^2
9) In this figure there's a circle in which more 2 circles are inscribed in such a way that the center of the bigger circle is in the circumferences of both the circles.
Area of the bigger circle =
[tex]\frac{\pi {d}^{2} }{4} = \frac{\pi \times 16 \times 16}{4} = 64\pi \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 200.96 {m}^{2} [/tex]
The small circles have equal diameter ( i.e. 8m) . So , Area of the small circles =
[tex]2( \frac{\pi {d}^{2} }{4} ) = 2( \frac{\pi \times {8}^{2} }{4} ) = 32\pi \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 100.48 \: {m}^{2} [/tex]
Area of the shaded region =
200.96 - 100.48 = 100.48 m^2
10) In the figure , the radius of the circle is 9 in.
So the area of a quarter =
[tex] \frac{\pi {r}^{2} }{4} = \frac{\pi \times {(9)}^{2} }{4} = \frac{81\pi}{4} = 63.58 \: {in}^{2} [/tex]
Also in that circle a right angled triangle is formed. Area of that right angles triangle =
[tex] \frac{1}{2} \times 9 \times 9 = 40.5 \: {in}^{2} [/tex]
So the area of the region shaded =
63.58 - 40.5 = 23.08 in^2
