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Lina20 [59]
3 years ago
15

What is the area of a sector with a central angle of 2π/9 radians and a diameter of 20.6 mm? use 3.14 for π and round your answe

r to the nearest hundredth. enter your answer as a decimal in the box.
this is just to help people:)

Mathematics
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

37.03 sq. mm.

Step-by-step explanation:

A sector is "part" of a circle. The formula for area of a sector (in radians) is:

Area of sector = \frac{1}{2}r^2 \theta

Where

r is the radius (half of diameter)

\theta  is the central angle of the sector

In this problem, the diameter is given as 20.6, so radius would be:

Radius (r) = 20.6/2 = 10.3

The central angle is given as  \frac{2\pi}{9} radians

Now, we substitute and find the value for the area:

A=\frac{1}{2}r^2 \theta\\A=\frac{1}{2}(10.3)^2 (\frac{2\pi}{9})\\A=\frac{1}{2}(106.09)(\frac{2(3.14)}{9})\\A=53.045*0.698\\A=37.03

Thus,

Area of sector = 37.03 sq. mm.

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Step-by-step explanation:

7 0
2 years ago
A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la
SOVA2 [1]

Answer:

8\pi\text{ square cm}

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

S=2\pi rh

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

4^2 = (2r)^2 + h^2   ( see in the below diagram ),

16 = 4r^2 + h^2

16 - 4r^2 = h^2

\implies h=\sqrt{16-4r^2}

Thus, the surface area of the cylinder,

S=2\pi r(\sqrt{16-4r^2})

Differentiating with respect to r,

\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})

=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})

=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})

Again differentiating with respect to r,

\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})

For maximum or minimum,

\frac{dS}{dt}=0

2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0

-8r^2 + 16 = 0

8r^2 = 16

r^2 = 2

\implies r = \sqrt{2}

Since, for r = √2,

\frac{d^2S}{dt^2}=negative

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

S = 2\pi (\sqrt{2})(\sqrt{16-8})

=2\pi (\sqrt{2})(\sqrt{8})

=2\pi \sqrt{16}

=8\pi\text{ square cm}

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