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nevsk [136]
3 years ago
11

How do you solve for the segment area in a circle?

Mathematics
1 answer:
Aleks04 [339]3 years ago
6 0
The area of the entire sector of DEF = 60 / 360 * PI * radius^2
sector area = 1 / 6 * 3.14159265... * 20^2
sector area = <span> <span> <span> 209.4395102393 </span> </span> </span>

segment area = sector area - triangle DEF Area
triangle DEF Area = (1/2) * 20 * sine 60 * 20
triangle DEF Area = (1/2) * 0.86603 * 400
triangle DEF Area = <span><span><span>(1/2) * 346.412 </span> </span> </span>
triangle DEF Area = <span> <span> <span> 173.206 </span> </span> </span>

segment area = <span> <span> 209.4395102393 </span> -173.206
</span>
segment area = <span> <span> <span> 36.2335102393 </span> </span> </span>
segment area = 36.23 m

Source:
http://www.1728.org/circsect.htm



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The students in Mr. Wilson's Physics class are making golf ball catapults. The
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Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

Part b) How high above the ground did the ball fly?

Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

To determine the y-coordinate of the vertex substitute the value of x in the quadratic equation and solve for y

y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

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The ball flew above the ground about 8.3 feet

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we know that

A  reasonable domain is the distance between the two x-intercepts

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0 \leq x \leq 48.6\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 48.6 feet

A  reasonable range is all real numbers greater than or equal to zero and less than or equal to the y-coordinate of the vertex

so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

All real numbers greater than or equal to 0 feet and less than or equal to 8.3 feet

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