Problem 8
The triangles are congruent, which means their corresponding pieces must also be congruent.
Focus on the second equation above. We can further say
angle B = angle E
E = B
12x+30 = 50
12x = 50-30
12x = 20
x = 20/12
x = 10/6
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Now move to the equation involving C and F to get
F = C
(y/2) - 10 = 40
y/2 = 40+10
y/2 = 50
y = 2*50
y = 100
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And finally,
D = A
sqrt(z) = 90
z = 90^2
z = 8100
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Problem 11
The radius of the circle is r = 5.
The area of the circle is A = pi*r^2 = pi*5^2 = 25pi
The diameter of the circle is 2*r = 2*5 = 10, which is the side length of the square because the circle is fit perfectly inside the square.
Since the square side length is 10, the area is 10^2 = 100
Subtract the two areas: large-small = square - circle = 100 - 25pi
We subtract the areas because we're basically kicking out the circle area when we don't shade it. Think of it like sawing out that portion of a wooden block perhaps.
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Problem 19
Part (a)
The point P is directly below the center and it is on the circle.
The location of point P is (18,0) because the x coordinates of both points are the same (so that P is directly below the center) and P resides on the x axis. All points on the x axis have y coordinate 0.
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Part (b)
The radius of the circle is r = 10 which is the distance from the center to point P.
The area of the circle is therefore A = pi*r^2 = pi*10^2 = 100pi
Answer: 100pi
