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Sub 505 (Easy)|   Geometry|      
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Bunuel
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 01:15
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

A. \(128 - 4\pi\)
B. \(128 - 8\pi\)
C. \(128 - 12\pi\)
D. \(128 - 16\pi\)
E. \(128 - 32\pi\)


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E
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 01:23
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Bunuel

Are you sure about all choices? It might be some rework on choices maybe?

Because diameter of each would be 8 and radius is 4. According to that it might be 128-32pi

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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 02:05
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Bunuel are you sure, Options are correct.

Getting answer 128-32π which is not in options.

As rectangle is 16x8. So the radius of the circle will be 4.
The area of one circle will be πr^2 i.e. 16π
for 2 circles, area will be 32π

Remaining area will be = 128 - 32π

IMO-E
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 04:17
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Area of the rectangle = 8*16 = 128

Radius of both the circles = 4, area = 2*pi*r^2 = 32*pi

So, leftover = 128−32π E
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 04:22
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the remaining area after the paper is cut will be = area of the rectangle - 2 circles area

area of the rectangle will be = 16* 8 = 128

since each circle have a diameter of 8, hence their radius will be 4 ,

hence area of both the circles = 2 * pi * (4)^2 = 32 PI

hence the area after the 2 circles have been cut is 128-32 PI

hence Option E




Bunuel

Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

A. \(128 - 4\pi\)
B. \(128 - 8\pi\)
C. \(128 - 12\pi\)
D. \(128 - 16\pi\)
E. \(128 - 32\pi\)


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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 16:30
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

Given:
Rectangle size - 16-by-18
Circles are congruent and adjacent circles

We know that the area of rectangle is l*w = 16*8 = 128

We also know that the sum of diameter of both the circles = length of rectangle
2d = 16
d = 8

Diameter of each circle is 8, so the radius is 4

Area of each circle is \(\pi\)r^2 = 16\(\pi\)

Sum of area of both the circles = 32\(\pi\)

Area of the paper remaining after the circles have been cut out = Area of rectangle - Sum of area of both the circles

Area of paper remaining = 128-32\(\pi\)

Ans: E
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 05 Nov 2020, 18:45
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The radius of circle (based on the fig) = 4

Area of rectangle = 16*8 = 128
Area of 2 circles = 2* pi * 4^2 = 32*pi

Remaining area = 128 - 32 pi

IMO E
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. Th 06 Nov 2020, 00:04
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Two congruent, adjacent circles are cut out of a 16-by-8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?

Area of rectangle = 16 * 8 = 128

Area of one circle = πr^2
here r=4
so, area = 16π

Area of two circles = 32π

Area of the paper remaining after the circles have been cut out = 128 - 32π

Option E
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