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In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 02:31
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In the figure above, two security lights, L1 and L2, are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chainlink fence. What is the total length s of fence, in feet, illuminated by the two lights? (A) 260 (B) 240 (C) 220 (D) 200 (E) 180 Are You Up For the Challenge: 700 Level QuestionsAttachment:
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Re: In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 09:50
Bunuel wrote: In the figure above, two security lights, L1 and L2, are located 100 feet apart. Each illuminates an area of radius 100 feet, and both are located 60 feet from a chainlink fence. What is the total length s of fence, in feet, illuminated by the two lights?
(A) 260 (B) 240 (C) 220 (D) 200 (E) 180
Join \(L_1\) and \(L_2\) Step I  Take \(L_1L_2EC\). It is a parallelogram so \(L_1L_2=EC=100\) Step II  Drop a perpendicular at D, so \(\triangle {L_2DE}\) is right angled triangle with two sides in ratio 60:100, so third side will be 80 as it is a 3:4:5 triangle, and BE=80+80=160. Step III  Find overlap or BC. \(BE=BC+CE.....160=BC+100...BC=60\) WE have to find AE.\(AE=AB+BC+CE=100+100+60=260\) OR AE=2*160overlap=2*16060=260 A
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In the figure above, two security lights, L1 and L2, are located 100
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Updated on: 11 Nov 2019, 10:55
The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60  80  100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be more than 160 + 160  80 = 240. We can only choose A. Attachment:
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Re: In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 08:03
TestPrepUnlimited wrote: The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60  80  100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160  80 = 240. We can only choose A. Attachment: 2cir.png TestPrepUnlimitedsee highlighted part ; option B is 240



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Re: In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 09:51
TestPrepUnlimited wrote: The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60  80  100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160  80 = 240. We can only choose A. Attachment: 2cir.png At least 240 means answer can be 240 or 260?? So A or B possible
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In the figure above, two security lights, L1 and L2, are located 100
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Updated on: 11 Nov 2019, 11:11
Archit3110 wrote: TestPrepUnlimited wrote: The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60  80  100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160  80 = 240. We can only choose A. Attachment: 2cir.png TestPrepUnlimitedsee highlighted part ; option B is 240 Archit3110Only possibility exists is A, since B would mean one circle passing through point of intersection of perpendicular from fence(chord) to center of other circle(L1 or L2). This would gravely deform the diagram. On the other side, if you are good at drawing circles to scale, you can observe that in the 60  80  100 triangle for L1 the perimeter of circle with center L2 bisects the side 80 in roughly a ratio of 1:3. Thus the common part of fence ~ \(\frac{3}{4} * 80 = 60\). Trying to quote 'Ron' Hence A.
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Originally posted by lnm87 on 11 Nov 2019, 10:20.
Last edited by lnm87 on 11 Nov 2019, 11:11, edited 1 time in total.



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In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 10:58
Thank you for pointing that out, I have fixed my wording. The intent was to avoid doing any other calculations by seeing the overlap is certainly less than 80, so the fence length > 240. Archit3110 wrote: TestPrepUnlimited wrote: The radius is 100 since their distance apart = radius = 100 ft. The distance of 60 ft represents the perpendicular length from the fence to any of the centers. So we can confirm each circle illuminates 80 * 2 = 160 ft of the fence using the 60  80  100 triangle. Finally, we need to subtract the overlap from the two fences. Note however the overlap cannot be more than 80, so the total fence length must be at least 160 + 160  80 = 240. We can only choose A. Attachment: 2cir.png TestPrepUnlimitedsee highlighted part ; option B is 240
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Re: In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 23:15
The length of the fence covered by the light is: S = 80+100+80=260 feet.
Answer is (A)
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Re: In the figure above, two security lights, L1 and L2, are located 100
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11 Nov 2019, 23:15






