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11 Nov 2019, 02:31
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A
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67% (02:23) correct
33%
(02:48)
wrong
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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 chain-link 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 Questions
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2cir.png [ 3.14 KiB | Viewed 5260 times ]
11 Nov 2019, 09:50
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Bunuel
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*160-overlap=2*160-60=260
A
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2cir.png [ 8.8 KiB | Viewed 4946 times ]
General Discussion
Updated on: 11 Nov 2019, 10:55
Originally posted by TestPrepUnlimited on 11 Nov 2019, 06:42.
Last edited by TestPrepUnlimited on 11 Nov 2019, 10:55, edited 1 time in total.
Last edited by TestPrepUnlimited on 11 Nov 2019, 10:55, edited 1 time in total.
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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.
double circles.png [ 8.27 KiB | Viewed 4972 times ]
Attachment:
double circles.png [ 8.27 KiB | Viewed 4972 times ]
