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25 Apr 2013, 03:32
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
50% (02:09) correct
50%
(02:38)
wrong
based on 110
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In triangle JKL shown above, B divides side KL in such a way that KB : BL = 1 : 3 and C divides side JL in such a way that JC : CL = 1:2. What is the area of triangle BCL?
(1) The area of triangle JKL is 96
(2) The length of side KL is 24
It is very easy to solve this problem if you know the formula of triangle's area with sine. Any suggestions how to solve it quickly without trigonometry?
Source: GoGMAT
Attachment:
problem2.png [ 40.42 KiB | Viewed 4098 times ]
25 Apr 2013, 03:50
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In triangle JKL shown above, B divides side KL in such a way that KB : BL = 1 : 3 and C divides side JL in such a way that JC : CL = 1:2. What is the area of triangle BCL?
(1) The area of triangle JKL is 96.
Notice that since KB:BL=1:3, then BL=3/4*KL and since JC:CL=1:2, then CL=2/3*JL.
\(area_{JKL}=\frac{1}{2}*(height)*KL=96\).
\(area_{JBL}=\frac{1}{2}*(height)*BL=\frac{1}{2}*(height)*(\frac{3}{4}*KL)=\frac{3}{4}*(\frac{1}{2}*(height)*KL)=\frac{3}{4}*area_{JKL}=72\).
Similarly, \(area_{BCL}=\frac{2}{3}*area_{JBL}=48\). Sufficient.
(2) The length of side KL is 24. Clearly insufficient.
Answer: A.
Hope it's clear.
General Discussion
25 Apr 2013, 04:02
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Very clear! Thanks!
Anyway with trigonometry the solution is much faster:)
Anyway with trigonometry the solution is much faster:)
