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# In triangle JKL shown above, B divides side KL in such a way

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Tutor
Joined: 20 Apr 2012
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In triangle JKL shown above, B divides side KL in such a way  [#permalink]

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25 Apr 2013, 03:32
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Difficulty:

85% (hard)

Question Stats:

51% (01:59) correct 49% (02:12) wrong based on 108 sessions

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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?

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Joined: 02 Sep 2009
Posts: 49275
Re: In triangle JKL shown above, B divides side KL in such a way  [#permalink]

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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.

Hope it's clear.
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Tutor
Joined: 20 Apr 2012
Posts: 99
Location: Ukraine
GMAT 1: 690 Q51 V31
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Re: In triangle JKL shown above, B divides side KL in such a way  [#permalink]

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25 Apr 2013, 04:02
Very clear! Thanks!

Anyway with trigonometry the solution is much faster:)
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Re: In triangle JKL shown above, B divides side KL in such a way  [#permalink]

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27 Mar 2014, 04:57
1
Bunuel wrote:

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.

Hope it's clear.

Hi,

I have a different method.

Let angle L = Ø
As we know area of triangle = (1/2) * Product of two sides * sin(included angle)
Let KB = x, BL = 3x, KL = 4x
Let JC = y, CL = 2y, JL = 3y

St1: The area of triangle JKL is 96.

area(JKL) = (1/2)* KL * JC * sinØ =
(1/2)*4x*3y*sinØ = 96 --- {1}
area(BCL) = (1/2)*BL*CL*sinØ =
(1/2)*3x*2y*sinØ = area(BCL) ---{2}

dividing eqn {1} by {2}, we get area(BCL) = 48

Sufficient!

ST2: length of side KL = 24. We need three factors to define a triangle (two sides and one angle).

Insufficient!

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Re: In triangle JKL shown above, B divides side KL in such a way  [#permalink]

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05 Jan 2018, 12:22
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Re: In triangle JKL shown above, B divides side KL in such a way &nbs [#permalink] 05 Jan 2018, 12:22
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