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The area of ∠ JKL is 65 sq.units [#permalink]
21 Sep 2017, 11:42
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QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281The area of ∠ JKL is 65 sq.units
Quantity A 
Quantity B 
\(KL\) 
\(LM\) 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given. D is the answer Here is my reasoning let me know if it is correct  We know area of triangle = 1/2 base X altitude ie for triangle JKL; KL * 10/2 = 65 or KL = 13 Now consider triangle JKM By Pythagoras theorem  KM = √ (26 ^ 210 ^ 2) = 24 or LM = 2413 = 11 (since LM = KMKL ) so Quantity A> Quantity B.
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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
22 Sep 2017, 00:45
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Your method is correct! Thanks for pointing out fixed the errors.
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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
15 Dec 2017, 15:27
Hi sir, I had a doubt while reading your explanation regarding this question.My doubt is that ,by considering the triangle JKL how can you take base as 10?i ma kindly requesting you for help to clear my doubt.



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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
16 Dec 2017, 01:03
divyavivek wrote: Hi sir, I had a doubt while reading your explanation regarding this question.My doubt is that ,by considering the triangle JKL how can you take base as 10?i ma kindly requesting you for help to clear my doubt. The base is not 10 units the height is 10 units. The height of the triangle JKL and JKM are both the same 10 units.
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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
13 Mar 2018, 16:30
Answer: D as JMK is a right side triangle we have: 10^2 + (KL+LM)^2 = 26^2 So KL + LM = 24 Area of triangle JKL is 65. So height * KJ /2 = 65 > height = 130/26 = 5 base is KJ and height is the one from angle L to the line KJ On the other hand, MJL is also a right side triangle. We have 10^2 + ML^2 = jk^2 ML^2 = JK^2  10^2 with the given information we can't find out wanted values and the correct answer is D
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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
14 Mar 2018, 01:19
Area of triangle JKL is 65. So height * KJ /2 = 65 .... This is incorrect.Area of triangle JKL = height * KL /2 = 65
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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
01 Sep 2018, 09:21
Answer is A right?



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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
18 Jan 2019, 18:16
divyavivek wrote: Hi sir, I had a doubt while reading your explanation regarding this question.My doubt is that ,by considering the triangle JKL how can you take base as 10?i ma kindly requesting you for help to clear my doubt. Base KL Height 10



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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
18 Jan 2019, 18:17
IshanGre wrote: Answer is A right? I agree.



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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
30 Sep 2019, 15:31
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An easier way to do this is to compare the areas and heights, since they are related as the two smaller triangles comprise of the largest triangle, and the height of all the triangles are equal to 10.
The largest triangle's, triangle JKM's, missing side MK can be solved for using 10^2 + (MK)^2 = 26^2, where MK is the base. Solving for MK results in 24. Now, calculate the area of triangle JKM using (1/2)bh, where b = 24 and h = 10. The area is 120.
It is stated that triangle JKL's area is 65. Per the drawing of the picture, we see that the area of triangle JKL + area of triangle JLM = area of triangle JKM, since the two smaller triangles make up the largest triangle, this means 65 + area of triangle JLM = 120. Area of triangle JKM thus is 55.
Now, we know the area of both smaller triangles (55 and 65), and since the 1/2 and h is the same for each of these smaller triangles, the only difference is the base that results in the difference in the size of the area. Therefore, the base for triangle JKL, which has the larger area of 65, must be larger. The base for triangle JKL is KL.
Choice A is the answer.



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Re: QUANTITATIVE SECTION 7 Ques 4 of 20 ID: Q0281 GGREClub [#permalink]
15 Oct 2019, 12:51
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sandy wrote: Your method is correct!
Thanks for pointing out fixed the errors. not sure why you're supporting pranab01's answer D. The correct answer is A..oddly confusing solution



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Re: The area of ∠ JKL is 65 sq.units [#permalink]
15 Oct 2019, 12:55
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Sandy is no longer part of the team, unfortunately I would say The answer is A. Ask if you do need further explanations Regards
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Re: The area of ∠ JKL is 65 sq.units [#permalink]
15 Oct 2019, 12:56
Carcass wrote: Sandy is no longer part of the team, unfortunately I would say The answer is A. Ask if you do need further explanations Regards thanks for clarifying!



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The area of JKL is 65 square meters [#permalink]
08 Apr 2020, 02:00
The area of \(JKL\) is 65 square meters
Quantity A 
Quantity B 
The length of side \(KL\) 
The length of side \(LM\) 
A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given. Attachment:
#greprepclub The area of JKL is 65 square meters.jpg [ 20.34 KiB  Viewed 1397 times ]
Last edited by Carcass on 09 Apr 2020, 01:01, edited 2 times in total.
Updated



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Re: Geometry  Triangles [#permalink]
08 Apr 2020, 15:07
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First, we have to find the length of side KM(KL+LM):
\(26^2 = KM^2 + 10^2\)
KM = 24
Now, we can find the area of the outer \(\triangle\)JKM:
\(\frac{1}{2}\)(10)(24)= 120
Then, subtract the area of the \(\triangle\)JKM from the given area of the \(\triangle\)JKL to the get the area of the \(\triangle\)JLM
12065= 55
The area of \(\triangle\)JLM= \(\frac{1}{2}\)(10)(LM) = 55
LM = 11
Now, we can find the length of KL:
KL= KMLM = 2411
KL = 13
Since KL > LM, the correct answer is A
Last edited by Faisal99 on 09 Apr 2020, 01:27, edited 3 times in total.



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Re: The area of JKL is 65 square meters [#permalink]
09 Apr 2020, 01:03
Please follow the rules posting on the board https://greprepclub.com/forum/rulesfor ... 1083.htmlAnd how to format a question the easy way https://greprepclub.com/forum/howtopo ... 12752.htmlAs I did above. Thank you Regards
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Re: The area of JKL is 65 square meters [#permalink]
02 Aug 2020, 23:11
Its easy to solve once you identify that JM is length and not width. The question is, how to assume or consider JM as length and not width?
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Re: The area of JKL is 65 square meters
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