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What is the area of a triangle created by the intersections
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Updated on: 27 Aug 2013, 01:53
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What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20 A. 42 B. 54 C. 66 D. 72 E. 96
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Originally posted by guerrero25 on 26 Aug 2013, 22:37.
Last edited by guerrero25 on 27 Aug 2013, 01:53, edited 1 time in total.




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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 09:34
guerrero25 wrote: What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20
A. 42 B. 54 C. 66 D. 72 E. 96 Diagram and solution
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Asif vai.....




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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 01:07
Using the equations provided, we could find the three vertices of the triangle as: (4,5), (20,5) and (4,17) ( vertices (20,5) and (4,17) could be deducted by substituating x=4 and Y =5 respectively into the equation) Now, the Area of the Triangle will be = 1/2 * ( 204) *(175) =96 Hence, the answer is /SW



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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 10:12
Hello Asif, Your visual representation helps big time in this instance, but your picture is very unproportional and definitely not "drawn to scale." For example, look at the difference shown between x= 0, x = 4, and x =20. That's why it's really hard to "visualize" the triangle we are trying to solve for in your diagram. The triangle should be much bigger (not small like you have pictured). The math is obviously the same, but I'm just nit picking on your drawing, sorry! lol The one thing that I don't understand is where you got 3x + 4y = 80 from. Can you please explain? If you substitute the points they provided in the problem (x=4, y=5), you would get 32=80, which is obviously incorrect so the equation for this line does not make sense. Thanks, ~ Im2bz2p345



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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 12:43
Im2bz2p345 wrote: Hello Asif, Your visual representation helps big time in this instance, but your picture is very unproportional and definitely not "drawn to scale." For example, look at the difference shown between x= 0, x = 4, and x =20. That's why it's really hard to "visualize" the triangle we are trying to solve for in your diagram. The triangle should be much bigger (not small like you have pictured). The math is obviously the same, but I'm just nit picking on your drawing, sorry! lol The one thing that I don't understand is where you got 3x + 4y = 80 from. Can you please explain? If you substitute the points they provided in the problem (x=4, y=5), you would get 32=80, which is obviously incorrect so the equation for this line does not make sense. Thanks, ~ Im2bz2p345 1st of all you have to know about patterns of gmat questions. Not drawn to scale is the format gmat uses 95%. And here is no need to draw according to scale ,because gmat will not provide us graph paper. All we have to do is evaluate the the base and perpendicular. thats it. no one will bestow you more marks for your large drawings, you have to do fast thats it. The one thing that I don't understand is where you got 3x + 4y = 80 from. Can you please explain? If you substitute the points they provided in the problem (x=4, y=5), you would get 32=80, which is obviously incorrect so the equation for this line does not make sense.  for this: if three lines pass through one same point then they are unable to form a triangle. but you are insisting on that. x=4 and y=5 both values will never satisfies 3x+4y = 80, because all three lines didn't pass through one common point here. y= 3/4 x + 20 or, 4y = 3x + 80 or, 3x+4y = 80 (got it where it came from ??? ) You need to work on basic concepts. Go through some math foundation gmat books. Any question?
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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 13:42
Asifpirlo wrote: 1st of all you have to know about patterns of gmat questions. Not drawn to scale is the format gmat uses 95%. And here is no need to draw according to scale ,because gmat will not provide us graph paper. All we have to do is evaluate the the base and perpendicular. thats it. no one will bestow you more marks for your large drawings, you have to do fast thats it. The one thing that I don't understand is where you got 3x + 4y = 80 from. Can you please explain? If you substitute the points they provided in the problem (x=4, y=5), you would get 32=80, which is obviously incorrect so the equation for this line does not make sense.  for this: if three lines pass through one same point then they are unable to form a triangle. but you are insisting on that. x=4 and y=5 both values will never satisfies 3x+4y = 80, because all three lines didn't pass through one common point here. y= 3/4 x + 20 or, 4y = 3x + 80 or, 3x+4y = 80 (got it where it came from ??? ) You need to work on basic concepts. Go through some math foundation gmat books. Any question? Thank you for the response Asif! I agree with your point that drawing to scale is not important on the GMAT exam.. but we are not in an exam center right now Hence, I thought for explaining the problem, it might have been useful to show points more accurately  similar to how I expect the OG13 answers to be shown accurately in their book's explanations. This just my preference and opinion, obviously you disagree and think it's important to "go fast" even when explaining things to others. Although, like I said earlier: not drawing to scale doesn't change anything about the question or method to solving the problem. gmat will not provide us graph paperThey actually do give you graph "paper!" Here is a picture of the scratch pad used for the actual exam with a graph drawn for you as well ( http://www.afewgoodminds.com/wpcontent ... 00x150.jpg). As you stated though, I'd rather draw fast and move on quickly if I know how to solve the question. Thanks for explaining your manipulation of the yintercept form equation provided in the original question. Not sure why you didn't label the line with how it is stated in the original problem, but the yintercept is supposed to 20 and it looks like the line is crossing more at y=10 (since it's about double the y=5 distance). This is why I asked. Thank you for taking the time to respond. All the best my friend! ~ Im2bz2p345



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Re: What is the area of a triangle created by the intersections
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27 Aug 2013, 17:34
Im2bz2p345 wrote: Asifpirlo wrote: 1st of all you have to know about patterns of gmat questions. Not drawn to scale is the format gmat uses 95%. And here is no need to draw according to scale ,because gmat will not provide us graph paper. All we have to do is evaluate the the base and perpendicular. thats it. no one will bestow you more marks for your large drawings, you have to do fast thats it. The one thing that I don't understand is where you got 3x + 4y = 80 from. Can you please explain? If you substitute the points they provided in the problem (x=4, y=5), you would get 32=80, which is obviously incorrect so the equation for this line does not make sense.  for this: if three lines pass through one same point then they are unable to form a triangle. but you are insisting on that. x=4 and y=5 both values will never satisfies 3x+4y = 80, because all three lines didn't pass through one common point here. y= 3/4 x + 20 or, 4y = 3x + 80 or, 3x+4y = 80 (got it where it came from ??? ) You need to work on basic concepts. Go through some math foundation gmat books. Any question? Thank you for the response Asif! I agree with your point that drawing to scale is not important on the GMAT exam.. but we are not in an exam center right now Hence, I thought for explaining the problem, it might have been useful to show points more accurately  similar to how I expect the OG13 answers to be shown accurately in their book's explanations. This just my preference and opinion, obviously you disagree and think it's important to "go fast" even when explaining things to others. Although, like I said earlier: not drawing to scale doesn't change anything about the question or method to solving the problem. gmat will not provide us graph paperThey actually do give you graph "paper!" Here is a picture of the scratch pad used for the actual exam with a graph drawn for you as well ( http://www.afewgoodminds.com/wpcontent ... 00x150.jpg). As you stated though, I'd rather draw fast and move on quickly if I know how to solve the question. Thanks for explaining your manipulation of the yintercept form equation provided in the original question. Not sure why you didn't label the line with how it is stated in the original problem, but the yintercept is supposed to 20 and it looks like the line is crossing more at y=10 (since it's about double the y=5 distance). This is why I asked. Thank you for taking the time to respond. All the best my friend! ~ Im2bz2p345 hmmmm right my friend..... this diagram is for concepts but not for practical uses... that's why i did so. You know one thing, the way we practice finally we have to perform just in that way....
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Re: What is the area of a triangle created by the intersections
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15 Jun 2017, 21:12
guerrero25 wrote: What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20
A. 42 B. 54 C. 66 D. 72 E. 96 You don't necessarily need a to draw a diagram in order to solve this question simply, x=4 means that there is a line in the a 4 quadrant plane in which all the x values are 4 a vertical line. The same holds true for y=5 there is a line in plane in which all y values are 5 a horizontal line. Therefore, this triangle is a 90 degree triangle the intersection of a perfectly horizontal and perfectly vertical line creates a 90 degree angle. Next, we just plug in values in order to find mutual points on our vertical line and diagonal line (3/4x +20) and our horizontal line and diagonal line again (3/4x +20) 5= 3/4 x +20 15 = 3/4x 60= 3x 20= x (but we aren't done here) Our two perfectly horizontal and vertical lines intersect at (4,5) and we've just calculated that there is a point (20 , 5) so the actual length our triangle side will be 204 = 16 We repeat the same step for the other side of the triangle y= 3 (4)/4 + 20 y= 12/4 + 20 y = 3 + 20 y = 17 (but again, we are not done just yet we've just calculated that there is a point our x line (4, 17) so we find the length of the other side, again, with reference to the intersection of lines x and y (4,5) 175 = 12 And because this is a right triangle we can simply apply the formula below Base 1 x Base 2/ 2 = Area 12 x 16/ 2 = Area 192/2= 96 And although it seems like the equation y = 3/4x +20 would be the slope it isn't actually because the square root of 192 is 14. Thus "E"



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Re: What is the area of a triangle created by the intersections
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31 Jul 2017, 06:13
Nunuboy1994 wrote: guerrero25 wrote: What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20
A. 42 B. 54 C. 66 D. 72 E. 96 You don't necessarily need a to draw a diagram in order to solve this question simply, x=4 means that there is a line in the a 4 quadrant plane in which all the x values are 4 a vertical line. The same holds true for y=5 there is a line in plane in which all y values are 5 a horizontal line. Therefore, this triangle is a 90 degree triangle the intersection of a perfectly horizontal and perfectly vertical line creates a 90 degree angle. Next, we just plug in values in order to find mutual points on our vertical line and diagonal line (3/4x +20) and our horizontal line and diagonal line again (3/4x +20) 5= 3/4 x +20 15 = 3/4x 60= 3x 20= x (but we aren't done here) Our two perfectly horizontal and vertical lines intersect at (4,5) and we've just calculated that there is a point (20 , 5) so the actual length our triangle side will be 204 = 16 We repeat the same step for the other side of the triangle y= 3 (4)/4 + 20 y= 12/4 + 20 y = 3 + 20 y = 17 (but again, we are not done just yet we've just calculated that there is a point our x line (4, 17) so we find the length of the other side, again, with reference to the intersection of lines x and y (4,5) 175 = 12 And because this is a right triangle we can simply apply the formula below Base 1 x Base 2/ 2 = Area 12 x 16/ 2 = Area 192/2= 96 And although it seems like the equation y = 3/4x +20 would be the slope it isn't actually because the square root of 192 is 14. Thus "E" Thank you for this lucid and step by step explanation. Was really helpful for someone yet gripping the basic



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What is the area of a triangle created by the intersections
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19 Aug 2017, 02:57
guerrero25 wrote: What is the area of a triangle created by the intersections of the lines x=4, y=5 and y=−3/4x+20
A. 42 B. 54 C. 66 D. 72 E. 96 I need urgent help in this question I know 3 points (4,0) , (0,5) and (20,17) When I am applying formula of area of triangle i.e Area = 1/2∗(x1y2−y1x2)+(x2y3−y2x3)+(x3y1−y3x1) Then the answer is wrong....WHY?????



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Re: What is the area of a triangle created by the intersections
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19 Apr 2018, 13:18
guerrero25 wrote: What is the area of a triangle created by the intersections of the lines x = 4, y = 5 and y = −(3/4)x + 20
A. 42 B. 54 C. 66 D. 72 E. 96 Let's first sketch the lines x = 4 and y = 5 To find the point where y = (3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (3/4) 4 + 20 = 17 So the point of intersection is (4, 17) To find the point where y = (3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (3/4)x + 20 When we solve for x, we get x = 20 So the point of intersection is (20, 5) Add this information to our sketch: From here, we can determine the length of the right triangle's base and height: Area = (1/2)(base)(height) = (1/2)(16)(12) = 96 Answer: E Cheers, Brent
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