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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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05 Nov 2014, 05:16
honchos wrote: Bunuel wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5). To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle. So ANY point on circumference of a circle with diameter \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 454590: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 306090 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). Of course there could be many other combinations. Back to the original question: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?(1) A < 3 > two vertices are on the Xaxis and the third vertex is on the Yaxis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,100) and the area would be more than 15. So not sufficient. (2) The triangle is right. > Obviously as the third vertex is on the Yaxis, the right angle must be at the third vertex. Which means the hypotenuse is on Xaxis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient. Answer: B. If we want to know how the area could be calculated with the help of statement 2, here you go: One of the approaches: The equation of a circle is \((x  a)^2 + (yb)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius. We know: \(r=2.5\), as the hypotenuse is 5. \(a=1.5\) and \(b=0\), as the center is on the Xaxis, at the point \((1.5, 0)\), half the way between the (1, 0) and (4, 0). We need to determine intersection of the circle with Yaxis, or the point \((0, y)\) for the circle. So we'll have \((01.5)^2 + (y0)^2 =2.5^2\) \(y^2=4\) > \(y=2\) and \(y=2\). The third vertex is either at the point \((0, 2)\) OR \((0,2)\). In any case \(Area=2*\frac{5}{2}=5\). Bunuel, Lets say base is  (1,0), (4,0) and (0,A): Third vertices Base =5, Height =A 1/2 X Base X height = 1/2 X 5 X A IF A <3 Then Area, A < 7.5 This answers the questions. Hence A is the solution. Statement (1) by itself is insufficient. If A is close to 0, the area of the triangle is small, but if A is a large negative (for example, A = 1000), the area of the triangle is large. Please reread/study solutions provided on previous pages.
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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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05 Nov 2014, 12:05
Oh I missed the fact that A can be negative also. Got the Point.
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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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09 Feb 2015, 01:31
VeritasPrepKarishma wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First, your question  No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 454590 triangle? The sides will not be 3 and 4. But in this question, you have something more. You know the line on which your third vertex of the triangle will fall. Attachment: The attachment Ques2.jpg is no longer available There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient. Quote: Dear Karishma, from the fogure above the triangle ABC is right triangle lets say that A=(1,0) B=(4,0) C= (0,A) my
question is why we say that AB is the basic and CD is the hight (middle point of AB) while it shuld be in right triangle
that CA is the basic and CB is the hight I really confused here why the basic and the hight are different. For example if
the triangle ABC is 906030 we will say that the basic is AC and the hight is CB.I mean why we try to find the CD
instead of CB Also if I have right triangle and I have both CD and CB what will be the hight in this case.
Dear Karishma, I apologize for the large number of questions but I really like your explanation.
Any given triangle has 3 sides. Any one of these sides can be the base. The corresponding altitude will be the height. In a right triangle, the altitude is one of the sides, that's all. Attachment:
Ques3.jpg [ 11.92 KiB  Viewed 1434 times ]
Look at a triangle with corresponding heights of the same color. Each side has its own altitude. The area of the triangle can be found by Area = (1/2)*base*height. You can take any baseheight pair. The area will be the same. Similarly, look at the right triangle. It also has 3 pairs of baseheight. Just that the height in 2 cases corresponds to the sides of the triangle. Again, you can find the area using any of the 3 pairs.
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If vertices of a triangle have coordinates (1,0), (4,0),
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11 Feb 2015, 12:49
Sharing an alternate solution.. we need to show if the height is more than 6. lets assume its 6, then calculate hypotenuse resulting two right triangles with 6,1 =\(\sqrt{37}\) and hypotenuse with 6,4 =\(\sqrt{52}\), clearly \(\sqrt{(37+52)}\) > 25, our assumption is wrong and angle at A is acute, if it becomes right it will only decrease height. Statement 2 sufficient.



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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12 Jul 2015, 01:03
My approach: Question can be simplified to : Area > 15? => 0.5*b*h>15 (Here base is 4+1 =5 and h =A)+>0.5*5*A>15=>A>6 so we have to find is the value of A lies between +6 and 6. From (1) A <3 i.e. A can be less than 6 also. Hence Not SufficientFrom(2) Triangle is right: Applying pythogarous theorem: 4^2+A^2+1^2+A^2=5^2; Simplifying A^2=9=>A=3 < 6. Sufficeint Answer is B
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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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31 Dec 2015, 09:59
Bunuel wrote: ramgmat wrote: Hi Bunuel, Please tell me where I am going wrong. Calculated using the matrix formula to solve the area of the triangle. \(1/2 [1 (1A) + 4 (A1) +0(11)] >15\) \(A>7\) Option 1 says A< 3 Hence Statement A is sufficient. Am I missing something here? This is a valid approach if you are familiar with the formula which gives the area based on the coordinates of the three vertices of a triangle. If the vetices of a triangle are: \(A(a_x, a_y)\), \(B(b_x, b_y)\) and \(C(c_x,c_y)\) then the area of ABC is: \(area=\frac{a_x(b_yc_y)+b_x(c_ya_y)+c_x(a_yb_y)}{2}\). So if we consider: \(A(1,0)\), \(B(4,0)\), and \(C(0,A)\) then the area would be: \(area=\frac{1(0A)+4(A0)+0(00)}{2}\) > \(area=\frac{5A}{2}\). Question: is \(area=\frac{5A}{2}>15\) > is \(A>6\). Statement (1) says A>3, which is not sufficient to say whether \(A>6\). P.S. You made some errors in calculation and also didn't put the area formula in . Hope it helps. Hi Bunuel, I understood till here A >6. we need to identify if A makes tis true so using st1 we cant say this. But I didn't understand how st2 is sufficient here. Please clarify Thanks



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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21 Nov 2016, 07:04
Hi Bunuel could you please add the chart since I can not follow solution. thanx



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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23 Mar 2017, 04:18
Bunuel wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5). To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle. So ANY point on circumference of a circle with diameter \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 454590: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 306090 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). Of course there could be many other combinations. Back to the original question: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?(1) A < 3 > two vertices are on the Xaxis and the third vertex is on the Yaxis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,100) and the area would be more than 15. So not sufficient. (2) The triangle is right. > Obviously as the third vertex is on the Yaxis, the right angle must be at the third vertex. Which means the hypotenuse is on Xaxis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient. Answer: B. If we want to know how the area could be calculated with the help of statement 2, here you go: One of the approaches: The equation of a circle is \((x  a)^2 + (yb)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius. We know: \(r=2.5\), as the hypotenuse is 5. \(a=1.5\) and \(b=0\), as the center is on the Xaxis, at the point \((1.5, 0)\), half the way between the (1, 0) and (4, 0). We need to determine intersection of the circle with Yaxis, or the point \((0, y)\) for the circle. So we'll have \((01.5)^2 + (y0)^2 =2.5^2\) \(y^2=4\) > \(y=2\) and \(y=2\). The third vertex is either at the point \((0, 2)\) OR \((0,2)\). In any case \(Area=2*\frac{5}{2}=5\). Bunuel Thank you for the fantastic explanation. But I have a small doubt. Like you mentioned 'the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex', so the 3rd vertex should have been at 2.5, but when we calculated it came out to be 2.Can you please clear on this



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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23 Mar 2017, 05:33
KARISHMA315 wrote: Bunuel wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5). To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle. So ANY point on circumference of a circle with diameter \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 454590: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 306090 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). Of course there could be many other combinations. Back to the original question: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15?(1) A < 3 > two vertices are on the Xaxis and the third vertex is on the Yaxis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,100) and the area would be more than 15. So not sufficient. (2) The triangle is right. > Obviously as the third vertex is on the Yaxis, the right angle must be at the third vertex. Which means the hypotenuse is on Xaxis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient. Answer: B. If we want to know how the area could be calculated with the help of statement 2, here you go: One of the approaches: The equation of a circle is \((x  a)^2 + (yb)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius. We know: \(r=2.5\), as the hypotenuse is 5. \(a=1.5\) and \(b=0\), as the center is on the Xaxis, at the point \((1.5, 0)\), half the way between the (1, 0) and (4, 0). We need to determine intersection of the circle with Yaxis, or the point \((0, y)\) for the circle. So we'll have \((01.5)^2 + (y0)^2 =2.5^2\) \(y^2=4\) > \(y=2\) and \(y=2\). The third vertex is either at the point \((0, 2)\) OR \((0,2)\). In any case \(Area=2*\frac{5}{2}=5\). Bunuel Thank you for the fantastic explanation. But I have a small doubt. Like you mentioned 'the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex', so the 3rd vertex should have been at 2.5, but when we calculated it came out to be 2.Can you please clear on this How did you come up with the highlighted part? That's not true. The radius is 2.5. The center is at (1.5, 0). The distance from the center to all three vertices is 2.5.
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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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23 Mar 2017, 05:45
Bunuel Thank you for the fantastic explanation. But I have a small doubt. Like you mentioned 'the third vertex must be one of two intersections of the circle with Yaxis. We'll get the two specific symmetric points for the third vertex', so the 3rd vertex should have been at 2.5, but when we calculated it came out to be 2.Can you please clear on this[/quote] How did you come up with the highlighted part? That's not true. The radius is 2.5. The center is at (1.5, 0). The distance from the center to all three vertices is 2.5.[/quote] Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or 4 Posted from my mobile device Posted from my mobile device Posted from my mobile device



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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23 Mar 2017, 06:01
KARISHMA315 wrote: Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or 4 Have you tried to draw it? Hope that the below image will help: Attachment:
Untitled.png [ 11.25 KiB  Viewed 792 times ]
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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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23 Mar 2017, 08:06
Bunuel wrote: KARISHMA315 wrote: Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or 4 Have you tried to draw it? Hope that the below image will help: Attachment: Untitled.png Thanks a lot Bunuel, got it now. Excellent explanation



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If vertices of a triangle have coordinates (1,0), (4,0),
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Updated on: 28 Aug 2017, 13:30
[wrapimg=][/wrapimg] VeritasPrepKarishma wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First, your question  No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 454590 triangle? The sides will not be 3 and 4. But in this question, you have something more. You know the line on which your third vertex of the triangle will fall. Attachment: Ques2.jpg There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient. Hi Experts. .. I understand why S1 is not sufficient for 2, how do you know only one angle is 90 % In the 1st diagram from karishma, why cant the reds be 90 degrees as well In fact, in a inscribed circle, if the hypothesis is the diameter, the triangle has many 90 degree angles ...



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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28 Aug 2017, 13:19
VeritasPrepKarishma wrote: gmat620 wrote: If vertices of a triangle have coordinates (1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks. First, your question  No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 454590 triangle? The sides will not be 3 and 4. But in this question, you have something more. You know the line on which your third vertex of the triangle will fall. Attachment: Ques2.jpg There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient. Karishma  could you please explain why there is only 1 such right angle. ..in a semicircle, with the hypotenuse as the diameter of the circle, there are many 90 degree angle ... Also even if it the vertex was on the y axis + plus the circle with the diameter as 5, why does this mean, there is only 1 triangle with a 90 degree angle



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Re: If vertices of a triangle have coordinates (1,0), (4,0),
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29 Sep 2018, 03:32
hello, can we solve the second st. by similar triangles?



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If vertices of a triangle have coordinates (1,0), (4,0),
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01 Apr 2019, 09:07
Base of triangle = 1+4 = 5 (1) If A = 2, then Area of triangle is 1/2*5*2 = 5 < 15  (answer for the question is NO),... If A= 10, Area of triangle is 1/2*5*10 = 25 (answer for the question is YES) => INSUFFICIENT (2) Let call point at (1,0) = B, (4,0) = C, A is always at vertical axis so AB = sqrtroot (OA^2 +1 ) (Triangle OAB is right at O) AC = sqrtroot (OA^2 + 16) (triangle OAC is right at O) Because stm2 said that triangle ABC is right so we can calculate its area in 2 ways: Area of ABC = 1/2 * AB * AC = 1/2 * BC * OA => 1/2 * sqrtroot (A^2 + 1 ) * sqrtroot (A^2 + 16) = 1/2 * 5 * OA => OA^4 + 17OA + 16 = 25*OA^2 => OA^4  8OA + 16 =0 => (OA^2  4)^2 = 0 => OA^2 = 4 => OA = 2 or OA = 2.. Hence, Area of triangle ABC = 1/2*5*2 = 5 which is always smaller than 15=> SUFFICIENT. Thus, B is the answer
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If vertices of a triangle have coordinates (1,0), (4,0),
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