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If vertices of a triangle have coordinates (-2,2), (3,2) and
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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02 Sep 2010, 04:45
Vertices of a triangle have coordinates \((-2, 2)\), \((3, 2)\), and \((x, y)\). What is the area of the triangle?

Given two points \(A(-2, 2)\) and \(B(3, 2)\). The question asks to find the area of triangle \(ABC\), where \(C(x,y)\). Refer to the diagram below:



(1) \(|y-2|=1\).

Either \(y=3\) or \(y=1\), hence vertex \(C\) could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). In any case, the area of \(ABC\) will be the same. Using the formula \(area=\frac{1}{2}*base*height\), we have \(base=AB=5\), and the height would be 1 for any point \(C\) (see two possible locations of \(C\): \(C_1\) and \(C_2\). The heights of \(ABC_1\) and \(ABC_2\) are the same and equal to 1). Therefore, the area is \(area=\frac{1}{2}*base*height=\frac{5}{2}\). This statement is sufficient.

(2) The angle at the vertex \((x, y)\) equals 90 degrees.

This statement implies that \(ABC\) is a right triangle with a hypotenuse \(AB\). Consider \(AB\) as the diameter of a circle. In this case, \(C\) could be anywhere on the circle, and it will form a right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle). However, the height of \(ABC\) will be different for different locations of point \(C\), resulting in different areas (see two possible locations of \(C\): \(C_3\) and \(C_4\). The heights of \(ABC_3\) and \(ABC_4\) are different). This statement is not sufficient.


Answer: A

Hope it helps.

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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04 Aug 2010, 07:42
I think i understand why (2) is not sufficient. If we take the points (-2,2), (3,2) as a diameter of a circle and the point (x,y) lies on the perimeter of the circle, then any triangle formed is a right-angled triangle but with different height, therefore producing different areas. What do you think?
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If vertices of a triangle have coordinates (-2,2), (3,2) and
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07 Jun 2013, 23:42
Bunuel


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.
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Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks
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If vertices of a triangle have coordinates (-2,2), (3,2) and
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08 Jun 2013, 02:16
imhimanshu
Bunuel


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

Can someone please explain the Red Part. Why height of C2 will be 1.
Thanks
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Consider triangle \(ABC_1\): the height is the distance (perpendicular) from \(C_1\) to AB, which is 1;
Consider triangle \(ABC_2\): the height is the distance (perpendicular) from \(C_2\) to AB, which is 1.

Hope it's clear.

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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12 Jul 2013, 06:30
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.
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Hi,

From the 2nd statemen, the hypotenuse is 5 units in length since the triangle is a right angle triangle that is 90 degree at point C.. then that basically means that the other 2 sides(base and height of the triangle) will be be either 3 and 4 since 3,4 and 5 forms a Pythagorean triple.. that means area which is half * base * height will always be (3*4)/2 .. That would mean that 2nd statement is also sufficient.. I do get your point about drawing a circle but how would you justify the 3 lengths according to Pythagoras theorem ?
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If vertices of a triangle have coordinates (-2,2), (3,2) and
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12 Jul 2013, 08:01
debshali
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).
Attachment:
render.php (1).png

(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it heps.

Hi,

From the 2nd statemen, the hypotenuse is 5 units in length since the triangle is a right angle triangle that is 90 degree at point C.. then that basically means that the other 2 sides(base and height of the triangle) will be be either 3 and 4 since 3,4 and 5 forms a Pythagorean triple.. that means area which is half * base * height will always be (3*4)/2 .. That would mean that 2nd statement is also sufficient.. I do get your point about drawing a circle but how would you justify the 3 lengths according to Pythagoras theorem ?
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You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 5 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 3:4:5. Or in other words: if \(a^2+b^2=5^2\) DOES NOT mean that \(a=3\) and \(b=4\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=5^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=3\) and \(b=4\).

For example: \(a=1\) and \(b=\sqrt{24}\) or \(a=2\) and \(b=\sqrt{21}\) ...

Hope it's clear.

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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Bunuel! is something else! amazing solution.
Is there any other way to do it? By doing some math... maybe..if yes please explain...
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If vertices of a triangle have coordinates (-2,2), (3,2) and
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08 Jul 2015, 23:37
Shreks1190
Bunuel! is something else! amazing solution.
Is there any other way to do it? By doing some math... maybe..if yes please explain...
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Here it is...

Quote:
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

A) |y-2|=1
B) angle at the vertex (x,y) equals to 90 degrees
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Point to Observe: (-2,2) and (3,2) have equal y-co-ordinate i.e. they are on the same horizontal line which can be considered as base of triangle [Base=3-(-2)=5 units] for calculation of area using Area=(1/2)Base*Height

Statement 1: |y-2|=1

i.e. y = 1 or 3
i.e. This point will be 1 unit above both other points or 1 unit below both other points
In each case the Height will be 1 unit and hence will result in equal area. Hence,
SUFFICIENT

Statement 2: angle at the vertex (x,y) equals to 90 degrees

i.e. Three points are making a right angle triangle with hypotenuse = 5 which has various possibilities of lengths of other two sides therefore, Various various values of Area of Triangle. Hence,
NOT SUFFICIENT

Answer: Option A

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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18 Oct 2016, 19:08
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) Angle at the vertex (x,y) equals to 90 degrees
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Please find answer as attached

Answer: option A
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If vertices of a triangle have coordinates (-2,2), (3,2) and
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07 Aug 2017, 07:47
Dear Bunuel,

I want to thank you. Your explanations are so flawless and simple.


Bunuel
noboru
If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

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If vertices of a triangle have coordinates (-2,2), (3,2) and
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18 Mar 2020, 08:54
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If vertices of a triangle have coordinates (-2,2), (3,2) and (x,y), what is the area of the triangle?

(1) |y-2|=1
(2) angle at the vertex (x,y) equals to 90 degrees

Given two points A(-2,2) and B(3,2). Question: Area ABC=?, where C(x,y).


(1) |y-2|=1 --> \(y=3\) or \(y=1\) --> vertex C could be anywhere on the blue line \(y=3\) or anywhere on the red line \(y=1\). But in ANY case the are of ABC will be the same --> \(area=\frac{1}{2}*base*height\) so \(base=AB=5\) and the height would be 1 for any point C (see two possible locations of C: C1 and C2, the heights of ABC1 and ABC2 are the same and equal to 1) --> \(area=\frac{1}{2}*base*height=\frac{5}{2}\). Sufficient.

(2) angle at the vertex C(x,y) equals to 90 degrees --> ABC is a right triangle with hypotenuse AB --> consider AB to be diameter of a circle and in this case C could be anywhere on the circle and it will be right angle (if the diameter of the circle is also the inscribed triangle’s side, then that triangle is a right triangle), thus height of ABC will be different for different location of point C, resulting the different areas (see two possible locations of of C: C3 and C4, heights of ABC3 and ABC4 are different). Not sufficient.

Answer: A.

Hope it helps.

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Great Insight and terrific explanation
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