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Trapezoid OPQR has one vertex at the origin. What is the ar

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Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 06 Aug 2017, 10:23
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 19 Sep 2017, 09:23
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To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 20 Feb 2018, 15:41
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IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?


Draw a line between point P to line R. We get a triangle and a parallelogram.
Triangle has a base of a/2 and height of a. Area of triangle is (a^2)/4
Parallelogram has a base of a and a height of a/2. Area of parallelogram is (a^2)/2
combine triangle and parallelogram we get (3a^2)/4
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 21 Feb 2018, 23:56
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just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 22 Feb 2018, 04:13
How to get the area of the above triangle though? I don't know OP and therefore don't know the height.

achirarulz wrote:
just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 22 Feb 2018, 10:48
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To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.


gremather wrote:
How to get the area of the above triangle though? I don't know OP and therefore don't know the height.

achirarulz wrote:
just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 27 Jun 2018, 11:59
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achirarulz wrote:
just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?


It's not square!

just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole rectangle) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 09 May 2019, 08:11
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We can solve this by using parallel lines.
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 19 Feb 2020, 00:25
achirarulz wrote:
just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?


How do you calculate area of triangle above trapezoid?
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 04 May 2020, 02:24
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pprakash786 wrote:
achirarulz wrote:
just take the area of two triangle above and below trapezoid and subtract it from area of whole square

(area of whole square) 2a^2 - (area of triangle above trapezoid) a^2/4 - (are of triangle below trapezoid) a^2

2a^2 - a^2/4 - a^2 = 3a^2/4

IlCreatore wrote:
To find coordinates of point P, I exploit the fact that the two basis of a trapezoid are parallel and thus since OR slope is (a-0)/(2a-0)=1/2, the slope of QP must be the same, i.e. (a-y)/(a-x) = 1/2, where (x,y) are the coordinates of P. Since P is on the x-axis, its x is equal to 0 and thanks to this information we can solve the equation (a-y)/a=1/2 that gives a value of y = a/2.

Given that should I use OP as height to compute the area? Because it does not seem to be perpendicular to the larger base. I also tried to think about dividing the figure in a triangle and a parallelogram but it doesn't work.

Any hint?




How do you calculate area of triangle above trapezoid?



After getting the length of OP (a/2), we know that the full length of the Y axis is a from (a,a) coordinate so the remaining length is a/2. This is one perpendicular side of the upper triangle. The other perpendicular side is a, from the (a,a) coordinate. So the area = a/2 * (a) = a^2/2.
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 16 Aug 2020, 11:25
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Quote:
How do you calculate area of triangle above trapezoid?


All trapezoids have one pair of parallel sides, so we can deduce that the slope of the line from \(O\) to \(R\) has to have the same slope from \(P\) to \(Q\).

Since \(O\) is the origin, the slope of the line \(OR\) is:

\(\frac{a-0}{2a-0}\)

\(\frac{a}{2a}\)

\(\frac{1}{2}\)

This means that the slope of the line \(PQ\) has a slope of \(\frac{1}{2}\).

To find the area of the above triangle, we would need the base and its height. Looking at the figure, we know the height (the distance from the y-axis to \(Q\)) is \(a\). Now we need to find the base.

We can use the slope formula to find it, since we know that the slope from the \(P\) to point \(Q\) has to be \(\frac{1}{2}\).

Since the point \(P\) is on the y axis, let \(P\) = (0,\(y\)).

Using the slope formula we get:

\(\frac{y-a}{0-a} = \frac{1}{2}\)

\(2y - 2a = -a\)

\(2y = a\)

\(y = \frac{a}{2}\)

So Point \(P\) = (0,\(\frac{a}{2}\))

That must mean that the base of the triangle is \(\frac{a}{2}\).

So now we solve for the area of the above triangle:

\(\frac{1}{2}* a * \frac{a}{2}\) =

\(\frac{a^{2}}{4}\)

And there's the area of the above triangle.
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink] New post 29 Sep 2020, 00:29
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Basically
Slope is = Rise/Run. For parallel lines it is same.

For line OR --> for the run 2a rise is a.
For line QP --> for a run of a i know the rise should be a/2 (half of run) but I have to reach to a rise of "a" therefore I must start from a/2 which is the point P.

Cheers
Re: Trapezoid OPQR has one vertex at the origin. What is the ar   [#permalink] 29 Sep 2020, 00:29
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