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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # Trapezoid OPQR has one vertex at the origin. What is the ar  Question banks Downloads My Bookmarks Reviews Important topics
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Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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Attachment: #GREpracticequestion Trapezoid OPQR has one vertex at the origin.jpg [ 17.11 KiB | Viewed 2928 times ]

Trapezoid OPQR has one vertex at the origin. What is the area of OPQR ?

A. $$\frac{a^2}{4}$$

B. $$\frac{a^2}{2}$$

C. $$\frac{3a^2}{4}$$

D. $$\frac{3a^2}{2}$$

E. $$2a^2$$
[Reveal] Spoiler: OA

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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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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? Intern Joined: 17 Feb 2018
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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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 Intern Joined: 05 Oct 2017
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
2
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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?
Manager Joined: 15 Feb 2018
Posts: 53
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Kudos [?]: 25 , given: 33

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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? Intern Joined: 05 Oct 2017
Posts: 10
Followers: 0

Kudos [?]: 8  , given: 1

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
2
KUDOS
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? Intern Joined: 14 Jun 2018
Posts: 36
Followers: 0

Kudos [?]: 9  , given: 100

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
1
KUDOS
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 Intern Joined: 27 Jan 2019
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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We can solve this by using parallel lines.
Attachments shot3.jpg [ 287.86 KiB | Viewed 2523 times ]

Manager  Joined: 20 Jun 2019
Posts: 135
GRE 1: Q158 V162 Followers: 0

Kudos [?]: 83 , given: 29

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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? Manager Joined: 04 Apr 2020
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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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. Manager Joined: 05 Aug 2020
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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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. Intern Joined: 18 Apr 2020
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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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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