It is currently 24 Mar 2019, 09:05

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:
Moderator
Joined: 18 Apr 2015
Posts: 5888
Followers: 96

Kudos [?]: 1154 [1] , given: 5471

Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  06 Aug 2017, 10:23
1
KUDOS
Expert's post
00:00

Question Stats:

42% (03:10) correct 57% (02:59) wrong based on 28 sessions

Attachment:

#GREpracticequestion Trapezoid OPQR has one vertex at the origin.jpg [ 17.11 KiB | Viewed 277 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

_________________
Director
Joined: 03 Sep 2017
Posts: 521
Followers: 1

Kudos [?]: 344 [2] , given: 66

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  19 Sep 2017, 09:23
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.

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
Posts: 31
Followers: 0

Kudos [?]: 24 [1] , given: 2

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  20 Feb 2018, 15:41
1
KUDOS
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
Posts: 10
Followers: 0

Kudos [?]: 7 [2] , given: 1

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  21 Feb 2018, 23:56
2
KUDOS
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
Followers: 1

Kudos [?]: 17 [0], given: 33

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

Kudos [?]: 7 [2] , given: 1

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  22 Feb 2018, 10:48
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 [?]: 7 [1] , given: 100

Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]  27 Jun 2018, 11:59
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
Re: Trapezoid OPQR has one vertex at the origin. What is the ar   [#permalink] 27 Jun 2018, 11:59
Display posts from previous: Sort by