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Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
06 Aug 2017, 10:23
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#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. \(2a^2\)
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Re: Trapezoid OPQR has one vertex at the origin. What is the ar [#permalink]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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]
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 (a0)/(2a0)=1/2, the slope of QP must be the same, i.e. (ay)/(ax) = 1/2, where (x,y) are the coordinates of P. Since P is on the xaxis, its x is equal to 0 and thanks to this information we can solve the equation (ay)/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
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27 Jun 2018, 11:59





