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FGHJ is a rectangle, such that FJ = 40 and FJ > FG. Point M is the midpoint of FJ, and a Circle C is constructed such that M is the center and FJ is the diameter. Circle C intersects the top side of the rectangle, GH, at two separate points. Point P is located on side GH. What is the area of triangle FJP?

Statement #1: One of the two intersections of Circle C with side GH is point P, one vertex of the triangle FJP.

Statement #2: One of the two intersections of Circle C with side GH is point R, such that RH = 7


Geometry is beautiful, and the GMAT loves it. This question is one of a set of 10 GMAT DS practice questions on geometry. For the other problems, as well as the OE for this question, see:
GMAT Data Sufficiency Geometry Practice Questions

Mike :-)




hi,

Important points, which will be clearer when you draw the figure..
1) FJ and GH are parallel, EQUAL and opposite sides.
2) so GH =20...
3) FJ becomes the DIA of the circle as the radius is half of FJ or 20...
4) from point 3 above angle FPJ becomes 90....
5) The intersection on GH will be at same distance from vertices G and H, that is P will be at same distance from G as R will be from H..

Let see the statements-

Statement #1: One of the two intersections of Circle C with side GH is point P, one vertex of the triangle FJP.
P could be anywhere depending on the other dimension FG of rectangle...
Insuff

Statement #2: One of the two intersections of Circle C with side GH is point R, such that RH = 7
Yes we can find the area of angle FRJ from this , BUT do we know where P lies?..
We may not know exact point P lies, but it lies ON the same line as R..
This means the height of FRJ and FPJ will be SAME and the BASE FJ is also same..
so area of FRJ and FPJ will be same
Suff

B


how can I find height of triangle FRJ? please clarify.
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robu
chetan2u
mikemcgarry
FGHJ is a rectangle, such that FJ = 40 and FJ > FG. Point M is the midpoint of FJ, and a Circle C is constructed such that M is the center and FJ is the diameter. Circle C intersects the top side of the rectangle, GH, at two separate points. Point P is located on side GH. What is the area of triangle FJP?

Statement #1: One of the two intersections of Circle C with side GH is point P, one vertex of the triangle FJP.

Statement #2: One of the two intersections of Circle C with side GH is point R, such that RH = 7


Geometry is beautiful, and the GMAT loves it. This question is one of a set of 10 GMAT DS practice questions on geometry. For the other problems, as well as the OE for this question, see:
GMAT Data Sufficiency Geometry Practice Questions

Mike :-)




hi,

Important points, which will be clearer when you draw the figure..
1) FJ and GH are parallel, EQUAL and opposite sides.
2) so GH =20...
3) FJ becomes the DIA of the circle as the radius is half of FJ or 20...
4) from point 3 above angle FPJ becomes 90....
5) The intersection on GH will be at same distance from vertices G and H, that is P will be at same distance from G as R will be from H..

Let see the statements-

Statement #1: One of the two intersections of Circle C with side GH is point P, one vertex of the triangle FJP.
P could be anywhere depending on the other dimension FG of rectangle...
Insuff

Statement #2: One of the two intersections of Circle C with side GH is point R, such that RH = 7
Yes we can find the area of angle FRJ from this , BUT do we know where P lies?..
We may not know exact point P lies, but it lies ON the same line as R..
This means the height of FRJ and FPJ will be SAME and the BASE FJ is also same..
so area of FRJ and FPJ will be same
Suff

B


how can I find height of triangle FRJ? please clarify.

Lets concentrate on FRJ, right angled at R and drop a Perpendicular from R on FJ at say T....
since R Is 7 m from G, T will be 7 m from F or J, say F..
so FT = 7 and TJ = 40-7 = 33...
All three triangles, Bigger FRJ and smaller FRT & FJT are similar and when you know two sides, you can find other sides..

Take FRT and FRJ....
\(\frac{FT}{RT} = \frac{RT}{TJ}......................FT*TJ = (RT)^2..............7*33 = (RT)^2.................RT = \sqrt{7*33}\)...
RT is nothing but the height
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