Gmat_mission wrote: ↑Sun Apr 26, 2020 12:06 pm
gpp-sgf_img3.png
In the diagram, \(HJLM\) is a square, and \(GH = 10.\) Find the area of trapezoid \(GHJK.\)
A. \(50+25\sqrt3\)
B. \(50+50\sqrt3\)
C. \(75+12.5\sqrt3\)
D. \(75+25\sqrt3\)
E. \(75+50\sqrt3\)
[spoiler]OA=D[/spoiler]
Source: Magoosh
We see that triangle GMH is a 30-60-90 right triangle and since GH = 10, we have GM = 10/2 = 5 and HM = 5√3 (notice that HM is the height of trapezoid GHJK).
Since HJLM is a square, HM = 5√3 means HJ = ML = 5√3 (notice that HJ is one of the bases of the trapezoid).
Since triangles GMH and KLJ are congruent, GM = KL = 5. Therefore, GK, the other base of the trapezoid, is 5 + 5√3 + 5 = 10 + 5√3.
Since the area of a trapezoid is ½ x (sum of the two bases) x height, the area of trapezoid GHJK is:
½ x (5√3 + 10 + 5√3) x 5√3 = ½ x (10 + 10√3) x 5√3 = ½ x (50√3 + 150) = 25√3 + 75
Answer: D
