GMAT Geometry Problems with Solutions

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1.In the adjoining figure ABCD is a parallelogram and E, F are the centroids of ABD and BCD respectively, then EF equals:

(a) AE
(b) BE
(c) CE
(d) DE

Ans.(a)
Sol. As E is the centroid of triangle ABD and AO is one of its medians.
⇒ AE: EO = 2: 1 ∴EO=1/3 OA,
Similarly, FO =1/3 OC
∴ EO + OF =1/3 OA+1/3 OC=1/3 AC=AE
∴ EF = AE

2. In a trapezium ABCD, if E and F be the mid-points of the diagonals AC and BD respectively. Then EF =?

(a) 1/2 AB
(b) 1/2 CD
(c) 1/2 (AB+CD)
(d) 1/2 (AB-CD)

Solution :

3.D and E are two points on side AC and BC respectively of ∆ABC such that DE = 18 cm, CE = 5 cm, ∠DEC=90°. If tan⁡∠ABC=3.6 then AC: CD =

(a) BC : 2CE
(b) 2CE : BC
(c) 2BC = CE
(d) CE = 2BC

 

Solution :

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