ShahadatHJ wrote:
In the Parallelogram, E is the midpoint of AC and EF is parallel to AB. If the area of ABCD is 36 sq. cm. and EX = 6. What is the area of the triangle ABX in sq. cm? (Picture is not drawn to scale)
A. 16
B. 12
C. 15
D. 09
E. None
Of course you can solve it otherwise, but easiest would be to take ABCD as a square.
So Each side is 6.
AC=6, and AE = 6/3 = 2.
Hence area of ABX=\(\frac{1}{2}*6*3=9\)
OR
EF divides the parallelogram in two equal parts, so area of each is 18 sqm.
Now area of ABX = area of AEX + area of AFX, as the height is same being between two same parallel lines and base too is equal.
AB = EF = EX+XF
So area of ABX = \(\frac{18}{2}=9\)
D
, as the height is same being between two same parallel lines and base too is equal.
Shouldn't it be BFX instead of AFX? Are we considering AB the base for these 2 triangles or AE and BF as bases?
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