A) 8 sq. units
B) 10 sq. units
C) 12 sq. units
D) 16 sq. units
Correct Answer: A
Solution :
The coordinates of the vertices D, E, G and F are \[(3,3),\]\[(6,3),\]\[(7,1)\]and \[(2,1)\]. |
(7,1) and (2,1). Now, |
\[DE=\sqrt{{{(6-3)}^{2}}+{{(3-3)}^{2}}}=\sqrt{9+0}=3\] units |
\[FG=\sqrt{{{(7-2)}^{2}}+{{(1-1)}^{2}}}=\sqrt{25+0}=5\] units |
and height of trapezium = 2 units [from figure] |
\[\therefore \] Area of trapezium |
\[DEGF=\frac{1}{2}\times (DE+FG)\times height\] |
\[=\frac{1}{2}\times (3+5)\times 2=8sq.\]unit |
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