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Byju's Answer

Standard X

Mathematics

Area of a Quadrilateral

A2,1, B4, -1,...

Question

$A (2, 1), B (4, - 1), C (3, 2)$ are the vertices of a triangle. Through $A$ and $C$ lines are drawn parallel to $B C$ and $A B$ respectively to intersect at $D$ . Then the area in square units of the quadrilateral $A B C D$ is

2

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4

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6

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8

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Solution

The correct option is B $4$
Clearly , ABCD is a parallelogram
$∴$ the area of ABCD $= 2 ($ area of $Δ A B C)$
Area of $△ A B C$ $= \frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & 1 4 & - 1 & 1 3 & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$= \frac{1}{2} (- 6 - 1 + 11) = 2$
So, the area of parallelogram $= 4$ sq.units.

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Question 6
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