The mid points of the sides of a triangle ABC along with any of the vertices as the fourth point make a parallelogram of area equal to

$a r (△ A B C)$

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$\frac{1}{2} a r (△ A B C)$

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$\frac{1}{3} a r (△ A B C)$

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$\frac{1}{4} a r (△ A B C)$

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Solution

The correct option is B
$\frac{1}{2} a r (△ A B C)$

If P, Q and R the midpoints of the sides of a $△ A B C$ , area of any parallelogram formed by the midpoints and one vertex of the given triangle has area $= \frac{1}{2} a r e a △ A B C$

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Similar questions

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The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to

(a) $\frac{1}{2} (a r △ A B C)$

(b) $\frac{1}{3} (a r △ A B C)$

(d) $(a r △ A B C)$

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