Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the midpoints of sides of $△ A B C .$ Then the area of $△ P Q R$ is

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Solution

Area of $△ A B C$ = 24 sq. units.

$∵$ Q and R mid points of AB and AC
$∴$ RQ || BC and $R Q = \frac{1}{2} B C$
Similarly, $P Q | | A B a n d P Q = \frac{1}{2} A B$
And, RP || AC and $R P = \frac{1}{2} A C$
$∴$ area of $△ P Q R = \frac{1}{4}$ Area of $△ A B C$
$= \frac{1}{4} \times 24 = 6$ sq. units

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Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the midpoints of sides of △ABC. Then the area of △PQR is

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