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Question

# Let ABC be a triangle of area 24 sq. units and PQR be the triangle formed by the mid-points of the sides of Δ ABC. Then the area of ΔPQR is (a) 12 sq. units (b) 6 sq. units (c) 4 sq. units (d) 3 sq. units

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Solution

## Given: (1) The Area of ΔABC = 24 sq units. (2) ΔPQR is formed by joining the midpoints of ΔABC To find: The area of ΔPQR Calculation: In ΔABC, we have Since Q and R are the midpoints of BC and AC respectively. PQ || BA PQ || BP Similarly, RQ || BP. So BQRP is a parallelogram. Similarly APRQ and PQCR are parallelograms. We know that diagonal of a parallelogram bisect the parallelogram into two triangles of equal area. Now, PR is a diagonal of APQR. ∴ Area of ΔAPR = Area of ΔPQR ……(1) Similarly, PQ is a diagonal of PBQR ∴ Area of ΔPQR = Area of ΔPBQ ……(2) QR is the diagonal of PQCR ∴ Area of ΔPQR = Area of ΔRCQ ……(3) From (1), (2), (3) we have Area of ΔAPR = Area of ΔPQR = Area of ΔPBQ = Area of ΔRCQ But Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC 4(Area of ΔPBQ) = Area of ΔABC ∴ Area of ΔPBQ Hence the correct answer is option (b).

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