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
12 sq. units
6 sq. units
4 sq. units
3 sq. units
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 ||gmAPQR.
∴ Area of ΔAPR = Area of ΔPQR ……(1)
PQ is a diagonal of ||gm PBQR
∴ Area of ΔPQR = Area of ΔPBQ ……(2)
QR is the diagonal of ||gm 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
Area of ΔAPR + Area of ΔPQR + Area of ΔPBQ + Area of ΔRCQ = Area of ΔABC
4(Area of ΔPBQ) = Area of ΔABC
`= 1/4 `Area of ΔABC
∴ Area of ΔPBQ `= 1/4 (24)`
= 6 sq units