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 Δ*PQR* is

#### Options

12 sq. units

6 sq. units

4 sq. units

3 sq. units

#### 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 ||^{gm}APQR.

∴ Area of ΔAPR = Area of ΔPQR ……(1)

Similarly,

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

But

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