Question

ABC is a triangle. Through the vertices A,C and B lines PQ ,QR and RP are drawn parallel to the side BC BA and AC respectively, forming a ΔPQR. If ar.(ΔPQR) ar.(ΔPQR) = $12^2$ cm, then the area of ΔABC =

Answer 1

Answer:

Clearly, ABCQ and ARBC are parallelograms.

∴BC=AQ and BC=AR

⇒AQ=AR

⇒A is the mid-point of QR.

Similarly, B and C are the mid-points of PR and PQ respectively.

∴AB=

2

1

PQ,BC=

2

1

QR and CA=

2

1

PR

⇒PQ=2AB,QR=2BC and PR=2CA

⇒PQ+QR+RP=2(AB+BC+CA)

⇒ Perimeter of ΔPQR=2 (Perimeter of ΔABC)