Question

In triangle ABC ,P, Q,and R are the midpoint of sides AB, bc and AC respectively if ar (PBQR) =36 cm2 find ar(ABC)

Answer 1

Answer: $72\text{ square cm}$

Step-by-step explanation:

Here ABC is a triangle,

In which P, Q,and R are the midpoint of sides AB, BC and AC respectively.

Since, line joining to the mid points of two sides is parallel to third side,

Therefore, PR ║ BQ and PB ║ RQ

⇒ Figure, PBQR is a parallelogram,

Since, the diagonal of parallelogram equally divides the area of the parallelogram,

⇒ Area of triangle PQR = 1/2 × area of parallelogram PBQR = 18 square cm

Also, by the mid point theorem,

1/4 × Area of triangle ABC = Area of triangle PQR

⇒  Area of triangle ABC = 4 × Area of triangle PQR = 4 × 18 = 72 square cm

Answer 2

Answer:

Area of triangle ABC is $72 cm^2$

Step-by-step explanation:

Given  triangle ABC ,P, Q,and R are the midpoint of sides AB, bc and AC respectively if ar (PBQR) =36 cm2. we have to find the ar(ABC).

In ΔABC, P and Q are the mid points of sides AB and BC ∴ by mid-point theorem PQ is parallel to AC and equal to half of AC.

⇒ PQ=AR and PQ||AR and also PQ=RC and PQ||RC

⇒ QPAR and PQRC both are parallelogram

As One diagonal of a parallelogram divides the parallelogram into 2 congruent triangles of equal area.

⇒ ar(PQR)=ar(APR) and ar(PQR)=ar(QRC)   →   (1)

Similarly, In ΔABC, P and R are the mid points of sides AB and AC  ∴ by mid-point theorem PR is parallel to BC and equal to half of BC

⇒ PR=BQ and PR||BQ

⇒ PRBQ is a parallelogram

As One diagonal of a parallelogram divides the parallelogram into 2 congruent triangles of equal area.

⇒ $ar(PQR)=ar(APR)=\frac{1}{2}ar(PBQR)=18cm^2$   →  (2)

From eq (1) and (2),  $ar(APR)=ar(PBQ)=ar(PQR)=ar(QRC)=18cm^2$

$Now, ar(ABC)=ar(APR)+ar(PBQ)+ar(PQR)+ar(QRC)$

                     $=18+18+18+18=72cm^2$