Question

P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also show that $ar\left(||gmPQRS\right)=\frac{1}{2}ar\left(||gmABCD\right)$.

Answer 1

For this question we can use the theorem that a line joining the mid points of 2 sides of the triangle is parallel to the third side and equal to half of its length.



P and Q are mid-points of the sides AB and BC of triangle ABC.

=> PQ is parallel to side AC of triangle ABC and of length = (1/2)AC.

R and S are mid-points of the sides CD and DA of triangle ACD
=> RS is parallel to side AC of triangle ACD and of length = (1/2)AC

=> PQ and RS which are the opposite sides of the quadrilateral PQRS are of equal length and both being parallel to AC are parallel to each other.
=> quadrilateral PQRS is a parallelogram.



Also since, triangle(PQB)~triangle (ABC)



ar(PQB)/ar(ABC) = $P{Q}^2/B{C}^2$ = 1/4

area(PQB) = 1/4 x ar(ABC)

Similarly, ar(SDR) = 1/4 x ar(ADC)

ar(CRQ) = 1/4 x ar(CDB)

ar(ASP) = 1/4 x (ADB)



ar(PQRS) = ar(ABCD) - ar(PQB) - ar(SDR) - ar(CRQ) - ar(ASP)



ar(PQRS) = ar(ABCD) - 1/4 x (ar(ABC)+ar(ADC)+ar(CDB)+ar(ADB))

ar(PQRS) = ar(ABCD) - 1/4x (2 x ar(ABCD))

ar(PQRS) =1/2 ar(ABCD)

Hence, proved