Question

P,Q,R and S are the midpoints of the sides AB, BC, CD and DA respectively of the parallelogram ABCD. Show that PQRS is a parallelogram whose area is half that of the parallelogram ABCD.

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) = PQ2 / BC2 = 1/4

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

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

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

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

 

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

 

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

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

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

Hence, proved

Answer 2

Answer:

Ryan 5tff

Step-by-step explanation:

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