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CBSE - IX - Mathematics - Areas of Parallelograms and Triangles

if P,Q,R,S be respectively the midpoints of the sides AB,BC,CD and DA of a quadrilateral ABCD. Show that PQRS is a parallelogram such that ar(PQRS)=1/2 ar(ABCD)

Asked by 23rd February 2013, 7:27 PM
Answered by Expert


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
Answered by Expert 23rd February 2013, 9:41 PM

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