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1) ABCD is a paralellogram. X and Y are mid points of BC and CD respectively.Prove that area (AXY) = 3/8 area(paralellogram ABCD)

2) In a triangle ABC P and Q are respectively the mid points of AB and BC and R is the mid point of AP. Prove that:

1) area(PRQ) =1/2 area(ARC)

2) area(RQC) = 3/8 area(ABC)

3) area(PBQ) = area(ARC)

1) ABCD is a paralellogram. X and Y are mid points of BC and CD respectively.Prove that area (AXY) = 3/8 area(paralellogram ABCD)

2) In a triangle ABC P and Q are respectively the mid points of AB and BC and R is the mid point of AP. Prove that:

1) area(PRQ) =1/2 area(ARC)

2) area(RQC) = 3/8 area(ABC)

3) area(PBQ) = area(ARC)

(1)

given: ABCD is a parallelogram. X and Y are the mid-points of BC and CD.

TPT: area(ΔAXY)=3/8 area(parallelogram ABCD)

Proof: construction: join AC and BD.

in ΔADC , AY is the median. since median divides the triangle into two equal areas.

area(ΔAYC)=1/2area(ΔADC)=1/4 area(parallelogram ABCD) .........(1)

since diagonals of a parallelogram divides it into two equal areas

similarly in ΔABC, area(ΔAXC)=1/2 area(ΔABC)=1/4 area(parallelogram ABCD) .............(2)

from (1) and (2) area(quadrilateral AXCY) = area(ΔAYC)+area(ΔAXC)=(1/4+1/4) area(parallelogram ABCD)

therefore area(quadrilateral AXCY)=1/2area(parallelogram ABCD) .........(3)

area(ΔAXY)=area(quadrilateral AXCY)-area(ΔXCY)

[since area(ΔXCY)=1/4area(ΔBCD)=area(parallelogram ABCD)]

area(ΔAXY)=1/2 area(parallelogram ABCD)-1/8area(parallelogram ABCD)

=3/8 area(parallelogram ABCD)

(2)

given: ABC is a triangle, P and Q are the mid-points of AB and AC. R is the mid-point of AP.

TPT:

1) area(PRQ) =1/2 area(ARC)

2) area(RQC) = 3/8 area(ABC)

3) area(PBQ) = area(ARC)

proof:

let the area of triangle ABC be x.

(i) area (ΔPRQ)=1/2 area(APQ)[since RQ is a median and median divides triangle into equal areas]

= area(ΔABC)=............(1)

area (ΔARC)=1/2 area(ΔAPC) [in the ΔACP , CR is the median]

=1/2*1/2 area(ΔABC) [since CP is the median of ABC]

=1/4*x=x/4 ............(2)

from (1) and (2) area(ΔPRQ)=1/2 area(ΔARC)

(ii)

area(ΔRQC)=1/2 area(ΔARC) [since RQ is the median of triangle ARC]

=1/2*1/2area(ΔAPC)=1/4*1/2 area(ΔABC)=1/8x

(iii)

area(ΔPBQ)=area(ΔPQC)[the triangle with same base and between same set of parallel lines ]

=1/2 area(ΔAPC) [since PQ is the median of triangle APC]

=1/2*1/2 area(ΔABC)

=1/4 area(ΔABC)=x/4

from (2) area (ΔARC)=x/4

hence area(ΔPBQ)=area(ΔARC)

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