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)