ABCD is a parallelogram and X is the mid point of AB. If ar(AXCD) =24cm²,then prove that ar(ABC) =16cm².
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Let the height of the parallelogram be 'h'
ar(ACX)=1/2 × AX ×h
ar(ABCD)=AB×h
As, 2AX=AB, we can conclude that
ar(ACX)=1/4×ar(ABCD) (1)
Also, diagonal divides parallelogram into two triangles of equal area.
Therefore, ar(ACD)=1/2×ar(ABCD) (2)
Adding (1) and (2)
ar(ACX)+ar(ACD)=3/4×ar(ABCD)
ar(AXCD)=3/4×ar(ABCD)
24=3/4×ar(ABCD)
32=ar(ABCD)
As, ar(ABC)=1/2×ar(ABCD)
ar(ABC)=1/2×32
ar(ABC)=16cm^2
ar(ACX)=1/2 × AX ×h
ar(ABCD)=AB×h
As, 2AX=AB, we can conclude that
ar(ACX)=1/4×ar(ABCD) (1)
Also, diagonal divides parallelogram into two triangles of equal area.
Therefore, ar(ACD)=1/2×ar(ABCD) (2)
Adding (1) and (2)
ar(ACX)+ar(ACD)=3/4×ar(ABCD)
ar(AXCD)=3/4×ar(ABCD)
24=3/4×ar(ABCD)
32=ar(ABCD)
As, ar(ABC)=1/2×ar(ABCD)
ar(ABC)=1/2×32
ar(ABC)=16cm^2
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