Answers

2015-03-26T20:02:53+05:30
We know that the area of a triangle is half of that of a parallelogram if they are on the same base and between the same parallels.
In case triangle XCB, base is half. Therefore, area of triangle XCB
= 1/2 x 1/2 ar(ABC)
=1/4 ar(ABC)

Now ar(AXCD) = ar(ABCD) - ar(XCB)
= 1 - 1/4
= 3/4
As area of AXCD is 3/4ar(ABCD), area of ABCD
= 24 x 4/3
= 32cm²

We know that a diagonal divides a parallelogram into two triangles of equal area.
Thus area of ABC = 1/2 ar(ABCD)
or, ar(ABC) = 1/2 x 32
or, ar(ABC) = 16cm² 

Hope that helps !!
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2015-03-26T20:03:19+05:30
GIVEN
X is the m.p. of AB
ar.AXCD=24cm^2

TO FIND
the ar.ABC

SO,
ar.AXC=1/2 ar.ABC (CX is the median as X is the m.p.)
ar.ABC=ar.ADC (AC is the diagonal and the diagonal divides the //gm into 2                                      cong.triangles)
Ar.AXCD=1/2 ar.ADC+ADC
AR.AXC=1/3 ADC=8cm^2
hence ar.ABC=8*2=16cm^2

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