Question

ABCD is a parallelogram and X is the midpoint of AB. If ar(ADC) = 24cm^2 then find ar(AXCD).

Answer 1

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 

I hope it will be helpful for you

Answer 2

Heya mate, this side,

Here is your answer, ⬇️⬇️

${ANSWER :-}$
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$<i>$ Here, we have,
ar(ADC) = 24 cm^2

Now, Here, in parallelogram ABCD,
AC is the diagonal

Thus, AC divides ABCD into two triangles which have equal area.
Thus, ar(ADC) = ar(ABC). .......... (1)
_ _ _ _ _ _ _ _

Now, ar(ABCD) = ar(ADC) + ar(ABC)
Or, ar(ABCD) = ar(ADC) + ar(ADC) [from(1)]
Or, ar(ABCD) = 2 * ar(ADC).

Thus, ar( ABCD) = 2 * ar(ADC)
Or, ar(ABCD) = 2 * 24 = 48 cm^2
_ _ _ _ _ _ _ _

Now, take point Y on DC such that, Y is the mid point of DC.

Now, join X and Y to form XY.

Now, we have that the line joining the mid points of two side of a parallelogram divides it into two two parallelogram of equal area.

So, ar(ADYX) = ar(BCYX). ............ (2)
_ _ _ _ _ _ _ _

Again, ar(ABCD) = ar(ADYX) + ar(BCYX)
Or, ar(ABCD) = ar(BCYX) + ar(BCYX). [from(2)]
Or, ar(ABCD) = 2 * ar(BCYX)
Or, 1/2 * ar(ABCD) = ar(BCYX).

So, ar(BCYX) = ar(ADYX) =
= 1/2 * ar(ABCD)
= 1/2 * 48 = 24cm^2. .................... (3)

_ _ _ _ _ _ _ _

Again, in parallelogram BCYX,
XC is a diagonal.

So, ar(XYC) = ar(XBC). ................. (4)

And, ar(BCYX) = ar(XYC) + ar(XBC)
Or, ar(BCYX) = ar(XYC) + ar(XYC) [from(4)]
Or, ar(BCYX) = 2 * ar(XYC)
Or, 1/2 * ar(BCXY) = ar(XYC)

Now, ar(XYC) = 1/2 * ar(BCXY)
= 1/2 * 24 = 12cm^2. ................. (5)

_ _ _ _ _ _ _ _

Now, ar(AXCD) = ar(ADYX) + ar(XYC)
= (24 + 12) cm^2 [from (3) and (5)]
= 36 cm^2 .
_ _ _ _ _ _ _ _

➡️ Hence, the area of quadrilateral AXCD is 36 cm^2.

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$<b>$ Hope this helps,.

If helps, please mark it as brainliest ^_^

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REGARDS, ARNAB ✌️

THANK YOU