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(B) We have $ABCD$ as a parallelogram and $X$ as the mid-point of $AB$. Now,$	ext{ar}(ABCD) = 	ext{ar}(AXCD) + 	ext{ar}(	riangle XBC) \dots(1)$Sin

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(B) We have $ABCD$ as a parallelogram and $X$ as the mid-point of $AB$. Now,$	ext{ar}(ABCD) = 	ext{ar}(AXCD) + 	ext{ar}(	riangle XBC) \dots(1)$Since the diagonal $AC$ of a parallelogram divides it into two triangles of equal area,we have $	ext{ar}(ABCD) = 2 	imes 	ext{ar}(	riangle ABC) \dots(2)$Since $X$ is the mid-point of $AB$,the median $CX$ of $	riangle ABC$ divides it into two triangles of equal area. Thus,$	ext{ar}(	riangle XBC) = \frac{1}{2} 	ext{ar}(	riangle ABC) \dots(3)$Substituting $(1)$ and $(3)$ into $(2)$,we get: $2 	imes 	ext{ar}(	riangle ABC) = 24 + \frac{1}{2} 	ext{ar}(	riangle ABC)$$2 	imes 	ext{ar}(	riangle ABC) - \frac{1}{2} 	ext{ar}(	riangle ABC) = 24$$\frac{3}{2} 	ext{ar}(	riangle ABC) = 24$$	ext{ar}(	riangle ABC) = \frac{24 	imes 2}{3} = 16 	ext{ cm}^2$.Since $16 	ext{ cm}^2 
eq 24 	ext{ cm}^2$,the given statement is False.

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

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Write True or False and justify your answer:$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.