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(D) In $\Delta ADF$ and $\Delta ECF$,we have:$\angle DAF = \angle CEF$ [Alternate interior angles]$AD = CE$ [Since $AD = BC$ and $BC = CE$ (Given)]$\a

Vedclass · Accepted Answer

$12$

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(D) In $\Delta ADF$ and $\Delta ECF$,we have:$\angle DAF = \angle CEF$ [Alternate interior angles]$AD = CE$ [Since $AD = BC$ and $BC = CE$ (Given)]$\angle ADF = \angle ECF$ [Alternate interior angles]Therefore,$\Delta ADF \cong \Delta ECF$ [By $ASA$ congruence rule]Thus,$DF = CF$ [$CPCT$].In $\Delta BCD$,$BF$ is the median because $DF = CF$.Since a median divides a triangle into two triangles of equal area,$	ext{ar}(\Delta BDF) = 	ext{ar}(\Delta BCF) = \frac{1}{2} 	ext{ar}(\Delta BCD)$.Given $	ext{ar}(\Delta DFB) = 3 \, 	ext{cm}^2$,so $	ext{ar}(\Delta BCD) = 2 	imes 3 = 6 \, 	ext{cm}^2$.Since the area of a triangle is half the area of a parallelogram on the same base and between the same parallels,$	ext{ar}(\Delta BCD) = \frac{1}{2} 	ext{ar}(	ext{parallelogram } ABCD)$.$6 = \frac{1}{2} 	ext{ar}(	ext{parallelogram } ABCD)$.$	ext{ar}(	ext{parallelogram } ABCD) = 12 \, 	ext{cm}^2$.

$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).

Similar Questions

$ABCD$ is a trapezium in which $AB \parallel DC$,$DC = 30 \, cm$ and $AB = 50 \, cm$. If $X$ and $Y$ are,respectively,the mid-points of $AD$ and $BC$,prove that $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

Area of a trapezium $=$ .........

In $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.

$(1)$ In $\Delta ABC$,$AD$ is an altitude. If $BC = 8 \text{ cm}$ and $AD = 5 \text{ cm}$,then $\text{ar}(\Delta ABC) = \dots \text{ cm}^2$.
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In a parallelogram $ABCD$,the base $AB$ and the corresponding altitude $DM$ are $16 \, cm$ and $12 \, cm$,then the area of $ABCD =$ ................ $cm^{2}$.

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