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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

$12$

Vedclass · Answer

(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

Which of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.

In parallelogram $PQRS$,$PQ = 15 \, cm$. Altitudes $SM$ and $SN$ are corresponding to bases $PQ$ and $QR$ respectively. If $SM = 6 \, cm$ and $SN = 10 \, cm$,find $QR$ and the perimeter of $PQRS$.

In the figure,if parallelogram $ABCD$ and rectangle $ABEM$ are of equal area,then:

Write True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$

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