In the figure,if parallelogram ABCD and recta | vedclass

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(B) Given that the area of parallelogram $ABCD$ is equal to the area of rectangle $ABEM$. Both share the same base $AB$ and lie between the same paral

Vedclass · Accepted Answer

Perimeter of $ABCD >$ Perimeter of $ABEM$

Vedclass · Answer

(B) Given that the area of parallelogram $ABCD$ is equal to the area of rectangle $ABEM$. Both share the same base $AB$ and lie between the same parallel lines $AB$ and $MC$.
Since the area of a parallelogram is given by $	ext{base} 	imes 	ext{height}$,and the height is the perpendicular distance between the parallel lines,both figures have the same height $AM = BE$.
In a right-angled triangle $AMD$,the hypotenuse $AD$ is always greater than the perpendicular side $AM$ $(AD > AM)$.
Similarly,in the right-angled triangle $BCE$,the hypotenuse $BC$ is greater than the side $BE$ $(BC > BE)$.
Since $AM = BE$,we have $AD > AM$ and $BC > BE$.
The perimeter of rectangle $ABEM = 2(AB + AM)$.
The perimeter of parallelogram $ABCD = AB + BC + CD + DA = AB + BC + AB + AD = 2AB + BC + AD$.
Since $BC > BE$ and $AD > AM$,it follows that $2AB + BC + AD > 2AB + BE + AM = 2AB + 2AM = 2(AB + AM)$.
Therefore,the perimeter of parallelogram $ABCD >$ perimeter of rectangle $ABEM$.

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

Similar Questions

In the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.

$ABCD$ 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 $\text{cm}^2$).

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 $\text{ar}(ABC) = 24 \text{ cm}^2$.

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 $\operatorname{ar}(\triangle BDE) = \frac{1}{4} \operatorname{ar}(\triangle ABC).$

In the figure,the area of parallelogram $ABCD$ is:

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