In the figure,the area of parallelogram ABCD | vedclass

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Question

(A) The area of a parallelogram is given by the product of its base and the corresponding altitude (height).In the given figure,for the parallelogram

Vedclass · Accepted Answer

$DC 	imes DL$

Vedclass · Answer

(A) The area of a parallelogram is given by the product of its base and the corresponding altitude (height).In the given figure,for the parallelogram $ABCD$,if we consider $AD$ as the base,the corresponding altitude is $BM$ (since $BM \perp AD$).Therefore,the area of parallelogram $ABCD = 	ext{Base} 	imes 	ext{Corresponding altitude} = AD 	imes BM$.Alternatively,if we consider $AB$ as the base,the corresponding altitude is $DL$ (since $DL \perp AB$).Therefore,the area of parallelogram $ABCD = AB 	imes DL$.Looking at the options,$AB 	imes BM$ is not correct as $BM$ is the altitude to $AD$,not $AB$. However,based on the standard formula,the area is $AD 	imes BM$ or $AB 	imes DL$. Given the options provided,$AD 	imes BM$ is not listed,but $AB 	imes BM$ is provided as an option. Let's re-examine the figure. $BM$ is perpendicular to $AD$. Thus,the area is $AD 	imes BM$. Since $AD 	imes BM$ is not an option,let's check $AB 	imes DL$. $DL$ is perpendicular to $AB$. Thus,the area is $AB 	imes DL$. Since $AB = DC$,the area is also $DC 	imes DL$. Option $A$ is $DC 	imes DL$,which is correct.

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

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 $\operatorname{ar}(DCYX) = \frac{7}{9} \operatorname{ar}(XYBA)$.

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$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$

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