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

State whether each of the following statements is true or false:
$(1)$ Area of a parallelogram $= \text{base} \times \text{corresponding altitude}$.
$(2)$ Area of a rhombus $= \frac{1}{2} \times \text{Product of its diagonals}$.
$(3)$ Area of a square $= (\text{Side})^2$.

If $P$ is any point on the median $AD$ of a $\triangle ABC$,then $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$. State whether this statement is True or False.

In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.

In the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$

The area of the parallelogram $ABCD$ is $90 \, cm^{2}$ (see figure). Find:
$(i) \; ar(ABEF)$
$(ii) \; ar(ABD)$
$(iii) \; ar(BEF)$

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