Question

A field in the form of a parallelogram has one of its diagonals 42m long and the

perpendicular distance of this diagonal from either of the outlying vertices is 80 cm. Find

the area of the field.


please answer​

Answer 1

$\frak{Given}\begin{cases}\sf{It's \;a \;\bf{parallelogram.}}\\\sf{One \;Diagonal=\bf{42\:cm.}}\\\sf{Perpendicular\:distance=\bf{80\:cm.}}\end{cases}$

$\underline{\textit{\textbf{To\;Find:}}}$

• The area of the parallelogram.

${\bf{\underbrace{\underline{Required\:Solution:-}}}}$

» Consider the attachment as the figure of this question.

So,

• AC = diagonal = 42 cm .
• BE & DF = perpendicular distance = 80 cm .

» To find the area of the parallelogram, we have to break this figure into two parts.

• First part = ∆ ABC
• Second part = ∆ ADC

» We would seperately find the area of both these figures and then add them so as to get the final answer.

$\underline{\frak{Finding\;the \;area\:of\:\sf{Triangle\:ABC\:and\:ADC:}}}$

In ∆ABC ,

• AC = Base of the triangle = 42 cm
• BE = Perpendicular Height = 80 cm

So,

$\colon \implies \sf \: Area _{ \: (\triangle \: ABC)} = \frac{1}{2} \times AC \times BE \\ \\ \colon \implies \sf \: Area _{ \: ( \triangle \: ABC)} = \frac{1}{ \cancel2} \times \cancel{42 } \times 80 \: {cm}^{2} \\ \\ \colon \implies \sf \: Area _{ \: ( \triangle \: ABC)} = 21 \times 80 \: {cm}^{2} \\ \\ \colon \dashrightarrow \underline{ \boxed{ \tt \pink{ \: Area _{ \: ( \triangle \: ABC)} = 1680 \: {cm}^{2} }}}$

Similarly,

In ∆ ADC ,

• AC = Base of the triangle = 42 cm.
• DF = Perpendicular height = 80 cm.

So,

$\colon \rarr \sf \: Area _{ \:( \triangle \: ADC)} = \frac{1}{2} \times AC \times DF\\ \\ \colon \rarr \sf \: Area _{ \: ( \triangle \: ADC)} = \frac{1}{ \cancel2} \times \cancel{42} \times 80 \: {cm}^{2} \\ \\ \colon \dashrightarrow \underline{ \boxed{ \tt\pink{Area _{ \: ( \triangle \: ADC)} = 1680 \: {cm}^{2} }}}$

Now,

$\star\sf\;Area\;of\;Parallelogram=Area\:of\:\triangle\:ABC+Area\:of\:\triangle\:ADC$

$\colon \leadsto \sf \: Area \: of \:ABCD = 1680 \: {cm}^{2} + 1680 \: {cm}^{2} \\ \\ \colon \dashrightarrow \underline{ \boxed{ \bf{ \red{Area \: of \: ABCD = 3360 \: {cm}^{2} }}}}$

$\therefore\underline{\sf{\;Area\;of\;Parallelogram=\bf{3360\:cm^2}}}$