Given :-
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- Base of parallelogram = 10 cm
- Height of Parallelogram = 7 cm
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To find:
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we know:-
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[tex] \bigstar \boxed{ \rm Area \: of \: parallelogram = base \times height}[/tex]
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So:-
[tex] \dashrightarrow\sf Area \: of \: parallelogram = base \times height \\ [/tex]
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[tex] \dashrightarrow\sf Area \: of \: parallelogram =10 \times 7 \\ [/tex]
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[tex] \dashrightarrow\sf Area \: of \: parallelogram =70 {cm}^{2} \\ [/tex]
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[tex] \therefore \underline{\bf Area \: of \: parallelogram is \: equal \: to \red{ 70 {cm}^{2}}}[/tex]
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Know More:-
[tex]\small\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf \small{Formulas\:of\:Areas:-}}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}[/tex]
