Question

What is the area of a parallelogram of base 11em and corresponding height 6cm? ​

Answer 1

$\large {\dag \; {\underline{\underline{\blue{\pmb{\textbf{\textsf{ \; Solution :- }}}}}}}}$

Formulae Used :-

The Area formula of a Parallelogram =

$\begin{gathered}\boxed{\rm{ area = Base × Height }} \\ \end{gathered}$

Base = 11 cm

height = 6 cm

Area =

$⟼11 \times 6 \\ = 66$

The Area of parallelogram is 66 CM²

More Formulae •:-

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf 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} </u></strong></p><p><strong><u>[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf 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} † </u></strong></p><p><strong><u>[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf 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} † F</u></strong></p><p><strong><u>[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf 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} † F </u></strong></p><p><strong><u>[tex]\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf 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} † F$

Answer 2

Question :-

• Find the Area of Parallelogram, whose Base is 11 cm and the Height is 6 cm .

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Answer :-

• Area of Parallelogram is 66 cm² .

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Explanation :-

• Here, Base of Parallelogram is given 11 cm . Height is given 6 cm . And, we have to calculate the Area of Parallelogram .

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➻ Formula Required :-

• $\sf{Area \: of \: Parallelogram = B \times H}$

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➻ Where ,

• B denotes to the Base .
• H denotes to the Height .

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➻ By substituting the given values :-

$\dag \: \: \sf{ Area \: _{Parallelogram} = Base \times Height}$

$\Longrightarrow \: \: \sf{Area \: _{Parallelogram} \: = \: 11 \: \times \: 6}$

$\Longrightarrow \: \: \sf{Area \: _{Parallelogram} \: = \: 66 \: {cm}^{2} }$

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Hence :-

• Area = 66 cm² .

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$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \\ \underline{ { \pmb {\sf \red{ \dag \: \: More \: Formulas \: \: \dag}}}} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Square = Side \times Side} \\ \\ \\ \footnotesize\bigstar \: \sf{Area \: of \: Rectangle = Lenght \times Breadth} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Triangle = \frac{1}{2} \times Base \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Parallelogram = Base \times Height} \\ \\ \\ \footnotesize \bigstar \: \sf{Area \: of \: Trapezium = \frac{1}{2} \times [ \: A + B \: ] \times Height } \\ \\ \\ \footnotesize \bigstar \: \sf {Area \: of \: Rhombus = \frac{1}{2} \times Diagonal \: 1 \times Diagonal \: 2}\end{array}}\end{gathered}\end{gathered}$