Question

Area of a parallelogram whose base is 35 cm. and height is 20. 5 cm is
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Answer 1

$\sf Given \begin{cases} & \sf{Base\:of\:parallelogram = \bf{35\:cm}} \\ & \sf{Height\:of\:parallelogram = \bf{20.5\:cm}} \end{cases}$

To find : Area of a parallelogram ?

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$\dag\;{\underline{\frak{As\:we\:know\:that\:,}}}$

$\star\;{\boxed{\sf{\pink{Area_{\;(parallelogram)} = Base \times Height}}}}$

$:\implies\sf 35 \times 20.5 \: cm$

$:\implies{\underline{\boxed{\frak{\purple{717.5\:cm^2}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{\; Area\;of\;parallelogram\;is\;717.5\;cm^2.}}}$

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$\qquad\quad\boxed{\bf{\mid{\overline{\underline{\bigstar\: More \ to \ know\: :}}}}\mid}$⠀

• Area of triangle = $\sf \dfrac{1}{2}$ × base × height

• Area of rhombus = $\sf \dfrac{1}{2} \times \sf \ d_1 \times d_2$

• Area of equilateral triangle = $\sf \: \dfrac{\sqrt{3}}{4} \: a^2$

• Height of equilateral triangle = $\sf \: \dfrac{\sqrt{3}}{2} \: a$

Answer 2

Correct Question-:

• Area of a parallelogram whose base is 35 cm and height is 20. 5 cm is __________ .

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ Area_{(Parallelogram)}  \: = \: 717.5 cm^{2} }}}}}$

$\dag{\sf{\large { EXPLANATION-:\:}}}$

• $\frak{Given \:\: -:} \begin{cases} \sf{The\:base\:of\:Parallelogram \:\:is\:= \frak{35cm}} & \\\\ \sf{Height \:of\:Parallelogram \: \:=\:\frak{20.5cm}}\end{cases}$

• $\frak{To \:Find\: -:} \begin{cases} \sf{The\:Area\:of\:Parallelogram .}\end{cases}$

$\dag{\sf{\large { Solution-:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ Area_{(Parallelogram)}  \: = \: Base \times Height }}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{The\:base\:of\:Parallelogram \:\:is\:= \frak{35cm}} & \\\\ \sf{Height \:of\:Parallelogram \: \:=\:\frak{20.5cm}}\end{cases}$

$\dag{\sf{\large { Now-:\:}}}$

• $\implies{\sf{\large { \:\:35 cm\times 20.5 cm\: }}}$

• $\implies{\sf{\large { \:\:717.5cm^{2} \: }}}$

$\dag{\sf{\large { Hence-:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ Area_{(Parallelogram)}  \: = \: 717.5 cm^{2} }}}}}$

$\star{\sf{\large { Figure \:of\:Parallelogram-:}}}$

• $\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1,1)(1,1)(6,1)\put(0.4,0.5){\bf D}\qbezier(1,1)(1,1)(1.6,4)\put(6.2,0.5){\bf C}\qbezier(1.6,4)(1.6,4)(6.6,4)\put(1,4){\bf A}\qbezier(6,1)(6,1)(6.6,4)\put(6.9,3.8){\bf B}\end{picture}$

$\boxed{\sf{\underline {\large {| More \:to\:Know-:|}}}}$

• Formulas of area :

• $\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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