[tex]{\huge{\boxed{\boxed{\purple{\mathcal{Question}}}}}}[/tex]
the area of a parallelogram and a square are the same if the perimeter of the square is 160 m and the height of the parallelogram is 20 m find the length of the corresponding base of the parallelogram
[tex]\huge\underline{\mathbb\red{❥︎A}\green{N}\mathbb\blue{S}\purple{W}\mathbb\orange{E}\pink{R}}\: [/tex]
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[tex]{\huge{\boxed{\boxed{\purple{\mathcal{Question}}}}}}[/tex]
the area of a parallelogram and a square are the same if the perimeter of the square is 160 m and the height of the parallelogram is 20 m find the length of the corresponding base of the parallelogram
[tex]\huge\underline{\mathbb\red{❥︎A}\green{N}\mathbb\blue{S}\purple{W}\mathbb\orange{E}\pink{R}}\: [/tex]
Given that,
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Perimeter of square = 160 m
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[tex]\begin{gathered}\dag\;{\underline{\frak{As\;we\;know\;that,}}}\\ \\\end{gathered} [/tex]
,
[tex]\begin{gathered}\star\;{\boxed{\sf{\pink{Perimeter_{\;(square)} = 4 \times side}}}}\\ \\\end{gathered} [/tex]
Therefore,
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[tex]\begin{gathered}:\implies\sf 4 \times side = 160\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies\sf side = \cancel{ \dfrac{160}{4}}\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies{\underline{\boxed{\frak{\purple{side = 40\;m}}}}}\;\bigstar\\ \\\end{gathered} [/tex]
[tex]\therefore\;{\underline{\sf{Thus,\;side\;of\; square\;is\; \bf{40\;m}.}}}[/tex]
.
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Now, Finding area of square,
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[tex]\begin{gathered}\star\;{\boxed{\sf{\pink{Area_{\;(square)} = side \times side}}}}\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies\sf Area_{\;(square)} = 40 \times 40\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies{\underline{\boxed{\frak{\purple{Area_{\;(square)} = 1600\;m^2}}}}}\;\bigstar\\ \\\end{gathered} [/tex]
[tex]\therefore\;{\underline{\sf{Area\;of\; square\;is\; \bf{1600\;m^2}.}}}[/tex]
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[tex]\begin{gathered}\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}\\ \\\end{gathered} [/tex]
[tex] \sf Area_{\;(parallelogram)} = Area_{\;(square)}Area
(parallelogram)[/tex]
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[tex]\begin{gathered}\star\;{\boxed{\sf{\pink{Area_{\;(parallelogram)} = Base \times Height}}}}\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}\sf Here \begin{cases} & \sf{Area = \bf{1600\;m^2}} \\ & \sf{Height = \bf{20\;m}} \end{cases}\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies\sf Base \times 20 = 1600\\ \\\end{gathered}
[/tex]
[tex]\begin{gathered}:\implies\sf Base = \cancel{ \dfrac{1600}{20}}\\ \\\end{gathered} [/tex]
[tex]\begin{gathered}:\implies{\underline{\boxed{\frak{\purple{Base = 80\;m}}}}}\;\bigstar\\ \\\end{gathered} [/tex]
[tex]\therefore\;{\underline{\sf{Length\;of\; corresponding\;base\;of\; parallelogram\;is\; \bf{80\;m}.}}}[/tex]
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