Question

Area of four walls of a room of dimensions 25m,35m,40m is
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plz I need it fast

Answer 1

Answer:

area of four walls=2 (l+b)×h

2(40+35)×25

2×75×25

150×25

3750sqm

Answer 2

Given: Ratio of length and breadth of a rectangular park is 5:3. & Area of park is 2160 m².

Need to find: Dimensions of rectangular park?

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❍ Let's consider area of four walls of a room be x.

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As we know that,

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$\begin{gathered}\star\:{\underline{\boxed{\frak{Lateral \: surface \: area \:_{\:(cuboid)} = 2(l + b)h}}}}\\\\\\ \bf{\dag}\:{\underline{\frak{Putting\:given\:values\:in\:formula,}}}\\\\\\ :\implies\sf 2(25 + 35)40 \\\\\\ :\implies\sf 2(60)40 \\\\\\ :\implies\sf 120(40) \\\\\\ :\implies{\underline{\boxed{\frak{\purple{x = 4800m^2}}}}}\:\bigstar\\\\\end{gathered}$

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Therefore,

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Area of cuboid, x = 4800m².

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$\therefore\:{\underline{\sf{Hence,\: Area \: of \: four \: walls \:is\:\bf{4800\:m^2}\: \sf{respectively}.}}}$

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$\underline{\underline{\maltese\:\:\textbf{ \textsf{Additional knowledge... }}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$