Question

The length breath and height of a room are 5m 4m and 3m respectively find the cost of washing the walls of the room and the ceiling at the rate of 12.50m square​

Answer 1

❁Question:-

• The length breath and height of a room are 5m 4m and 3m respectively find the cost of washing the walls of the room and the ceiling at the rate of 12.50m square

✿Given:-

• Length of room= 5m
• Breadth of room= 4m
• Height of room = 3m

❀To find:-

• Cost of white washing the room and the ceiling at the rate of Rs 12.50m²

۞Solution:-

$\underline{ \boxed{ \sf \: Area \: to \: be \: white \: washed = Area \: of \: cuboid \: -Area \: of \: base}}$

• Formula for area of cuboid =2(lb+bh+lh)
• Area of base = length ×breadth

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ : \implies} \: \: 2(lb + bh + hl) - lb$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ : \implies}\sf \: 2(lb) +2( bh) +2( hl) - lb$

$\implies \sf \: lb \: + 2(bh) + 2(lh)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \: {(5 \times 4) + (2 \times 4 \times 3) + (2 \times 5 \times 3) \brack} {m}^{2}$

$\\ \: \: \: \: \implies(20 + 24 + 30) {m}^{2}$

$\\ \: \: \: \: \: \: \implies \: 74 {m}^{2}$

⇒Cost of white washing per m²= Rs 7.50

⇒Cost of white washing 74m² area=Rs(74×7.50)

⇒Rs 555

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Learn more ✔:-

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

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Formulas of Surface Area

$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of \: surface \: Area:-}\\ \\ \star\sf cuboid=2(lb + bh + hl)\\ \\ \star\sf cube=6 {a}^{2} \\\\ \star\sf right \: circular \: cone=\pi \: r(l + r)\\\\ \star \sf sphere=4\pi {r}^{2} \\ \\ \star \sf hemisphere =3\pi {r}^{2} \ \\ \end {array}}\end{gathered}$