Question

A plastic box 1.5m long, 1.25m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:

(i) the area of the sheet required for making the box.

(ii) the cost of sheet for it, if a sheet measuring 1m2 cost Rs. 20.

Answer 1

Length of box = 1.5 m,Breadth = 1.25 m,Height = 0.65 m

⇒ Area to be white-washed =

Area of cuboid − Area of base

(i) Area of sheet requred =2lh+2bh+lb

___[Box is open]

=[2×1.5×0.65+2×1.25×0.65+1.5×1.25] m²

=(1.95+1.625+1.875)m² = 5.45 m²

ii) Cost of sheet per m² area = Rs. 20

Cost of sheet of 5.45m² area = Rs. 5.45×20

= Rs. 109

Answer 2

$\LARGE\mathfrak{\underline{✯\; Given :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Length of the box ( l )= 1.5 m}$

$\qquad\tt{:}\longrightarrow\large\textsf{Width of the box ( b ) = 1.25 m}$

$\qquad\tt{:}\longrightarrow\large\textsf{Depth of the box = 65 cm ( h ) = 0.65 m}$

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; To \; \; Find :-}}$

$\large\textsf{ }$

1. The area of the sheet required for making the box = ?
2. The cost of the sheet for it , if a sheet measuring 1m² cost ₹20 .

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; Formula :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{purple}{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }}$

• l = Length
• b = Width or Breadth
• h = Depth or Height

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; How \; \; To \; \; Solve :-}}$

$\large\textsf{ }$

• As we have been provided with the information that a box is made with open at the . We have to find the area of this box , so to find the area of this box first we have to find the Total Surface Area ( T.S.A. ) of the box and then we should subtract the Area of the top of the box box .
• By calculating the above steps we would get the Area of the sheet required to make a box with open top .
• Now once we get the Area of the sheet we can calculate the total cost of the sheet required to make the box .

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; Solution :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{ {\large\textsf{Area}}_{\large\textsf{( \; Sheet \; )}} = {\large\textsf{T.S.A.}}_{\large\textsf{( \; Box \; )}} - {\large\textsf{Area}}_{\normalsize\textsf{( \; Top of the box \; )}} }$

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = [2 ( lb + bh + hl )] - ( l × b ) }$

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = 2 [( 1.5 × 1.25 ) + ( 1.25 × 0.65 ) + ( 0.65 × 1.5 )] - ( 1.5 × 1.25 )}$

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = 2 ( 1.875 + 0.8125 + 0.975 ) - 1.875}$

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = 2 ( 3.6625 ) - 1.875}$

$\qquad\tt{:}\longrightarrow\large\textsf{\; \; \; = 7.325 - 1.875}$

$\qquad\tt{:}\longrightarrow\boxed{\large\mathfrak\textcolor{magenta}{\; \; \; = \; 5.45 \; m²}}$

$\large\textsf\textcolor{orange}{∴ Area of the sheet required = 5.45 m²}$

$\large\textsf{ }$

◈ ━━━━━━━ ⸙ ━━━━━━━ ◈

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{Cost of the sheet = 20 × 5.45}$

$\qquad\tt{:}\longrightarrow\boxed{\large\mathfrak\textcolor{magenta}{\; \; \; = ₹ 109}}$

$\large\textsf\textcolor{orange}{∴ Cost of the sheet = ₹ 109}$

$\large\textsf{ }$

__________________________________________________________

$\large\textsf{ }$

$\LARGE\mathfrak{\underline{✯\; More \; \; Formulas:-}}$

$\large\textsf{ }$

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf\textcolor{red}{( \; Cuboid \; )}} = \large\textsf{2h ( l + b )} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{2 ( lb + bh + hl )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cuboid \; )}} = \large\textsf{l×b×h} }$

◈ ━━━━━━━ ⸙ ━━━━━━━ ◈

$\large\textsf{ {\large\textsf{L.S.A.}}_{\large\textsf\textcolor{red}{( \; Cube \; )}} = \large\textsf{4×l²} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{6 × l²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cube \; )}} = \large\textsf{l²} }$

◈ ━━━━━━━ ⸙ ━━━━━━━ ◈

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf\textcolor{red}{( \; Cylinder \; )}} = \large\textsf{2 × πrh} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{2πr × ( r + h )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cylinder \; )}} = \large\textsf{πr²h} }$

◈ ━━━━━━━ ⸙ ━━━━━━━ ◈

$\large\textsf{ {\large\textsf{C.S.A.}}_{\large\textsf\textcolor{red}{( \; Cone \; )}} = \large\textsf{πrl} }$

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf{( \; Cone \; )}} = \large\textsf{πr × ( r + l )} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Cone \; )}} } \large\textsf{ = \cfrac{\large\textsf{1}}{\large\textsf{3}} }\large\textsf{× πr²h}$

◈ ━━━━━━━ ⸙ ━━━━━━━ ◈

$\large\textsf{ {\large\textsf{T.S.A.}}_{\large\textsf\textcolor{red}{( \; Sphere \; )}} = \large\textsf{4πr²} }$

$\large\textsf{ {\large\textsf{Volume}}_{\large\textsf{( \; Sphere \; )}} } \large\textsf{ = \cfrac{\large\textsf{4}}{\large\textsf{3}} }\large\textsf{× πr³}$

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