A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box. 1) The cost of sheet for it, if a sheet measuring Im² costs ₹20
Answers & Comments
please mark as brainliest !!
Step-by-step explanation:
Given: A plastic box 1.5 m long, 1.25 m wide and 65 cm deep. It is opened at the top. The cost of 1 m² sheet is ₹20.
To find: Area of the sheet required for making the box and its cost.
Since the box is opened at the top, it has only 5 surfaces, including the 4 walls and the base. The area of the sheet required for making the cuboidal box includes the 4 walls of the box and the base.
Hence, the area of the sheet can be obtained by adding the area of the base to the lateral surface area of the cuboidal box.
Lateral surface area of cuboid = 2(l + b)h
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened 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 costs ₹ 20.
The cost of the sheet to create the box will be equal to area of the sheet multiplied by rate of 1m² sheet.
Length, l = 1.5 m
Breadth, b = 1.25 m
Height, h = 65 cm = 65/100 m = 0.65 m
The area of the sheet required to make the box = lb + 2(l + b)h
= (1.5 m × 1.25m) + 2 × (1.25 m + 1.5 m) × 0.65 m
= 1.875 m² + 2 × 2.75 m × 0.65 m
= 1.875 m² + 3.575 m²
= 5.45 m²
Therefore, the cost of the sheet = Rate of the sheet × Area of the sheet
= ₹ 20 / m² × 5.45 m²
= ₹109
Therefore, the area of the sheet required for making the open box is 5.45 m² and the cost of the sheet is ₹109.
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Dimensions of plastic box are as follow:
Length of box, l = 1.5 m
Breadth of box, b = 1.25 m
Height of box, h = 65 cm = 0.65 m
Now, Area of sheet required for making the open box is evaluated as
[tex]\bf\: Area_{(Sheet\:required)} \\ [/tex]
[tex]\sf\: = \: 2(l + b) \times h + lb \\ [/tex]
[tex]\sf\: = \: 2(1.5+ 1.25) \times 0.65 + 1.5 \times 1.25 \\ [/tex]
[tex]\sf\: = \: 2(2.75) \times 0.65 + 1.875 \\ [/tex]
[tex]\sf\: = \: 3.575 + 1.875 \\ [/tex]
[tex]\sf\: = \: 5.45 \: {m}^{2} \\ [/tex]
Thus,
[tex]\implies\bf\:Area_{(Sheet\:required)} = 5.45 \: {m}^{2} \\ [/tex]
Further given that,
Cost of sheet measuring 1 m² = ₹ 20
So, Cost of sheet measuring 5.45 m² = 5.45 × 20 = ₹ 109
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ {More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\: \end{array} }}[/tex]