Find the area of the cardboard required to make an open box of length

1.	Find the area of the cardboard required to make an open box of length 28 cm, breadth 0.5 m and height is 0.2 m10. Find the area of the four walls of the room whose length, breadth and height are 8 m, 6 m and 5respectively. Ignoring the door, windows and other openings, find the cost of whitewashing the wa
Answer» Question 1 ${\frak{Dimension_{\;(box)}}} \begin{cases} & \text{Length = \bf{28 \;cm} } \\ & \text{Breadth = 0.5 m = \bf{50\; cm}} \\ & \text{Height= 0.2 m = \bf{20 \;cm}} \end{cases}\\ \\$ $\dag\;{\underline{\frak{We\;know\;that,}}}\\ \\$ $\star\;{\boxed{\sf{\pink{T.S.A_{\;(cuboid)} = 2(lb + bh + hl)}}}}\\ \\$ $\qquad\qquad\qquad:\implies\sf 2(28 \times 50 + 50 \times 20 + 20 \times 28)\\ \\$ $\qquad\qquad\qquad\qquad:\implies\sf 2(1400 + 560 + 1000)\\ \\$ $\qquad\qquad\qquad\qquad\qquad:\implies\sf 2(2460)\\ \\$ $\qquad\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{\pink{5920\;cm^2}}}}}\;\bigstar\\ \\$ ☯ Box is OPEN at top Therefore, AREA of top SURFACE is, $\\$ $\qquad\qquad\qquad\qquad:\implies\sf l \times b\\ \\$ $\qquad\qquad\qquad\quad:\implies\sf 28 \times 50\\ \\$ $\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{\purple{1400\;cm^2}}}}}\;\bigstar\\ \\$ ⠀⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━ Therefore, REQUIRED area of box is, $\\$ $\qquad\qquad\qquad:\implies\sf 5920 - 1400\\ \\$ $\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{4520\;cm^2}}}}\;\bigstar\\ \\$ $\therefore\;{\underline{\sf{Area\;of\; cardboard\; required\;to\;make\;box\;is\; \bf{4520\;cm^2}}}}$ ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Question 2 ${\frak{Dimension_{\;(room)}}} \begin{cases} & \text{Length = \bf{8\; m} } \\ & \text{Breadth = \bf{6\; m}} \\ & \text{Height= \bf{5\; m}} \end{cases}\\ \\$ ☯ We have to FIND, Area of four walls. So, $\\$ $\dag\;{\underline{\frak{We\;know\;that,}}}\\ \\$ $\star\;{\boxed{\sf{\pink{C.S.A_{\;(cuboid)} = 2(l + b)h}}}}\\ \\$ $\qquad\qquad\qquad:\implies\sf 2(8 + 6) \times 5\\ \\$ $\qquad\qquad\qquad\quad:\implies\sf 2 \times 14 \times 5\\ \\$ $\qquad\qquad\qquad\qquad:\implies\sf 28 \times 5\\ \\$ $\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\sf{\purple{140\;m^2}}}}}\;\bigstar\\ \\$ $\therefore\;{\underline{\sf{Area\;of\;four\;walls\;of\;room\;is\; \bf{140\;m^2}}}}$

Find the area of the cardboard required to make an open box of length 28 cm, breadth 0.5 m and height is 0.2 m10. Find the area of the four walls of the room whose length, breadth and height are 8 m, 6 m and 5respectively. Ignoring the door, windows and other openings, find the cost of whitewashing the wa

Answer»

Question 1

${\frak{Dimension_{\;(box)}}} \begin{cases} & \text{Length = \bf{28 \;cm} } \\ & \text{Breadth = 0.5 m = \bf{50\; cm}} \\ & \text{Height= 0.2 m = \bf{20 \;cm}} \end{cases}\\ \\$

$\dag\;{\underline{\frak{We\;know\;that,}}}\\ \\$

$\star\;{\boxed{\sf{\pink{T.S.A_{\;(cuboid)} = 2(lb + bh + hl)}}}}\\ \\$

$\qquad\qquad\qquad:\implies\sf 2(28 \times 50 + 50 \times 20 + 20 \times 28)\\ \\$

$\qquad\qquad\qquad\qquad:\implies\sf 2(1400 + 560 + 1000)\\ \\$

$\qquad\qquad\qquad\qquad\qquad:\implies\sf 2(2460)\\ \\$

$\qquad\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{\pink{5920\;cm^2}}}}}\;\bigstar\\ \\$

☯ Box is OPEN at top Therefore, AREA of top SURFACE is,

$\\$

$\qquad\qquad\qquad\qquad:\implies\sf l \times b\\ \\$

$\qquad\qquad\qquad\quad:\implies\sf 28 \times 50\\ \\$

$\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{\purple{1400\;cm^2}}}}}\;\bigstar\\ \\$

⠀⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━

Therefore,

REQUIRED area of box is,

$\\$

$\qquad\qquad\qquad:\implies\sf 5920 - 1400\\ \\$

$\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\frak{4520\;cm^2}}}}\;\bigstar\\ \\$

$\therefore\;{\underline{\sf{Area\;of\; cardboard\; required\;to\;make\;box\;is\; \bf{4520\;cm^2}}}}$

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Question 2

${\frak{Dimension_{\;(room)}}} \begin{cases} & \text{Length = \bf{8\; m} } \\ & \text{Breadth = \bf{6\; m}} \\ & \text{Height= \bf{5\; m}} \end{cases}\\ \\$