Question

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 m

10. Find the area of the four walls of the room whose length, breadth and height are 8 m, 6 m and 5
respectively. Ignoring the door, windows and other openings, find the cost of whitewashing the wa​

Answer 1

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,

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

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

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

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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

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

☯ We have to find, Area of four walls. So,

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

$\star\;{\boxed{\sf{\pink{C.S.A_{\;(cuboid)} = 2(l + b)h}}}}$

$\qquad\qquad\qquad:\implies\sf 2(8 + 6) \times 5$

$\qquad\qquad\qquad\quad:\implies\sf 2 \times 14 \times 5$

$\qquad\qquad\qquad\qquad:\implies\sf 28 \times 5$

$\qquad\qquad\qquad\quad:\implies{\underline{\boxed{\sf{\purple{140\;m^2}}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Area\;of\;four\;walls\;of\;room\;is\; \bf{140\;m^2}}}}$