Question

Michaela is making memory boxes. She wants to cover the boxes with sheets of decorative paper on all sides before filling them. This net represents a single memory box.

How much paper is needed to cover each memory box?

Enter your answer in the box.

Answer 1

Answer: 5640 sq. in.

Explanation:

The length and breath of the small rectangle is 35 in. and 30 in.

Area of a Rectangle = length * breadth

= 35 in. * 30 in.

= 1050 sq. in.

There are four boxes of the same size of 35 in. by 30 in.

So, 1050 * 4

= 4200 sq. in.

The small rectangles which are on the side have length and breadth of 24 in. and 30 in.

Area of a Rectangle = length * breadth

= 30 in. * 24 in.

= 720 sq. in.

There are 2 boxes of the same sizes.

= 720 * 2

= 1440 sq. in.

All the squares added will be = 4200 sq. in. + 1440 sq. in.

= 5640 sq. in.

Answer 2

5640 in² Paper is needed to cover each memory box

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GiveN ::

• We've 4 rectangles in the middle each measuring :-

• ➸ Length = 35 in
• ➸ Breadth = 30 in

ㅤ

• We've 2 rectangles in both sides each measuring :-

• ➸ Length = 24 in
• ➸ Breadth = 30 in

SolutioN ::

`\begin{gathered} \\ \; \underline{\boxed{\sf{Area_((Rectangle)) = (Length * Breadth) \: sq. \: unit}}} \; \pmb{\purple{\bigstar}} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longmapsto \; \sf{Area_((Middle \; Rectangle)) = (35 * 30) \; in^(2)} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longmapsto \; \sf{Area_((Middle \; Rectangle)) = 1050 \; in^(2)} \\ \\ \end{gathered}`

As we've 4 rectangles in middle so,

`\begin{gathered} \; :\longrightarrow \; \sf{(1050 * 4) \; in^(2)} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longrightarrow \; \boxed{\sf{4200 \; in^(2)}} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longmapsto \; \sf{Area_((Side \; Rectangle)) = (24 * 30) \; in^(2)} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longmapsto \; \sf{Area_((Middle \; Rectangle)) = 720 \; in^(2)} \\ \\ \end{gathered}`

As we've 2 rectangles in both sides so,

`\begin{gathered} \; :\longrightarrow \; \sf{(720 * 2) \; in^(2)} \\ \\ \end{gathered}`

`\begin{gathered} \; :\longrightarrow \; \boxed{\sf{1440 \; in^(2)}} \\ \\ \end{gathered}`

`\begin{gathered} \\ \; \dag \; \underline{\underline{\sf{Paper \; Needed :-}}} \\ \\ \end{gathered}`

`\begin{gathered} \; :\leadsto \; \sf{(4200 + 1440) \; in^(2)} \\ \\ \end{gathered}`

`\begin{gathered} \; :\leadsto \; \underline{\boxed{\frak{\pink{5640 \; in^(2)}}}} \; \pmb{\bigstar} \\ \\ \end{gathered}`

`\begin{gathered} \\ \\ {\underline{\pmb{\rule{170pt}{4pt}}}} \end{gathered}`

More To Know ←⁠(⁠>⁠▽⁠<⁠)⁠ﾉ

`\begin{gathered}\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{\boxed{\begin{array}{cc} \\ \star \: \bf{Area_((Rectangle)) = Length * Breadth} \\ \\ \star \: \bf{Perimeter_((Rectangle)) = 2(Length + Breadth)} \\ \\ \star \: \bf{Diagonal_((Rectangle)) = \sqrt{(Length)^(2) + (Breadth)^(2)}} \\ \\ \star \: \bf{Length_((Rectangle)) = (Area)/(Breadth)} \\ \\ \star \: \bf{Breadth_((Rectangle)) = (Area)/(Length)}\\ \: \end{array}}}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}`