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Canned fishare sold in bulk from the wholesale .They are packaged in boxes with different dimensions as shown below Thabong Secondary school intends to buy canned fish for the National School Nutrition Programme Canned fish diameter =76mm height=10cm. Box A length=36cm Breadth=25cm height=35cm .Box B length=50cm Breadth=20cm height=35cm (a)the number of cans that can fit along the lengths of the boxes (b) the number of cans needed along the width of the boxes (c)the number of cans that can fit on the bases of the boxes

Sagot :

The area of a shape is the amount of space the shape can take. The following are the results of the computations:

  • The first box can take 4 cans along its length, while the length of the second box can take 6 cans.
  • The first box can take 3 cans along its width, while the width of the second box can take 2 cans.
  • 19 cans can fit the base of the first box, while the base of the second box can take 22 cans.

A. Number of cans that can fit along the lengths of the boxes

Given that:

[tex]d = 76mm[/tex] ---- the diameter of the can

[tex]L_1 = 36cm[/tex] -- the length of the first box

[tex]L_2 = 50cm[/tex] -- the length of the second box

The number of cans (n) is calculated by:

[tex]n = \frac{Length}{diameter}[/tex]

For the first box:

[tex]n_1 = \frac{L_1}{d}[/tex]

[tex]n_1 = \frac{36cm}{76mm}[/tex]

Convert mm to cm

[tex]n_1 = \frac{36cm}{76cm\times 0.1}[/tex]

[tex]n_1 = \frac{36cm}{7.6cm}[/tex]

[tex]n_1 = 4.736...[/tex]

Remove the decimal part (do not approximate)

[tex]n =4[/tex]

For the second box

[tex]n_2 = \frac{L_2}{d}[/tex]

[tex]n_2 = \frac{50cm}{76mm}[/tex]

Convert mm to cm

[tex]n_2 = \frac{50cm}{7.6cm}[/tex]

[tex]n_2 = 6.578..[/tex]

Remove the decimal part (do not approximate)

[tex]n_2 = 6[/tex]

Hence, 4 cans can fit the along the lengths of the first box, while the second box can take 6 cans, along its length.

B. Number of cans that can fit along the widths of the boxes

Given that:

[tex]W_1 = 25cm[/tex] -- the width of the first box

[tex]W_2 = 20cm[/tex] -- the width of the second box

The number of cans (n) is calculated by:

[tex]n = \frac{Width}{diameter}[/tex]

For the first box:

[tex]n_1 = \frac{W_1}{d}[/tex]

[tex]n_1 = \frac{25cm}{76mm}[/tex]

Convert mm to cm

[tex]n_1 = \frac{25cm}{7.6cm}[/tex]

[tex]n_1 = 3.2894...[/tex]

Remove the decimal part (do not approximate)

[tex]n_1 =3[/tex]

For the second box

[tex]n_2 = \frac{W_2}{d}[/tex]

[tex]n_2 = \frac{20cm}{76mm}[/tex]

Convert mm to cm

[tex]n_2 = \frac{20cm}{7.6cm}[/tex]

[tex]n_2 = 2.631...[/tex]

Remove the decimal part (do not approximate)

[tex]n_2 = 2[/tex]

Hence, 3 cans can fit the along the width of the first box, while the second box can take 2 cans, along its width.

C. Number of cans that can fit on the bases of the boxes

First, we calculate the area of the base of the can.

The base of the can is circular; so the area is:

[tex]Area = \pi r^2[/tex]

Where:

[tex]r = \frac d2 = \frac{76mm}{2} =38mm = 3.8cm[/tex]

So, we have:

[tex]Area = 3.14 \times 3.8^2[/tex]

[tex]Area = 45.3416cm^2[/tex]

Next, calculate the areas of the base of the box

The boxes are rectangular ; so, the area is:

[tex]Area=Length \times Width[/tex]

For the first box;

[tex]A_1 =36cm \times 25cm[/tex]

[tex]A_1 =900 cm^2[/tex]

For the second box;

[tex]A_2 = 50cm \times 20cm[/tex]

[tex]A_2 = 1000cm^2[/tex]

So, the number of cans on the bases is:

[tex]n = \frac{Area\ of\ base}{Area\ of\ can}[/tex]

For the first box:

[tex]n_1 = \frac{900cm^2}{45.3416cm^2}[/tex]

[tex]n_1 = 19.85[/tex]

Remove the decimal part (do not approximate)

[tex]n_1 = 19[/tex]

For the second box;

[tex]n_2= \frac{1000cm^2}{45.3416cm^2}[/tex]

[tex]n_2 = 22.05[/tex]

Remove the decimal part (do not approximate)

[tex]n_2 = 22[/tex]

Hence, 19 cans can fit the base of the first box, while the base of the second box can take 22 cans.

Read more about areas at:

https://brainly.com/question/4988011