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RideAnS [48]
1 year ago
13

A cosmetics company makes small, cylindrical bars of soap, and wraps them in plastic prior to packing them in boxes for shipping

. Each bar of soap has a diameter of 5 cm and is 2 cm in height.a. How much soap material is used to make each bar of soap?b. How much plastic is required to wrap each bar of soap, assuming it is shrink-wrapped tightly around each bar?C.The company wants to be able to ship a minimum of 60 bars per box to save on shipping costs. What will be the dimensions of the box if the soap must be stacked no more than 5 bars high?d. Find the capacity of the box in litres.
Mathematics
1 answer:
Setler79 [48]1 year ago
4 0

a. In order to determine how much soap material is used for each bar of soap, consider that the shape of the bar is a cylinder, then, use the folowing formula for the volume of a cylinder:

V = π r² h

where r is the radius of the base and h is the height.

The diameter is 5 cm, then, the radius is r = d/2 = 5 cm/2 = 2.5 cm

The height is h = 2 cm

replace the previous values into the formula for V:

V = (3.14)(2.5 cm)²(2 cm)

V = 39.25 cm³

Hence, 39.25 cm³ of soap material is used to make each bar.

b. In this case it is necessary to calculate the suface area of the plastic used. As before, take into account that the bar has a cylindrical shape. Then, use the following formula for the surface area of a cylinder:

S = 2πr² + 2πrh

replace the values of the parameters to find S:

S = 2π(2.5 cm)² + 2π(2.5cm)(2cm)

S = 70.68 cm²

Hence, 70.68 cm² of plastic is required to wrap each bar

c. Take into account that the soap must be stacked no more than 5 bars high, and that the company wants to ship a minimum of 60 bars.

If there are 5 bars high x 3 bars length, in one face of the box you have 15 bars. Moreover, if there are 4 bars width, then, you have a total of 15x4 = 60 bars in one box, which is what the company wants.

The height of the box is given by 5 times the height of one bar (because there are 5 bars high). The length is given by 3 times the diameter of one bar and the width of the box is given by 4 times the diameter of one bar:

height of the box = 5(2 cm) = 10 cm

length of the box = 3(5 cm) = 15 cm

width of the box = 4(5 cm) = 20 cm

d. First, calculate the volume of the box, as follow:

V = h x l x w = (10 cm)(15 cm)(20 cm) = 3,000 cm³

take into account the conversion 1,000 cm³ = 1 L, then, the capacity of the box is:

V = 3,000 cm³ = 3L

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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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