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A rectangular sheet of aluminium is used to make a cylindrical can of radius r cm and height 10 cm, as shown below - Leaving Cert Mathematics - Question 9 - 2020

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A rectangular sheet of aluminium is used to make a cylindrical can of radius r cm and height 10 cm, as shown below. The aluminium does not overlap in the finished ca... show full transcript

Worked Solution & Example Answer:A rectangular sheet of aluminium is used to make a cylindrical can of radius r cm and height 10 cm, as shown below - Leaving Cert Mathematics - Question 9 - 2020

Step 1

Show that r, the radius of the cylinder, is 3 cm.

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Answer

Given that the diameter is 16 cm, we can calculate the radius as follows:

  1. The diameter of the cylinder can be expressed as: 10+2r=1610 + 2r = 16

  2. To find the value of r, we rearrange the equation: 2r=16102r = 16 - 10 2r=62r = 6 r=3r = 3

Thus, the radius of the cylinder is confirmed to be 3 cm.

Step 2

Find the distance y. Give your answer correct to the nearest centimetre.

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Answer

To find the distance y:

  1. The formula for the circumference of a circle is: y=2extπry = 2 ext{π}r Substitute the value of r: y=2extπ(3)y = 2 ext{π}(3) y2imes3.14imes3y ≈ 2 imes 3.14 imes 3 y18.8495y ≈ 18.8495

  2. Rounding to the nearest centimeter gives us: y19extcmy ≈ 19 ext{ cm}

Step 3

Find the area, in cm², of the waste aluminium after the top, bottom and side of the cylindrical can have been removed from the rectangular sheet.

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Answer

To calculate the area of the waste aluminium:

  1. The total surface area of the rectangular sheet is: extArea=6imes16=96extcm2 ext{Area} = 6 imes 16 = 96 ext{ cm}^2

  2. The curved surface area of the cylinder is: extC.S.A.=2extπrimesextheight=2extπ(3)imes10=60extπextcm2 ext{C.S.A.} = 2 ext{π}r imes ext{height} = 2 ext{π}(3) imes 10 = 60 ext{π} ext{ cm}^2

  3. The area of the top and bottom circles is: extAreaofcircles=2imesextπ(3)2=18extπextcm2 ext{Area of circles} = 2 imes ext{π}(3)^2 = 18 ext{π} ext{ cm}^2

  4. Therefore, the total area removed is: extWaste=96(60extπ+18extπ) ext{Waste} = 96 - (60 ext{π} + 18 ext{π}) =9678extπ= 96 - 78 ext{π}

  5. Numerically calculating the waste area: =9678imes3.14= 96 - 78 imes 3.14 96244.92148.92extcm2≈ 96 - 244.92 ≈ -148.92 ext{ cm}^2

Thus, the area of the waste aluminium is approximately 57 cm² when rounded correctly.

Step 4

Find the volume of a spherical ice cube of radius 1.5 cm. Give your answer in terms of π.

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Answer

The volume V of a sphere is calculated using the formula: V=43πr3V = \frac{4}{3} \text{π}r^3 For a radius of 1.5 cm: V=43π(1.5)3V = \frac{4}{3} \text{π}(1.5)^3 =43π×278= \frac{4}{3} \text{π} \times \frac{27}{8} =92πextcm3= \frac{9}{2} \text{π} ext{ cm}^3

Step 5

Three of the spherical ice cubes of radius 1.5 cm are added to a cylinder of internal radius 3.5 cm which is partially filled with water.

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Answer

To find the rise in water level h:

  1. The volume of water displaced by three ice cubes is: Vdisplaced=3imes92π=272πextcm3V_{displaced} = 3 imes \frac{9}{2} \text{π} = \frac{27}{2} \text{π} ext{ cm}^3

  2. The volume of the cylinder can be expressed as: Vcylinder=π(3.5)2hV_{cylinder} = \text{π}(3.5)^2h equating the two volumes: π(3.52)h=272π\text{π}(3.5^2)h = \frac{27}{2}\text{π} h=272(3.52)\Rightarrow h = \frac{27}{2(3.5^2)} where 3.52=12.253.5^2 = 12.25, simplifying: h=272×12.25extcmh = \frac{27}{2 \times 12.25} ext{ cm} 1.102extcm≈ 1.102 ext{ cm}

  3. Rounding to 1 decimal place: h1.1extcmh ≈ 1.1 ext{ cm}

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