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Volume of Composite Shapes>
Volume of Composite Shapes 1Volume of Composite Shapes 1
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Question 1 of 5
1. Question
Find the volume of the figure- `\text(Volume )=` (25200) `\text(cm)^3`
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Volume of a Rectangular Prism
`\text(Volume )=\text(length) times \text(width) times``\text(depth)`Labelling the given lengths
`\text(Smaller Rectangle)``\text(length)=30``\text(width)=20``\text(depth)=12``\text(Larger Rectangle)``\text(length)=60``\text(width)=25` `(55-``30``)``\text(depth)=12`First, find the area of the smaller rectangle`\text(Area)` `=` `\text(length)``times``\text(width)` `=` `30``times``20``=``600 \text(cm)^2` Next, find the area of the larger rectangle`\text(Area)` `=` `\text(length)``times``\text(width)` `=` `60``times``25``=``1500 \text(cm)^2` Next, add the area of the smaller rectangle and the area of the larger rectangle`=` `600``+``1500` Plug in the two areas `=` `2100 \text(cm)^2` Finally, multiply the area by the depth to find the volume`\text(Volume)` `=` `\text(area)``times``\text(depth)` Finding the volume `=` `2100``times``12` Plug in the known lengths `=` `25200 \text(cm)^3` The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=25200 \text(cm)^3` -
Question 2 of 5
2. Question
Find the volume of the figureRound your answer to the nearest whole numberUse `pi=3.141592654`- `\text(Volume )=` (31616, 31605, 31625) `\text(cm)^3`
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Volume of a Rectangular Prism
`\text(Volume )=``\text(length)``times``\text(breadth)``times``\text(height)`Volume of a Cone
`\text(Volume)=1/3 times pi times``\text(radius)^2``times``\text(height)`Labelling the given lengths
`\text(Cone)``\text(radius)=18``\text(height)=62``\text(Rectangular Prism)``\text(length)=46``\text(breadth)=46``\text(height)=5`First, find the area of the rectangle`\text(Area)` `=` `\text(length)``times``\text(breadth)` `=` `46``times``46``=``2116 \text(cm)^2` Next, multiply the area by the height to find the volume`\text(Volume)` `=` `\text(area)``times``\text(height)` Finding the volume `=` `2116``times``5` Plug in the known lengths `=` `10580 \text(cm)^3` Next, use the formula to find the volume of the coneUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `1/3 times pi times``\text(radius)^2``times``\text(height)` `=` `1/3 times 3.141592654 times``18^2``times``62` `=` `21036.10441 \text(cm)^3` Finally, add the volume of the cube and the volume of the cone`=` `10580``+``21036.10441` Plug in the two volumes `=` `31616.10441` `=` `31616 \text(cm)^3` Rounded to the nearest whole number The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=31616 \text(cm)^3`The answer will depend on which `pi` you use.In this solution we used: `pi=3.141592654`.Using Answer `pi=3.141592654` `31616 cm^3` `pi=3.14` `31605 cm^3` `pi=(22)/(7)` `31625 cm^3` -
Question 3 of 5
3. Question
Find the volume of the figureRound your answer to `2` decimal placesUse `pi=3.141592654`- `\text(Volume )=` (179.07, 178.98, 179.14) `\text(cm)^3`
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Volume of a Hemisphere
`\text(Volume)=1/2 times 4/3 times pi times``\text(radius)^3`Volume of a Cone
`\text(Volume)=1/3 times pi times``\text(radius)^2``times``\text(height)`Labelling the given lengths
`\text(radius)=?``\text(diameter)=6``\text(height)=13`We need to add volume of the hemisphere and the coneFirst, recall that the radius is equal to half of the diameter`\text(radius)` `=` `1/2 times ``6` `\text(radius)` `=` `3` Next, use the formula to find the volume of the hemisphereUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `1/2 times 4/3 times pi times``\text(radius)^3` `=` `1/2 times 4/3 times 3.141592654 times``3^3` `=` `56.54866 \text(cm)^3` Next, use the formula to find the volume of the coneUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `1/3 times pi times``\text(radius)^2``times``\text(height)` `=` `1/3 times 3.141592654 times``3^2``times``13` `=` `122.52211 \text(cm)^3` Finally, add the volume of the sphere and the volume of the cone`=` `56.54866``+``122.52211` Plug in the two volumes `=` `179.07078` `=` `179.07 \text(cm)^3` Rounded to `2` decimal places The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=179.07 \text(cm)^3`The answer will depend on which `pi` you use.In this solution we used: `pi=3.141592654`.Using Answer `pi=3.141592654` `179.07 cm^3` `pi=3.14` `178.98 cm^3` `pi=(22)/(7)` `179.14 cm^3` -
Question 4 of 5
4. Question
Find the volume of the figureNote: The cylinder is hollowRound your answer to `2` decimal placesUse `pi=3.141592654`- `\text(Volume )=` (2211.68, 2210.56, 2212.57) `\text(cm)^3`
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Volume of a Cylinder
`\text(Volume)=pi times``\text(radius)^2``times``\text(height)`Labelling the given lengths
`\text(Outer Cylinder)``\text(radius)=?``\text(diameter)=12``\text(height)=22``\text(Inner Cylinder)``\text(radius)=2``\text(height)=22`We need to find the volume of the figure, not including its hollow partFirst, recall that the radius is equal to half of the diameter`\text(radius)` `=` `1/2 times ``12` `\text(radius(Larger Circle))` `=` `6` Next, use the formula to find the area of the Outer CircleUse `pi=3.141592654` See `pi` explained`\text(Area)` `=` `pi times``\text(radius)^2` `=` `3.141592654 times``6^2` `=` `113.09733 \text(cm)^2` Next, find the area of the Inner CircleUse `pi=3.141592654` See `pi` explained`\text(Area)` `=` `pi times``\text(radius)^2` `=` `3.141592654 times``2^2` `=` `12.56637 \text(cm)^2` Now, subtract the area of the Inner Circle from the Outer Circle`=` `113.09733``-``12.56637` Plug in the two areas `=` `100.53096 \text(cm)^2` Finally, multiply the area by the height to find the volume`\text(Volume)` `=` `\text(area)``times``\text(height)` Finding the volume `=` `100.53096``times``22` Plug in the known lengths `=` `2211.68122` `=` `2211.68 \text(cm)^3` Rounded to `2` decimal places The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=2211.68 \text(cm)^3`The answer will depend on which `pi` you use.In this solution we used: `pi=3.141592654`.Using Answer `pi=3.141592654` `2211.68 cm^3` `pi=3.14` `2210.56 cm^3` `pi=(22)/(7)` `2212.57 cm^3` -
Question 5 of 5
5. Question
Find the volume of the figureRound your answer to two decimal places- `\text(Volume )=` (508.94, 508.68, 509.14) `\text(cm)^3`
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Volume of a Cylinder
`\text(Volume)=pi times``\text(radius)^2``times``\text(height)`Labelling the given lengths
`\text(Short Cylinders)``\text(radius)=?``\text(diameter)=12``\text(height)=2``\text(Long Cylinder)``\text(radius)=?``\text(diameter)=3``\text(height)=8`First, recall that the radius is equal to half of the diameter`\text(radius)` `=` `1/2 times ``12` `\text(radius(short cylinders))` `=` `6` ` ` ` ` ` ` `\text(radius)` `=` `1/2 times ``3` `\text(radius(long cylinder))` `=` `1.5` Next, use the formula to find the volume of the two short cylindersUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `3.141592654 times``\text(radius)^2``times``\text(height)` `=` `3.141592654 times``6^2``times``2` `=` `226.19467` Since there are two short cylinders, we multiply our answer by two.`\text(Volume)` `=` `226.19467 times 2` `=` `452.38934 \text(cm)^3` Now, use the formula to find the volume of the long cylinderUse `pi=3.141592654` See `pi` explained`\text(Volume)` `=` `3.141592654 times``\text(radius)^2``times``\text(height)` `=` `3.141592654 times``1.5^2``times``8` `=` `56.54866 \text(cm)^3` Finally, add the volume of the two short cylinders and the volume of the long cylinder`=` `452.38934``+``56.54866` Plug in the two volumes `=` `508.938` `=` `508.94 \text(cm)^3` Rounded to `2` decimal places The given measurements are in centimetres, so the volume is measured as centimetres cubed`\text(Volume)=508.94 \text(cm)^3`The answer will depend on which `pi` you use.In this solution we used: `pi=3.141592654`.Using Answer `pi=3.141592654` `508.94 cm^3` `pi=3.14` `508.68 cm^3` `pi=(22)/(7)` `509.14 cm^3`
Quizzes
- Volume of Shapes 1
- Volume of Shapes 2
- Volume of Shapes 3
- Volume of Shapes 4
- Volume of Composite Shapes 1
- Volume of Composite Shapes 2
- Surface Area of Shapes 1
- Surface Area of Shapes 2
- Surface Area of Shapes 3
- Surface Area and Volume Mixed Review 1
- Surface Area and Volume Mixed Review 2
- Surface Area and Volume Mixed Review 3
- Surface Area and Volume Mixed Review 4