Hard GCSE 3D solids, plans and elevations Questions

Challenging, exam-style GCSE 3D solids, plans and elevations questions with worked solutions. Stretch yourself on the hardest solids made from cubes, isometric drawing, surface area, volume of a cuboid problems.

solids made from cubesisometric drawingsurface areavolume of a cuboidplan viewfront elevation
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
A solid is made from centimetre cubes standing on a flat table. Its base is 33 squares wide and 33 squares deep. The numbers of cubes in the stacks, read from left to right, are 1,2,31, 2, 3 (front row), 3,2,13, 2, 1 (middle row) and 2,1,22, 1, 2 (back row). Which list gives the heights, in cubes, of the columns of the side elevation of the solid, from the front of the solid to the back?
Show worked solution

Worked solution

  1. Say what you see when you look at the solid from the side.

    side elevation=the view from the side\text{side elevation} = \text{the view from the side}

    Looking from the side you look along the width of the solid, so each ROW of the base gives one column of the elevation.

  2. Work out the height of each column of the side elevation.

    side elevation=3, 3, 2\text{side elevation} = 3,\ 3,\ 2

    Looking from the side, a column is as tall as the tallest stack in that row of the base, so take the largest number in each row: 3,3,23, 3, 2.

  3. Write the stack heights out as a grid, row by row from the front.

    1,2,33,2,12,1,21, 2, 3 \\ 3, 2, 1 \\ 2, 1, 2

    The base is 33 squares wide and 33 squares deep, so there are 99 stacks in all (a stack of 00 means that square is empty).

  4. Rule out using the columns of the base.

    3, 2, 3≠3, 3, 23,\ 2,\ 3 \ne 3,\ 3,\ 2

    Taking the largest number in each column of the grid gives the FRONT elevation, 3,2,33, 2, 3, not the side elevation.

  5. Rule out adding the cubes in each row.

    6, 6, 5≠3, 3, 26,\ 6,\ 5 \ne 3,\ 3,\ 2

    Adding a row counts cubes hidden behind one another. Only the tallest stack in the row is seen.

  6. Rule out reading the rows from the back.

    2, 3, 3≠3, 3, 22,\ 3,\ 3 \ne 3,\ 3,\ 2

    The question asks for the columns from the front of the solid to the back, so the front row must come first.

  7. Check the number of columns.

    columns=3\text{columns} = 3

    The base is 33 squares deep, so the side elevation has 33 columns — not 33, which is what the front elevation has.

  8. Check the tallest column.

    max⁡=3=3\max = 3 = 3

    The tallest column of the side elevation must be the tallest stack of the solid, 33 cubes.

  9. Work out the area of the side elevation.

    3+3+2=83 + 3 + 2 = 8

    The elevation covers 88 squares, so it has an area of 8 cm28\text{ cm}^2.

  10. Count the base squares that have at least one cube on them.

    covered squares=9 out of 9\text{covered squares} = 9 \text{ out of } 9

    The plan is the view from directly above, so a base square appears in the plan exactly when a cube stands on it: 99 of the 99 squares are covered.

  11. Check that the elevation cannot show every cube.

    8≤178 \le 17

    The solid has 1717 cubes but the side elevation shows only 88 squares, because the cubes behind hide each other.

  12. Check the front elevation reaches the same height.

    3=33 = 3

    Both elevations are views of the same solid, so their tallest columns must agree.

  13. Add up the cubes in each row of the base.

    1+2+3=6,3+2+1=6,2+1+2=51 + 2 + 3 = 6 , \quad 3 + 2 + 1 = 6 , \quad 2 + 1 + 2 = 5

    Row by row from the front, the stacks hold 6,6and56, 6 and 5 cubes.

  14. Add the row totals to get the number of cubes.

    6+6+5=176 + 6 + 5 = 17

    Altogether the solid uses 1717 cubes.

  15. State the side elevation.

    side elevation=3, 3, 2\text{side elevation} = 3,\ 3,\ 2

    From the front to the back the columns are 3,3,23, 3, 2.

Answer
side elevation columns: 3,3,2\text{side elevation columns: } 3, 3, 2
Question 2
6 markschallenging
A solid is made from centimetre cubes standing on a flat table. Its base is 33 squares wide and 22 squares deep. The numbers of cubes in the stacks, read from left to right, are 3,1,23, 1, 2 (front row) and 2,2,12, 2, 1 (back row). Which of these statements about the solid is correct?
Show worked solution

Worked solution

  1. Work out all three views, then test each statement.

    A=plan,front elevation,side elevationA = \text{plan}, \quad \text{front elevation}, \quad \text{side elevation}

    Only one statement can be right, so work out the area of each view for yourself and then read the options.

  2. Write the stack heights out as a grid, row by row from the front.

    3,1,22,2,13, 1, 2 \\ 2, 2, 1

    The base is 33 squares wide and 22 squares deep, so there are 66 stacks in all (a stack of 00 means that square is empty).

  3. Count the base squares that have at least one cube on them.

    covered squares=6 out of 6\text{covered squares} = 6 \text{ out of } 6

    The plan is the view from directly above, so a base square appears in the plan exactly when a cube stands on it: 66 of the 66 squares are covered.

  4. Write down the area of the plan.

    plan=6 cm2\text{plan} = 6\text{ cm}^2

    The plan is 66 unit squares, so its area is 6 cm26\text{ cm}^2.

  5. Work out the height of each column of the front elevation.

    front elevation=3, 2, 2\text{front elevation} = 3,\ 2,\ 2

    Looking from the front, a column is as tall as the TALLEST stack behind it, so take the largest number in each column of the grid: 3,2,23, 2, 2.

  6. Add the column heights to get the area of the front elevation.

    3+2+2=73 + 2 + 2 = 7

    Each square of the elevation is 1 cm21\text{ cm}^2, so the area is the total of the column heights, 7 cm27\text{ cm}^2.

  7. Work out the height of each column of the side elevation.

    side elevation=3, 2\text{side elevation} = 3,\ 2

    Looking from the side, a column is as tall as the tallest stack in that row of the base, so take the largest number in each row: 3,23, 2.

  8. Add the column heights to get the area of the side elevation.

    3+2=53 + 2 = 5

    The side elevation is 55 squares, so its area is 5 cm25\text{ cm}^2.

  9. Compare the three areas.

    plan=6,front=7,side=5\text{plan} = 6, \quad \text{front} = 7, \quad \text{side} = 5

    The three views have areas 6 cm26\text{ cm}^2, 7 cm27\text{ cm}^2 and 5 cm25\text{ cm}^2.

  10. Test the statement about the plan.

    plan=6 cm2\text{plan} = 6\text{ cm}^2

    Any option claiming a plan of anything other than 6 cm26\text{ cm}^2 is false.

  11. Test the statement about the front elevation.

    front elevation=7 cm2\text{front elevation} = 7\text{ cm}^2

    Any option claiming a front elevation of anything other than 7 cm27\text{ cm}^2 is false.

  12. Test the statement about the side elevation.

    side elevation=5 cm2\text{side elevation} = 5\text{ cm}^2

    Any option claiming a side elevation of anything other than 5 cm25\text{ cm}^2 is false.

  13. Check the widths of the two elevations.

    3 and 23 \text{ and } 2

    The front elevation is 33 columns wide and the side elevation is 22 columns wide, which is a quick way to spot a view that has been mixed up.

  14. Check the tallest column of each elevation.

    3=3=33 = 3 = 3

    Both elevations reach 33 cubes, the tallest stack in the solid.

  15. State the correct statement.

    side elevation=5 cm2\text{side elevation} = 5\text{ cm}^2

    The only true statement is that the side elevation has an area of 5 cm25\text{ cm}^2.

Answer
area of the side elevation=5 cm2\text{area of the side elevation} = 5\text{ cm}^2
Question 3
5 markschallenging
A solid is made from centimetre cubes standing on a flat table. Its base is 33 squares wide and 22 squares deep. The numbers of cubes in the stacks, read from left to right, are 2,3,12, 3, 1 (front row) and 1,2,11, 2, 1 (back row). Which of these statements about the solid is correct?
Show worked solution

Worked solution

  1. Work out all three views, then test each statement.

    A=plan,front elevation,side elevationA = \text{plan}, \quad \text{front elevation}, \quad \text{side elevation}

    Only one statement can be right, so work out the area of each view for yourself and then read the options.

  2. Write the stack heights out as a grid, row by row from the front.

    2,3,11,2,12, 3, 1 \\ 1, 2, 1

    The base is 33 squares wide and 22 squares deep, so there are 66 stacks in all (a stack of 00 means that square is empty).

  3. Count the base squares that have at least one cube on them.

    covered squares=6 out of 6\text{covered squares} = 6 \text{ out of } 6

    The plan is the view from directly above, so a base square appears in the plan exactly when a cube stands on it: 66 of the 66 squares are covered.

  4. Write down the area of the plan.

    plan=6 cm2\text{plan} = 6\text{ cm}^2

    The plan is 66 unit squares, so its area is 6 cm26\text{ cm}^2.

  5. Work out the height of each column of the front elevation.

    front elevation=2, 3, 1\text{front elevation} = 2,\ 3,\ 1

    Looking from the front, a column is as tall as the TALLEST stack behind it, so take the largest number in each column of the grid: 2,3,12, 3, 1.

  6. Add the column heights to get the area of the front elevation.

    2+3+1=62 + 3 + 1 = 6

    Each square of the elevation is 1 cm21\text{ cm}^2, so the area is the total of the column heights, 6 cm26\text{ cm}^2.

  7. Work out the height of each column of the side elevation.

    side elevation=3, 2\text{side elevation} = 3,\ 2

    Looking from the side, a column is as tall as the tallest stack in that row of the base, so take the largest number in each row: 3,23, 2.

  8. Add the column heights to get the area of the side elevation.

    3+2=53 + 2 = 5

    The side elevation is 55 squares, so its area is 5 cm25\text{ cm}^2.

  9. Compare the three areas.

    plan=6,front=6,side=5\text{plan} = 6, \quad \text{front} = 6, \quad \text{side} = 5

    The three views have areas 6 cm26\text{ cm}^2, 6 cm26\text{ cm}^2 and 5 cm25\text{ cm}^2.

  10. Test the statement about the plan.

    plan=6 cm2\text{plan} = 6\text{ cm}^2

    Any option claiming a plan of anything other than 6 cm26\text{ cm}^2 is false.

  11. Test the statement about the front elevation.

    front elevation=6 cm2\text{front elevation} = 6\text{ cm}^2

    Any option claiming a front elevation of anything other than 6 cm26\text{ cm}^2 is false.

  12. Test the statement about the side elevation.

    side elevation=5 cm2\text{side elevation} = 5\text{ cm}^2

    Any option claiming a side elevation of anything other than 5 cm25\text{ cm}^2 is false.

  13. Check the widths of the two elevations.

    3 and 23 \text{ and } 2

    The front elevation is 33 columns wide and the side elevation is 22 columns wide, which is a quick way to spot a view that has been mixed up.

  14. Check the tallest column of each elevation.

    3=3=33 = 3 = 3

    Both elevations reach 33 cubes, the tallest stack in the solid.

  15. State the correct statement.

    front elevation=6 cm2\text{front elevation} = 6\text{ cm}^2

    The only true statement is that the front elevation has an area of 6 cm26\text{ cm}^2.

Answer
area of the front elevation=6 cm2\text{area of the front elevation} = 6\text{ cm}^2
Question 4
5 markschallenging
The plan of a cuboid has area 30 cm230\text{ cm}^2, its front elevation has area 12 cm212\text{ cm}^2 and its side elevation has area 10 cm210\text{ cm}^2. Work out the volume of the cuboid.
Show worked solution

Worked solution

  1. Give the three edges of the cuboid letters.

    w=width,d=depth,h=heightw = \text{width}, \quad d = \text{depth}, \quad h = \text{height}

    The plan shows the width and the depth, the front elevation shows the width and the height, and the side elevation shows the depth and the height.

  2. Write each view as a product of two edges.

    wd=30,wh=12,dh=10wd = 30, \quad wh = 12, \quad dh = 10

    Every view of a cuboid is a rectangle, so each area is the product of the two edges you can see in that view.

  3. Multiply the three areas together.

    wd×wh×dh=w2d2h2=(wdh)2wd \times wh \times dh = w^2 d^2 h^2 = (wdh)^2

    Each of the three letters appears exactly twice in the product, so the answer is the square of the volume.

  4. Work out that product.

    30×12×10=360030 \times 12 \times 10 = 3600

    The three areas multiply to 36003600.

  5. Take the square root to find the volume.

    wdh=3600=60 cm3wdh = \sqrt{3600} = 60\text{ cm}^3

    The volume is the square root of 36003600, which is 60 cm360\text{ cm}^3.

  6. Find the height by dividing the volume by the plan.

    h=wdhwd=6030=2 cmh = \frac{wdh}{wd} = \frac{60}{30} = 2\text{ cm}

    Dividing the volume by the plan cancels the width and the depth, leaving the height, 2 cm2\text{ cm}.

  7. Find the depth by dividing the volume by the front elevation.

    d=wdhwh=6012=5 cmd = \frac{wdh}{wh} = \frac{60}{12} = 5\text{ cm}

    The front elevation carries the width and the height, so dividing leaves the depth, 5 cm5\text{ cm}.

  8. Find the width by dividing the volume by the side elevation.

    w=wdhdh=6010=6 cmw = \frac{wdh}{dh} = \frac{60}{10} = 6\text{ cm}

    The side elevation carries the depth and the height, so dividing leaves the width, 6 cm6\text{ cm}.

  9. Check the plan.

    w×d=6×5=30 cm2w \times d = 6 \times 5 = 30\text{ cm}^2

    The width and the depth give the stated plan of 30 cm230\text{ cm}^2.

  10. Check the front elevation.

    w×h=6×2=12 cm2w \times h = 6 \times 2 = 12\text{ cm}^2

    The width and the height give the stated front elevation of 12 cm212\text{ cm}^2.

  11. Check the side elevation.

    d×h=5×2=10 cm2d \times h = 5 \times 2 = 10\text{ cm}^2

    The depth and the height give the stated side elevation of 10 cm210\text{ cm}^2.

  12. Check the volume directly from the three edges.

    V=6×5×2=60 cm3V = 6 \times 5 \times 2 = 60\text{ cm}^3

    Multiplying the three edges gives 60 cm360\text{ cm}^3, agreeing with the square root found earlier.

  13. Note why the square root is needed.

    (wdh)2=3600,wdh=60(wdh)^2 = 3600, \quad wdh = 60

    Multiplying the three areas counts every edge twice, which is why the product is the square of the volume and not the volume itself.

  14. Check the surface area for consistency.

    2×(30+12+10)=104 cm22 \times (30 + 12 + 10) = 104\text{ cm}^2

    A cuboid has each of its three rectangles twice, so its surface area is 104 cm2104\text{ cm}^2.

  15. State the volume of the cuboid.

    V=60 cm3V = 60\text{ cm}^3

    The volume of the cuboid is 60 cm360\text{ cm}^3.

Answer
V=60 cm3V = 60\text{ cm}^3
Question 5
6 markschallenging
The plan of a cuboid has area 24 cm224\text{ cm}^2, its front elevation has area 32 cm232\text{ cm}^2 and its side elevation has area 12 cm212\text{ cm}^2. Work out the volume of the cuboid.
Show worked solution

Worked solution

  1. Give the three edges of the cuboid letters.

    w=width,d=depth,h=heightw = \text{width}, \quad d = \text{depth}, \quad h = \text{height}

    The plan shows the width and the depth, the front elevation shows the width and the height, and the side elevation shows the depth and the height.

  2. Write each view as a product of two edges.

    wd=24,wh=32,dh=12wd = 24, \quad wh = 32, \quad dh = 12

    Every view of a cuboid is a rectangle, so each area is the product of the two edges you can see in that view.

  3. Multiply the three areas together.

    wd×wh×dh=w2d2h2=(wdh)2wd \times wh \times dh = w^2 d^2 h^2 = (wdh)^2

    Each of the three letters appears exactly twice in the product, so the answer is the square of the volume.

  4. Work out that product.

    24×32×12=921624 \times 32 \times 12 = 9216

    The three areas multiply to 92169216.

  5. Take the square root to find the volume.

    wdh=9216=96 cm3wdh = \sqrt{9216} = 96\text{ cm}^3

    The volume is the square root of 92169216, which is 96 cm396\text{ cm}^3.

  6. Find the height by dividing the volume by the plan.

    h=wdhwd=9624=4 cmh = \frac{wdh}{wd} = \frac{96}{24} = 4\text{ cm}

    Dividing the volume by the plan cancels the width and the depth, leaving the height, 4 cm4\text{ cm}.

  7. Find the depth by dividing the volume by the front elevation.

    d=wdhwh=9632=3 cmd = \frac{wdh}{wh} = \frac{96}{32} = 3\text{ cm}

    The front elevation carries the width and the height, so dividing leaves the depth, 3 cm3\text{ cm}.

  8. Find the width by dividing the volume by the side elevation.

    w=wdhdh=9612=8 cmw = \frac{wdh}{dh} = \frac{96}{12} = 8\text{ cm}

    The side elevation carries the depth and the height, so dividing leaves the width, 8 cm8\text{ cm}.

  9. Check the plan.

    w×d=8×3=24 cm2w \times d = 8 \times 3 = 24\text{ cm}^2

    The width and the depth give the stated plan of 24 cm224\text{ cm}^2.

  10. Check the front elevation.

    w×h=8×4=32 cm2w \times h = 8 \times 4 = 32\text{ cm}^2

    The width and the height give the stated front elevation of 32 cm232\text{ cm}^2.

  11. Check the side elevation.

    d×h=3×4=12 cm2d \times h = 3 \times 4 = 12\text{ cm}^2

    The depth and the height give the stated side elevation of 12 cm212\text{ cm}^2.

  12. Check the volume directly from the three edges.

    V=8×3×4=96 cm3V = 8 \times 3 \times 4 = 96\text{ cm}^3

    Multiplying the three edges gives 96 cm396\text{ cm}^3, agreeing with the square root found earlier.

  13. Note why the square root is needed.

    (wdh)2=9216,wdh=96(wdh)^2 = 9216, \quad wdh = 96

    Multiplying the three areas counts every edge twice, which is why the product is the square of the volume and not the volume itself.

  14. Check the surface area for consistency.

    2×(24+32+12)=136 cm22 \times (24 + 32 + 12) = 136\text{ cm}^2

    A cuboid has each of its three rectangles twice, so its surface area is 136 cm2136\text{ cm}^2.

  15. State the volume of the cuboid.

    V=96 cm3V = 96\text{ cm}^3

    The volume of the cuboid is 96 cm396\text{ cm}^3.

Answer
V=96 cm3V = 96\text{ cm}^3

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