Problem
Volume of a composite solid
Two solids are drawn. Solid (1) is a single step. Solid (2) is a house with a roof. Find the volume of each.
Reasoning · Grade 6-1 Volume of Solid Figures
Your answer
Watch out The odd shapes look hard, but both are prisms: front area times depth.
How to solve
Strategy Identify Subproblems — Neither solid has a volume formula of its own, but both are prisms — one flat face dragged 5 cm backwards. So every awkward 3D shape here reduces to one flat area times 5. That is the subproblem split. For the step in (1) I use the complement: instead of adding the two blocks, start from the full 8 by 5 by 7 box that encloses it and subtract the bite that was taken out of the top right. For the house in (2) adding is easier than subtracting, because the roof and the walls are already a clean triangular prism sitting on a clean rectangular prism. Drawing the cutting line on the front face is what makes each split visible.
1STEP 1
Recognise both solids as prisms
Both are prisms: front area times depth.
(volume) = (area of the front face) × (depth)
A prism is one flat shape stacked over and over. Five stacked centimetre-thick slices, each with the same face area, is just that area times 5 — the same reason a box is length times width times height.
7.G.B.6Draw A DiagramBoth solids are prisms, so each volume is one flat face's area multiplied by how far the solid runs back.
Why?
A prism is the same flat shape repeated all the way along, so stacking identical slices is what fills it.
🧱A prism is its base stacked upA prism is one flat face repeated all the way up, so its volume is the base area multiplied by the height.
Why?
The awkward face can be handled as a plain rectangle with a notch taken out, and the pieces add back to the whole.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
2STEP 2
Solid (1): recover the two missing lengths of the step
The missing lengths in (1) are 4 cm each.
8 - 4 = 4 cm (width of the tall part), 7 - 3 = 4 cm (rise of the step)
On a step figure every edge is either horizontal or vertical, so opposite edges have to add up the same way. That lets two subtractions replace two measurements the picture never gave.
4.MD.A.3Draw A Diagram3STEP 3
Solid (1): fill in the notch and subtract it (the complement)
Filling and subtracting gives 200 cm³.
8 × 5 × 7 - 4 × 5 × 4 = 280 - 80 = 200 cm³
It is often easier to describe what is missing than what is there. The missing piece is a plain box, so one multiplication finds it, and subtraction does the rest.
5.MD.C.5Change Focus Count The Complement4STEP 4
Solid (1): confirm by splitting into two blocks instead
Splitting into two blocks also gives 200 cm³.
4 × 5 × 7 + 4 × 5 × 3 = 140 + 60 = 200 cm³ ✓
Adding the parts and subtracting the hole are two different routes to one number, so agreement is a real check. It also catches the commonest slip, using 7 cm instead of the 4 cm rise for the removed block.
5.MD.C.5Identify Subproblems5STEP 5
Solid (2): find the area of the rectangular part of the pentagon
The rectangular part of (2) is 8 cm².
4 × 2 = 8 cm²
The horizontal line where roof meets wall is the natural cut: it leaves two shapes whose areas are both already known formulas.
6.G.A.1Identify Subproblems6STEP 6
Solid (2): find the area of the roof triangle
The roof triangle is 4 cm².
4 × 2 ÷ 2 = 8 ÷ 2 = 4 cm²
The right-angle mark is doing important work: it certifies that the 2 cm really is the perpendicular height, which is the only height the triangle formula accepts. Where the peak sits along the top edge does not change the area at all.
6.G.A.1Identify Subproblems7STEP 7
Solid (2): multiply the pentagon's area by the depth
Times the depth gives 60 cm³.
(8 + 4) × 5 = 12 × 5 = 60 cm³, (4 × 2 ÷ 2) × 5 + 4 × 5 × 2 = 20 + 40 = 60 cm³
Adding the two face areas first and multiplying once, or multiplying each piece by 5 and then adding, give the same number — that is just the distributive property, and having both routes agree is a free check.
7.G.B.6Identify SubproblemsAnswer
200, 60 cm³
280 − 80 = 200, 12 × 5 = 60
The units are right: centimetres times centimetres times centimetres gives cubic centimetres in both answers. The sizes sit where they should. Solid (1) must be smaller than the 8 x 5 x 7 = 280 cubic centimetre box that encloses it and larger than the 8 x 5 x 3 = 120 cubic centimetre slab underneath it, and 200 falls between 120 and 280. Solid (2) must be smaller than the 4 x 5 x 4 = 80 cubic centimetre box around it and larger than the 4 x 5 x 2 = 40 cubic centimetre wall part alone, and 60 falls between 40 and 80 — in fact the roof adds exactly half of the wall's volume, which matches a triangle being half of its rectangle. Both solids were also computed two different ways, by difference and by sum for (1) and by whole-face area and by two prisms for (2), and both pairs agree.
Takeaway
Any solid whose shape never changes front to back is just one flat area dragged backwards — so find that area, by adding pieces or by subtracting the missing bite, and multiply by the depth.
- Recognise both solids as prisms
- Solid (1): recover the two missing lengths of the step
- Solid (1): fill in the notch and subtract it (the complement)
- Solid (1): confirm by splitting into two blocks instead
- Solid (2): find the area of the rectangular part of the pentagon
- Solid (2): find the area of the roof triangle
- Solid (2): multiply the pentagon's area by the depth
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