Composite Perimeter, Area and Volume

Worked example

1. 1

   $\frac{1}{2}\times 12\times 5=30$ — The triangle on top: half × base × height. Its height 5 is at right angles to the 12 cm base.

2. 2

   $12\times x=12x$ — The rectangle: width × height. The height is the unknown, so its area is 12x.

3. 3

   $12x+30=114$ — The two pieces make the whole shape, so their areas add up to the total, 114.

4. 4

   $12x=84$ — Take 30 from both sides so the x term is alone.

5. 5

   $x=7\text{ cm}$ — Divide both sides by 12. Check: 12 × 7 + 30 = 114.

Worked example

1. 1

   $10-4=6$ — The two top edges together span the full width, 10 cm. One is 4, so the inside top edge is 6.

2. 2

   $8-3=5$ — The two right-hand edges together make the full height, 8 cm. One is 3, so the inside step is 5.

3. 3

   $P=10+8+4+5+6+3=36\text{ cm}$ — Walk round all six outside edges once and add them.

4. 4

   $4\times 8=32$ — For the area, cut with a line straight down. The tall part is 4 wide and 8 high.

5. 5

   $6\times 3=18$ — The short part on the right is 6 wide and 3 high.

6. 6

   $32+18=50{\text{ cm}}^2$ — Add the two pieces. Check another way: 10 × 8 − 6 × 5 = 50 too.

Worked example

1. 1

   $9\times 7=63$ — The full rectangle round the floor, as if the corner were still there.

2. 2

   $4\times 3=12$ — The missing corner.

3. 3

   $63-12=51{\text{ m}}^2$ — Whole take the corner leaves the floor area.

4. 4

   $51÷12=4.25$ — Each tin covers 12 m², so divide to see how many tins the floor needs.

5. 5

   $5\text{ tins}$ — 4 tins cover only 48 m², less than 51. So round up, never down.

6. 6

   $5\times \$18=\$90$ — The cost is the number of tins × the price of one tin.

Worked example

1. 1

   $10\times 6=60$ — The rectangle: length × width.

2. 2

   $r=6÷2=3$ — The semicircle sits on the 6 cm side, so 6 is its diameter. Halve it to get the radius.

3. 3

   $\pi \times 3^2=9\pi$ — The area of the whole circle, $\pi r^2$.

4. 4

   $\frac{1}{2}\times 9\pi =4.5\pi$ — The semicircle is half a circle, so halve it.

5. 5

   $60+4.5\pi =74.1{\text{ cm}}^2$ — Add the pieces. Keep π in the calculator until this last line, then round to 3 s.f.

Worked example

1. 1

   $14\times 10=140$ — The whole rectangle, hole included.

2. 2

   $\pi \times 4^2=16\pi$ — The hole: $\pi r^2$ with radius 4.

3. 3

   $140-16\pi =89.7{\text{ cm}}^2$ — Whole take the hole leaves the shaded part. Round to 3 s.f.

Worked example

1. 1

   $10+6+10=26$ — The three straight outside sides. The 6 cm side under the semicircle is inside, so it is left out.

2. 2

   $\pi \times 6=6\pi$ — A full circle's edge is $\pi d$. The diameter is 6.

3. 3

   $\frac{1}{2}\times 6\pi =3\pi$ — Only half the circle is on the edge, so halve it.

4. 4

   $26+3\pi =35.4\text{ cm}$ — Add the straight part and the curved part. Round to 3 s.f.

Worked example

1. 1

   $\pi \times 3^2\times 10=90\pi$ — The cylinder: $\pi r^{2}h$, the circle's area times the height.

2. 2

   $\frac{4}{3}\times \pi \times 3^3=36\pi$ — A whole sphere of radius 3: $\frac{4}{3}\pi r^3$.

3. 3

   $\frac{1}{2}\times 36\pi =18\pi$ — The hemisphere is half a sphere, so halve it.

4. 4

   $90\pi +18\pi =108\pi$ — Add the two pieces. If the question says “in terms of π”, stop here.

5. 5

   $=339{\text{ cm}}^3$ — Otherwise work out 108 × π = 339.29… and round to 3 s.f.

Worked example

1. 1

   $\pi \times 4^2=16\pi$ — The flat circle at the bottom of the cylinder: $\pi r^2$.

2. 2

   $2\pi \times 4\times 9=72\pi$ — The curved side of the cylinder: $2\pi rh$, the label unrolled into a rectangle.

3. 3

   $2\pi \times 4^2=32\pi$ — The curved top: the hemisphere, $2\pi r^2$.

4. 4

   $16\pi +72\pi +32\pi =120\pi$ — Add the three outside surfaces. The circle where the two solids join is hidden inside, so it is not counted.

5. 5

   $=377{\text{ cm}}^2$ — 120 × π = 376.99…, which is 377 to 3 s.f. If they ask “in terms of π”, the answer is 120π.

Worked example

1. 1

   $12\times 12\times 10=1440$ — The box: length × width × height, in cm³.

2. 2

   $\pi \times 6^2\times 10=360\pi$ — The tin: $\pi r^{2}h$. Leave it as 360π so nothing is rounded too early.

3. 3

   $1440-360\pi =309.03\dots$ — The empty space is the box take the tin, about 309 cm³.

4. 4

   $\frac{309.03\dots }{1440}\times 100=21.5\%$ — A percentage is part ÷ whole × 100. The whole is the box, not the tin.