Area, Volume & Surface Area

Area is $\pi r^2$, so it depends on the square of the radius. Doubling $r$ multiplies the area by $2^2 = 4$. (The circumference, $2\pi r$, would only double.)

Circumference $= 2\pi r = 2\times 3.14\times 7 = 43.96$, which rounds to 44.0 cm. (Using $\pi r$ forgets the factor of 2; $\pi r^2$ is the area.)

$C = \pi d = \pi\times 10 = 10\pi$ cm. (Or $2\pi r$ with $r = 5$. $25\pi$ is the area $\pi r^2$.)

$A = \pi r^2 = \pi\times 5^2 = 25\pi$ cm². ($100\pi$ uses the diameter; $10\pi$ is the circumference.)

$C = 2\pi r$, so $r = \dfrac{C}{2\pi} = \dfrac{12\pi}{2\pi} = 6$ cm. The $\pi$ cancels, so there is no $\pi$ in the answer.

Outer rectangle $8\times 6 = 48$ cm², minus the cut-out $4\times 3 = 12$ cm²: $48 - 12 = 36$ cm². (48 is the outer rectangle only; adding gives 60.)

Garden area $= 20\times 15 = 300$ m²; patch area $= 6\times 6 = 36$ m². Fraction $= \dfrac{36}{300} = 0.12 = 12\%$. (Comparing the side lengths instead of the areas gives the wrong figure.)

Full circle $\pi r^2 = \pi\times 6^2 = 36\pi$, halved for a semicircle: $\dfrac{36\pi}{2} = 18\pi$ cm². ($36\pi$ forgets to halve.)

Rectangle $10\times 8 = 80$ cm². Semicircle $\dfrac{\pi r^2}{2} = \dfrac{\pi\times 4^2}{2} = \dfrac{16\pi}{2} = 8\pi$ cm². Total $(80 + 8\pi)$ cm². ($16\pi$ uses the full circle.)

Volume $= 50\times 40\times 30 = 60\,000$ cm³, which is $\dfrac{60\,000}{1000} = 60$ L. At 4 L per minute, time $= \dfrac{60}{4} = 15$ minutes. (Stopping at 60 forgets to divide by the flow rate.)

Volume is $s^3$, so it depends on the cube of the edge length. Doubling $s$ multiplies the volume by $2^3 = 8$. (The surface area, which depends on $s^2$, would only become 4 times larger.)

Along each edge you fit $\dfrac{8}{2}=4$, $\dfrac{6}{2}=3$ and $\dfrac{4}{2}=2$ cubes, so $4\times 3\times 2 = 24$. (Or box volume $192$ cm³ divided by cube volume $2^3 = 8$ cm³ gives $24$. Dividing by 2 instead of 8 gives the wrong 96.)

Cross-section $= \dfrac{1}{2}\times 4\times 3 = 6$ cm², times the length: $6\times 10 = 60$ cm³. (Forgetting the half gives 120.)

$V = \pi r^2 h = \pi\times 3^2\times 10 = 90\pi$ cm³. ($30\pi$ forgets to square; $9\pi$ forgets the height.)

The straight edges are the base and the two sides: $20 + 30 + 30 = 80$ cm. The top is a semicircular arc of diameter 20 cm, so its length is $\dfrac{1}{2}\pi d = \dfrac{1}{2}\pi(20) = 10\pi$ cm. Total perimeter $(80 + 10\pi)$ cm. (The top straight edge is replaced by the arc, so it is not counted; using $20\pi$ takes the full circumference.)

Walls $= 2(4\times 2.5) + 2(3\times 2.5) = 20 + 15 = 35$ m²; floor $= 4\times 3 = 12$ m². Total $35 + 12 = 47$ m². Each tin covers 6 m², so $\dfrac{47}{6} = 7.83\ldots$. You cannot buy part of a tin, so round up to 8 tins. (Rounding down to 7 leaves some wall unpainted.)

$SA = 2(lw + lh + wh) = 2(5\times 3 + 5\times 2 + 3\times 2) = 2(15 + 10 + 6) = 62$ cm². (30 is the volume; 31 forgets the factor of 2.)

Two ends $2\pi r^2 = 2\pi\times 25 = 50\pi$, lateral surface $2\pi r h = 2\pi\times 5\times 10 = 100\pi$. Total $50\pi + 100\pi = 150\pi$ cm². ($250\pi$ is the volume.)

The amount of water that fills a pool is a volume: $V = 10\times 6\times 1.5 = 90$ m³. (Surface area is for paint or tiles; perimeter is a length; the floor area ignores the depth.)

$V = 40\times 30\times 20 = 24\,000$ cm³. Convert: $\dfrac{24\,000}{1000} = 24$ L. (24,000 forgets to convert.)

Full circle $\pi r^2 = \pi\times 64 = 64\pi$, a quarter is $\dfrac{64\pi}{4} = 16\pi$ cm². ($32\pi$ is a half; $64\pi$ is the full circle.)

The paved region is a ring: big circle minus small. $\pi\times 7^2 - \pi\times 5^2 = 49\pi - 25\pi = 24\pi$ m². (Using $\pi(7-5)^2 = 4\pi$ is wrong, since $7^2 - 5^2 \ne (7-5)^2$.)

Poster $80\times 100 = 8000$ cm². Hole $\pi r^2 = 3.14\times 20^2 = 3.14\times 400 = 1256$ cm². Remaining $8000 - 1256 = 6744$ cm². (Adding gives 9256.)

$A = \pi r^2$, so $36\pi = \pi r^2$, giving $r^2 = 36$ and $r = \sqrt{36} = 6$ cm. (The $\pi$ cancels; 12 is the diameter.)

Trapezium $= \dfrac{1}{2}(6 + 12)(8) = \dfrac{1}{2}(18)(8) = 72$ m². Semicircle (radius $\dfrac{4}{2} = 2$): $\dfrac{\pi\times 2^2}{2} = \dfrac{4\pi}{2} = 2\pi$ m². Garden left: $(72 - 2\pi)$ m². ($(144 - 2\pi)$ forgets the half in the trapezium.)