Composite Area and Surface Area

For a fixed perimeter the circle always encloses the most area (the isoperimetric property). E.g. perimeter $4$: a square has area $1$, but a circle has area $\dfrac{4}{\pi} \approx 1.27$.

Height $= \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3}$ cm, so $A = \dfrac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}$ cm². (18 cm² used the side $6$ as the height.)

$\pi r^2 = 36\pi$ gives $r = 6$, so $C = 2\pi r = 12\pi$ cm. ($36\pi$ confused area with circumference.)

$A = \dfrac{1}{2}(a + b)h$, so $40 = \dfrac{1}{2}(6 + b)(5)$. Then $6 + b = \dfrac{2 \times 40}{5} = 16$, giving $b = 16 - 6 = 10$ cm. (16 cm forgot to subtract the known side; undo the $\tfrac{1}{2}$ and the height first.)

$\pi(5^2) - \pi(3^2) = 25\pi - 9\pi = 16\pi$ cm². ($4\pi$ used $(5 - 3)^2$; you must subtract the squared radii.)

Add the two parts: rectangle $8 \times 5 = 40$ cm² and triangle $\dfrac{1}{2}(8)(3) = 12$ cm², total $40 + 12 = 52$ cm². (40 cm² forgot the triangle.)

Square $10^2 = 100$ cm², circle $\pi \times 3^2 = 9\pi$ cm², so remaining $= 100 - 9\pi$ cm². ($100 + 9\pi$ added instead of subtracting; $100 - 6\pi$ used $2r$.)

Outer $10 \times 8 = 80$ cm², cut-out $6 \times 4 = 24$ cm², so the L-shape $= 80 - 24 = 56$ cm². (104 cm² added; 24 cm² is only the cut-out.)

Use the two-sides-and-included-angle formula $A = \dfrac{1}{2}ab\sin C$: $A = \dfrac{1}{2}(8)(5)\sin 30^\circ = \dfrac{1}{2}(40)(0.5) = 10$ cm². ($40$ cm² forgot the $\sin 30^\circ = 0.5$ factor.)

Rectangle $100 \times 60 = 6000$ m². The two semicircles combine into one full circle of radius 30 m: $\pi \times 30^2 = 900\pi$ m². Total $= 6000 + 900\pi$ m². ($6000 + 60\pi$ used $2r$; $6000 + 1800\pi$ counted the circle twice.)

$\dfrac{\theta}{360^\circ} \times \pi(6)^2 = 12\pi$, so $\dfrac{\theta}{360} \times 36 = 12$ and $\theta = 120^\circ$.

The diagonal, width and length form a right-angled triangle, so the length is $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. The area is $5 \times 12 = 60$ cm². ($65$ cm² wrongly multiplied the diagonal by the width.)

Arc $= \dfrac{72}{360} \times 2\pi(10) = \dfrac{1}{5} \times 20\pi = 4\pi$ cm. Add the two radii: perimeter $= 4\pi + 10 + 10 = (4\pi + 20)$ cm. ($4\pi$ alone is just the arc; the radii must be added.)

$L = \dfrac{\theta}{360^\circ} \times 2\pi r = \dfrac{90}{360} \times 2\pi \times 6 = \dfrac{1}{4} \times 12\pi = 3\pi$ cm. ($12\pi$ is the full circumference; arc length uses $2\pi r$, not $\pi r^2$.)

Three rectangle sides are exposed: $10 + 10 + 6 = 26$ cm (the joined 6 cm side is internal). The semicircle adds its arc only: $\dfrac{1}{2} \times 2\pi r = \pi r = 3\pi$ cm. Total $= 26 + 3\pi$ cm. ($26 + 6\pi$ used the diameter as the radius; $60 + 18\pi$ computed the area instead.)

Surface area is $6s^2$, so it scales with the square of the side: doubling $s$ multiplies the area by $2^2 = 4$. (×8 is how the volume scales, since $V$ depends on $s^3$.)

$6s^2 = 96$ gives $s^2 = 16$, so $s = 4$ cm and $V = s^3 = 4^3 = 64$ cm³. (16 cm³ stopped at $s^2$; you must cube the side.)

Total surface area $= \pi r l + \pi r^2 = \pi(3)(5) + \pi(3)^2 = 15\pi + 9\pi = 24\pi$ cm². ($15\pi$ is the curved surface only, forgetting the circular base.)

$SA = 2\pi r^2 + 2\pi r h = 2\pi(9) + 2\pi(3)(10) = 18\pi + 60\pi = 78\pi$ cm². ($60\pi$ is the lateral surface only; $18\pi$ is the two ends only.)

Big cube $6 \times 6^2 = 216$ cm². The small cube hides a $2 \times 2 = 4$ cm² patch of the top but adds its own 5 exposed faces $5 \times 4 = 20$ cm²: $216 - 4 + 20 = 232$ cm². ($236$ forgot to subtract the hidden patch.)

The cube's surface area is $6 \times 10^2 = 600$ cm². Removing the corner cube takes away three $2 \times 2$ squares ($12$ cm²) but reveals three new $2 \times 2$ squares ($12$ cm²), a net change of 0. So the surface area stays $600$ cm². (A corner removal does not change the surface area.)

Outer lateral $2\pi(5)(20) = 200\pi$ cm², inner lateral $2\pi(3)(20) = 120\pi$ cm², two annular ends $2 \times \pi(5^2 - 3^2) = 2 \times 16\pi = 32\pi$ cm². Total $= 200\pi + 120\pi + 32\pi = 352\pi$ cm². ($320\pi$ forgot the annular ends.)

The circle's radius is $5$ cm, so its area is $\pi(5)^2 = 25\pi \approx 78.5$ cm², while the square's area is $10^2 = 100$ cm². The circle covers $\dfrac{78.5}{100} = 78.5\%$. (The ratio is always $\dfrac{\pi}{4} \approx 78.5\%$ for an inscribed circle.)

Base $6^2 = 36$ cm², plus four triangular faces $4 \times \dfrac{1}{2}(6)(5) = 60$ cm². Total $= 36 + 60 = 96$ cm². (60 cm² forgot the base; 36 cm² is the base only.)

Cube surface area $6 \times 100 = 600$ cm². Subtract the two entry/exit circles $2 \times \pi(3^2) = 18\pi$ cm², then add the hole's lateral surface $2\pi(3)(10) = 60\pi$ cm²: $600 - 18\pi + 60\pi = 600 + 42\pi$ cm². ($600 + 60\pi$ forgot to subtract the circles; $600 - 60\pi$ subtracted the lateral surface.)