Area, Perimeter and Volume with Algebra

Volume of a cylinder =\pi r^2h, and the whole of 2x is squared, not just the x.

1. Volume of a cylinder =\pi r^2h
2. =\pi\times(2x)^2\times5x
3. (2x)^2=4x^2, not 2x^2
4. =\pi\times4x^2\times5x
5. Answer: 20\pi x^3 cm^3

A solid hemisphere has a curved surface and a flat circular face.

1. Curved surface =\frac12\times4\pi r^2=2\pi r^2
2. Flat circular face =\pi r^2
3. Total =2\pi r^2+\pi r^2
4. Answer: 3\pi r^2

The perimeter of a semicircle is the curved edge plus the straight edge (the diameter).

1. Curved edge =\frac12\times2\pi r=\pi r
2. Straight edge = diameter =2r
3. \pi r+2r=36
4. Factorise: r(\pi+2)=36
5. Answer: r=\dfrac{36}{\pi+2}

The area of the base is (2x)^2, not 2x^2.

1. Area of base =(2x)^2=4x^2
2. Volume of pyramid =\frac13\times 4x^2\times h=\frac{4x^2h}{3}
3. Volume of cube =x^3
4. \frac{4x^2h}{3}=x^3
5. 4x^2h=3x^3
6. h=\frac{3x^3}{4x^2}=\frac{3x}{4}

Set the two formulae equal to each other, then divide both sides by everything they have in common.

1. \frac43\pi r^3=4\pi r^2
2. Divide both sides by 4\pi r^2 (allowed because r\neq0)
3. \dfrac{r}{3}=1
4. Answer: r=3

Write the two unlabelled sides in terms of x, add up all six sides and solve for x.

1. Unlabelled vertical side =(x+5)-4=x+1
2. Unlabelled horizontal side =(2x+3)-x=x+3
3. Perimeter =(2x+3)+4+(x+3)+(x+1)+x+(x+5)
4. =6x+16
5. 6x+16=52, so x=6
6. Split into two rectangles: bottom 15\times4=60
7. Top: 6\times(11-4)=6\times7=42
8. Answer: 102 cm^2

Area of a trapezium =\frac12(a+b)h; multiply by the length to get the volume, then solve the quadratic.

1. Area of cross-section =\frac12(x+x+4)\times x
2. =\frac12(2x+4)x=x^2+2x
3. Volume =10(x^2+2x)=480
4. x^2+2x=48
5. x^2+2x-48=0
6. (x+8)(x-6)=0
7. x=-8 is impossible for a length
8. Answer: x=6

Area of a triangle =\frac12\times base \times height; set up an equation and remember x cannot be 0.

1. Area of triangle =\frac12(x+5)(2x)=x(x+5)=x^2+5x
2. Area of square =x^2
3. x^2+5x=3x^2
4. 2x^2-5x=0
5. x(2x-5)=0
6. x=0 is impossible for a length, so 2x-5=0
7. x=2.5

Write the volume of each part in terms of \pi and r; a hemisphere is half a sphere.

1. Hemisphere: \frac12\times\frac43\pi r^3=\frac23\pi r^3
2. Cone: \frac13\pi r^2\times2r=\frac23\pi r^3
3. Total: \frac43\pi r^3=288\pi
4. r^3=288\times\frac34=216
5. r=\sqrt[3]{216}
6. Answer: r=6

Find the slant height with Pythagoras' theorem, and remember the total surface area includes the circular base.

1. Slant height: l^2=(5x)^2+(12x)^2=169x^2
2. l=13x
3. Curved surface =\pi\times5x\times13x=65\pi x^2
4. Base =\pi(5x)^2=25\pi x^2
5. Total =65\pi x^2+25\pi x^2
6. Answer: 90\pi x^2 cm^2

Write both volumes in terms of \pi and x, then divide.

1. Sphere: \frac43\pi(3x)^3=\frac43\pi\times27x^3=36\pi x^3
2. One cone: \frac13\pi x^2\times2x=\frac23\pi x^3
3. n=36\pi x^3\div\frac23\pi x^3
4. =36\times\frac32
5. Answer: n=54

The two semicircles make one full circle, so use the perimeter to write y in terms of r first.

1. The two semicircles make a circle of circumference 2\pi r
2. Perimeter: 2y+2\pi r=400
3. y=200-\pi r
4. The rectangle is y m long and 2r m wide
5. A=2r(200-\pi r)
6. Answer: A=400r-2\pi r^2

Find the radius and the height of P above the x-axis, then subtract the triangle from the semicircle.

1. Radius =\sqrt{169}=13, so AB=26
2. At P: 25+y^2=169, so y^2=144
3. P is above the x-axis, so its height is 12
4. Semicircle =\frac12\pi\times13^2=\frac{169}{2}\pi
5. Triangle APB=\frac12\times26\times12=156
6. Answer: \dfrac{169}{2}\pi-156

Height =\dfrac{2\times\text{area}}{\text{base}}; then rationalise the denominator by multiplying by 2-\sqrt3.

1. Area =\frac12\times base \times height, so height =\dfrac{2\times\text{area}}{\text{base}}
2. Height =\dfrac{10+6\sqrt3}{2+\sqrt3}
3. Multiply top and bottom by 2-\sqrt3
4. Bottom: (2+\sqrt3)(2-\sqrt3)=4-3=1
5. Top: (10+6\sqrt3)(2-\sqrt3)=20-10\sqrt3+12\sqrt3-18
6. Answer: 2+2\sqrt3 cm

Both total surface areas include a circular base, so set them equal to find the slant height, then use Pythagoras' theorem.

1. Cone: \pi r^2+\pi rl
2. Hemisphere: \frac12\times4\pi r^2+\pi r^2=3\pi r^2
3. \pi r^2+\pi rl=3\pi r^2, so \pi rl=2\pi r^2
4. l=2r
5. Height^2=(2r)^2-r^2=3r^2
6. Answer: \sqrt3\,r cm

Set \pi rl equal to 1.5 times \pi r^2 to link l and r, then use the right-angled triangle inside the cone.

Part (a)

1. \pi rl=1.5\pi r^2
2. Divide by \pi r: l=1.5r
3. The radius, the height and the slant height make a right-angled triangle
4. The angle \theta at the base has \cos\theta=\dfrac{r}{l}=\dfrac{r}{1.5r}=\dfrac23
5. \theta=\cos^{-1}\left(\dfrac23\right)=48.18\ldots
6. Answer: 48.2^\circ

Part (b)

1. Pythagoras: l^2=r^2+h^2, so (1.5r)^2=r^2+(2\sqrt5)^2
2. 2.25r^2=r^2+20
3. 1.25r^2=20, so r^2=16 and r=4
4. l=1.5\times4=6
5. Total surface area =\pi r^2+\pi rl=16\pi+24\pi
6. Answer: 40\pi cm^2

The diagonals of a rhombus cross at right angles and cut each other in half, so the rhombus is four congruent right-angled triangles. Answer part (a) before attempting part (b).

Part (a)

1. Let AC=k, so BD=2k
2. The diagonals cut the rhombus into four right-angled triangles with shorter sides \frac{k}{2} and k
3. Area =4\times\frac12\times\frac{k}{2}\times k=k^2
4. k^2=64, so AC=8 and BD=16
5. AM=4 and BM=8, so AB^2=4^2+8^2=80
6. AB=\sqrt{80}=4\sqrt5
7. Perimeter =4\times4\sqrt5
8. Answer: 16\sqrt5 cm

Part (b)

1. A rhombus is a parallelogram, so area = base \times perpendicular height
2. Base AB=4\sqrt5 (from part (a))
3. 4\sqrt5\times height =64
4. Height =\dfrac{64}{4\sqrt5}=\dfrac{16}{\sqrt5}=7.155\ldots
5. Answer: 7.16 cm

Write the area of the ring as the larger circle's area minus the smaller circle's area, then divide through by \pi.

Part (a)

1. Area between the circles =\pi(x+3)^2-\pi x^2
2. Set equal to the smaller circle: \pi(x+3)^2-\pi x^2=\pi x^2
3. Divide by \pi: x^2+6x+9-x^2=x^2
4. x^2-6x-9=0
5. x=\dfrac{6\pm\sqrt{36+36}}{2}=\dfrac{6\pm 6\sqrt2}{2}
6. x=3\pm3\sqrt2
7. 3-3\sqrt2 is negative, so reject it
8. x=3+3\sqrt2

Part (b)

1. Radius of larger circle =x+3=6+3\sqrt2
2. (6+3\sqrt2)^2=36+36\sqrt2+18=54+36\sqrt2
3. Area =(54+36\sqrt2)\pi cm^2
4. (Check: the larger circle is twice the smaller one, 2\pi(3+3\sqrt2)^2=2\pi(27+18\sqrt2) ✓)

The total surface area is two circles plus the curved surface; after solving the quadratic inequality, make sure the height is positive.

1. Total surface area =2\pi x^2+2\pi x(2x-3)
2. =2\pi x^2+4\pi x^2-6\pi x=6\pi x^2-6\pi x
3. 6\pi x^2-6\pi x<36\pi
4. Divide by 6\pi: x^2-x<6
5. x^2-x-6<0, so (x-3)(x+2)<0
6. Critical values -2 and 3; the graph is U-shaped, so -2<x<3
7. The height must be positive: 2x-3>0, so x>\frac32
8. Answer: \frac32<x<3

The surface is the curved surface of the cylinder plus the curved surface of the hemisphere. Answer part (a) before attempting part (b).

Part (a)

1. Curved surface of the cylinder =2\pi rh
2. Curved surface of the hemisphere =\frac12\times4\pi r^2=2\pi r^2
3. 2\pi rh+2\pi r^2=54\pi
4. Divide by 2\pi: rh+r^2=27
5. Answer: h=\dfrac{27-r^2}{r}

Part (b)

1. Cylinder: \pi r^2h=\pi r^2\times\dfrac{27-r^2}{r}=\pi r(27-r^2)
2. =27\pi r-\pi r^3
3. Hemisphere: \frac12\times\frac43\pi r^3=\frac23\pi r^3
4. V=27\pi r-\pi r^3+\frac23\pi r^3
5. Answer: V=27\pi r-\frac13\pi r^3