The diagram above shows a solid structure that is formed from two identical cones and a cylinder. The cylinder is attached to the bases of the two cones. The cone has radius 3x cm3x\ \mathrm{cm}, height 4x cm4x\ \mathrm{cm} and slant height l cml\ \mathrm{cm}. The cylinder has radius 3x cm3x\ \mathrm{cm} and height $y\ \mathrm{cm}.

  1. (a)

    Express ll in terms of xx.

    Answer l=……………………l = \dots\dots\dots\dots\dots\dots\dots\dots

  2. (b)

    The volume of the cylinder is equal to the volume of one cone.

    Show that y=43xy = \frac{4}{3}x.

    Answer

  3. (c)

    The total surface area of the solid structure is (200−x)π cm2(200 - x)\pi\ \mathrm{cm}^2. Write an expression, in terms of xx, for the total surface area of the solid and show that it reduces to $38x^2 + x - 200 = 0.

    Answer

  4. (d)

    Solve the equation 38x2+x−200=038x^2 + x - 200 = 0, giving your solutions correct to 2 decimal places.

    Answer x=………………………x = \dots\dots\dots\dots\dots\dots\dots\dots\dots

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Worked solution

8(a)

1. Find the slant height using Pythagoras' theorem

The cone's radius, height and slant height form a right-angled triangle.

l2=(3x)2+(4x)2l^2=(3x)^2+(4x)^2
l2=9x2+16x2=25x2l^2=9x^2+16x^2=25x^2
l=5xl=5x

Answer

l=5x cml=5x\text{ cm}

8(b)

1. Write the volume of the cylinder and the volume of one cone in terms of $x$ and $y$

Radius of both solids =3x=3x.

Volume of cylinder:
πr2h=π(3x)2y=9πx2y\pi r^2h=\pi(3x)^2y=9\pi x^2y

Volume of one cone:
13πr2h=13π(3x)2(4x)=12πx3\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(3x)^2(4x)=12\pi x^3

2. Equate the two volumes and solve for $y$

Given that the volume of the cylinder is equal to the volume of one cone,
9πx2y=12πx39\pi x^2y=12\pi x^3

Divide both sides by 9πx29\pi x^2:
y=12πx39πx2=43xy=\frac{12\pi x^3}{9\pi x^2}=\frac{4}{3}x

Answer

y=43x cmy=\dfrac{4}{3}x\text{ cm}

8(c)

1. Find the curved surface area of the two cones

From earlier, l=5xl=5x and the radius is 3x3x.

Curved surface area of one cone:

πrl=π(3x)(5x)=15πx2\pi rl=\pi(3x)(5x)=15\pi x^2

For two identical cones:

2(15πx2)=30πx22(15\pi x^2)=30\pi x^2

2. Find the curved surface area of the cylinder and write the total surface area

From earlier, y=43xy=\dfrac{4}{3}x and the radius is 3x3x.

Curved surface area of cylinder:

2πrh=2π(3x)(43x)=8πx22\pi rh=2\pi(3x)\left(\frac{4}{3}x\right)=8\pi x^2

Total surface area of the solid:

30πx2+8πx2=38πx230\pi x^2+8\pi x^2=38\pi x^2

3. Equate the expression to the given total surface area and reduce to the required quadratic equation

Given that the total surface area is (200−x)π cm2(200-x)\pi\text{ cm}^2,

38πx2=(200−x)π38\pi x^2=(200-x)\pi

Divide by π\pi:

38x2=200−x38x^2=200-x

Bring all terms to one side:

38x2+x−200=038x^2+x-200=0

Answer

Total surface area =38πx2 cm2=38\pi x^2\text{ cm}^2.

Since the total surface area is also (200−x)π cm2(200-x)\pi\text{ cm}^2,

38πx2=(200−x)π38\pi x^2=(200-x)\pi

38x2+x−200=038x^2+x-200=0

8(d)

1. Identify the coefficients and substitute into the quadratic formula

For 38x2+x−200=038x^2+x-200=0,

a=38,b=1,c=−200a=38,\quad b=1,\quad c=-200

Using x=−b±b2−4ac2a,x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},

x=−1±12−4(38)(−200)2(38)x=\frac{-1\pm\sqrt{1^2-4(38)(-200)}}{2(38)}

2. Simplify the discriminant and calculate the two roots

x=−1±1+3040076x=\frac{-1\pm\sqrt{1+30400}}{76}

x=−1±3040176x=\frac{-1\pm\sqrt{30401}}{76}

x=2.280…orx=−2.307…x=2.280\ldots \quad \text{or} \quad x=-2.307\ldots

3. Round the solutions correct to 2 decimal places

Correct to 2 decimal places,

x=2.28orx=−2.31x=2.28 \quad \text{or} \quad x=-2.31

Answer

x=2.28x=2.28 or x=−2.31x=-2.31

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