上图显示了一个由两个相同的圆锥和一个圆柱组成的几何体。圆柱连接在两个圆锥的底面上。 圆锥的半径为 $3x\ \mathrm{cm}$，高度为 $4x\ \mathrm{cm}$，斜高为 $l\ \mathrm{cm}$。圆柱的半径为 $3x\ \mathrm{cm}$，高度为 $y\ \mathrm{cm}$。

1. Find the slant height using Pythagoras' theorem

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

$l^2=(3x)^2+(4x)^2$
$l^2=9x^2+16x^2=25x^2$
$l=5x$

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

Radius of both solids $=3x$.

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

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

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

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

Divide both sides by $9\pi x^2$:
$y=\frac{12\pi x^3}{9\pi x^2}=\frac{4}{3}x$

1. Find the curved surface area of the two cones

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

Curved surface area of one cone:

$\pi rl=\pi(3x)(5x)=15\pi x^2$

For two identical cones:

$2(15\pi x^2)=30\pi x^2$

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

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

Curved surface area of cylinder:

$2\pi rh=2\pi(3x)\left(\frac{4}{3}x\right)=8\pi x^2$

Total surface area of the solid:

$30\pi x^2+8\pi x^2=38\pi x^2$

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

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

$38\pi x^2=(200-x)\pi$

Divide by $\pi$:

$38x^2=200-x$

Bring all terms to one side:

$38x^2+x-200=0$

1. Identify the coefficients and substitute into the quadratic formula

For $38x^2+x-200=0$,

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

Using $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$

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

2. Simplify the discriminant and calculate the two roots

$x=\frac{-1\pm\sqrt{1+30400}}{76}$

$x=\frac{-1\pm\sqrt{30401}}{76}$

$x=2.280\ldots \quad \text{or} \quad x=-2.307\ldots$

3. Round the solutions correct to 2 decimal places

Correct to 2 decimal places,

$x=2.28 \quad \text{or} \quad x=-2.31$