Super-Hard SAT Math Question 96 | Free SAT Prep | Aniko

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Question 96·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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Cone	Volume	Surface Area
A	$100\pi$	$k\pi$
B	$12{,}500\pi$	$j\pi$

The table gives the volume of two similar right circular cones. If the radius of cone A is 5 cm, what is the value of $j-k$ ?

For a right circular cone:

Volume $= \frac{1}{3}\pi r^2 h$

Surface area $= \pi r^2 + \pi r\ell$ , where $\ell$ is the slant height

$\ell = \sqrt{r^2+h^2}$

When a problem involves similar 3D figures, separate it into two phases: (1) compute one figure’s needed measurement (here, surface area of cone A) using the given formulas, and (2) scale to the other figure using similarity rules. Remember: volume gives you the linear scale factor via a cube root, and surface area then follows from squaring that linear factor.

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Start with cone A

Use the given radius and volume of cone A in $V=\frac{1}{3}\pi r^2h$ to find its height.

Use a right triangle for the slant height

For a right cone, the radius and height are perpendicular legs, and the slant height is the hypotenuse: $\ell=\sqrt{r^2+h^2}$ .

Use similarity rules

For similar solids, volume scales with the cube of the linear scale factor, and surface area scales with the square.

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Solve for the height of cone A by graphing

Enter the equations:

$y=100$
$y=(25/3)x$

Find their intersection; the $x$ -coordinate is $h$ .

Compute slant height and $k$

Define:

$h=12$ (use the value you found)
$l=\sqrt{5^2+h^2}$
$k=25+5l$

Check that $l$ is an integer and that $k$ is the coefficient of $\pi$ in the surface area.

Find the linear scale factor from the volume ratio

Enter:

$s=(12500/100)^(1/3)$

This is the linear scale factor between the similar cones.

Scale the surface area and compute the difference

Enter:

$j=s^2\cdot k$
$j-k$

Match the computed value of $j-k$ to one of the answer choices.

Step-by-step Explanation

Find the height of cone A from its volume

Use $V=\frac{1}{3}\pi r^2h$ with $V=100\pi$ and $r=5$ :

\begin{aligned} 100\pi &= \frac{1}{3}\pi(5^2)h \\ 100 &= \frac{25}{3}h \\ 300 &= 25h \\ h &= 12 \end{aligned}

Find the slant height and surface area of cone A

Compute slant height:

\ell=\sqrt{r^2+h^2}=\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13

Surface area of cone A:

\begin{aligned} \text{SA}_A &= \pi r^2+\pi r\ell \\ &= \pi(25)+\pi(5)(13) \\ &= (25+65)\pi \\ &= 90\pi \end{aligned}

So $k=90$ .

Use the volume ratio to get the similarity scale factor

Because the cones are similar,

\frac{V_B}{V_A}=\left(\text{scale factor}\right)^3

Compute the volume ratio:

\frac{12{,}500\pi}{100\pi}=125=5^3

So the linear scale factor from A to B is $5$ .

Scale surface area and compute $j-k$

Surface area scales by the square of the linear factor, so

\frac{\text{SA}_B}{\text{SA}_A}=5^2=25

Since $\text{SA}_A=90\pi$ , then $\text{SA}_B=25(90\pi)=2250\pi$ , so $j=2250$ .

Finally,

j-k=2250-90=2160

Correct answer: 2160

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Step-by-step Explanation

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