SAT Nonlinear Functions Question 234 | Free SAT Prep | Aniko

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Question 234·Hard·Nonlinear Functions

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Equilateral triangle $T$ has side length $x$ centimeters. Equilateral triangle $S$ has an area that is $9\sqrt{3}$ square centimeters greater than the area of triangle $T$ . The function $g$ gives the perimeter of triangle $S$ , in centimeters. Which of the following defines $g$ ?

For equilateral triangle questions, quickly recall the key formulas: perimeter $P = 3s$ and area $A = \frac{\sqrt{3}}{4}s^2$ . Translate all the words into variables (here, $x$ for triangle $T$ and $s$ for triangle $S$ ), write an equation for the area relationship (area of $S$ equals area of $T$ plus the given amount), and solve that equation for the new side length in terms of $x$ . Finally, multiply the side length by 3 to get the perimeter function and match it to the answer choices, watching carefully for whether the variable appears inside a square or a square root.

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Relate perimeter to side length

For an equilateral triangle, the perimeter is three times the side length. What do you need to find about triangle $S$ in order to write $g(x)$ ?

Use the area formula for an equilateral triangle

Recall that the area of an equilateral triangle with side length $a$ is $A = \frac{\sqrt{3}}{4}a^2$ . Write the areas of triangles $T$ and $S$ using this formula.

Set up the area equation and solve for the side of S

Use the fact that the area of $S$ is $9\sqrt{3}$ greater than the area of $T$ to write an equation involving $x$ and the side length $s$ of $S$ . Then solve that equation for $s$ in terms of $x$ .

Go from side length to g(x)

Once you have $s$ expressed in terms of $x$ , how do you write the perimeter of $S$ as a function of $x$ ? Match that expression to one of the answer choices.

Desmos Guide

Enter the area relationship in Desmos

In one expression line, type the equation representing the areas of triangles $S$ and $T$ :

(\sqrt{3}/4)\,s^2 = (\sqrt{3}/4)\,x^2 + 9\sqrt{3}.

Solve the equation for s in terms of x

Use Desmos to solve this equation for $s$ (for example, by using its solve feature or rearranging algebraically and then entering the result). Desmos should display $s$ as a square root expression in terms of $x$ .

Create the perimeter function in Desmos

In a new line, define the perimeter of triangle $S$ as a function of $x$ by typing something like

P(x) = 3 \times (\text{the expression for } s \text{ you just found}).

Desmos will show you a simplified expression for $P(x)$ .

Compare with the answer choices

Look at the expression Desmos shows for $P(x)$ and compare its form to each of the four answer options. Choose the option whose expression exactly matches the Desmos perimeter function.

Step-by-step Explanation

Translate the situation into variables

Let the side length of triangle $T$ be $x$ centimeters (given), and let the side length of triangle $S$ be $s$ centimeters.

Because both triangles are equilateral:

Perimeter of $T$ is $3x$ .
Perimeter of $S$ is $3s$ .

The function $g(x)$ is supposed to give the perimeter of triangle $S$ in terms of $x$ , so ultimately we need to express $3s$ as a function of $x$ .

Write area formulas for equilateral triangles

The area $A$ of an equilateral triangle with side length $a$ is

A = \frac{\sqrt{3}}{4}a^2.

So for the two triangles:

Area of triangle $T$ is $A_T = \frac{\sqrt{3}}{4}x^2$ .
Area of triangle $S$ is $A_S = \frac{\sqrt{3}}{4}s^2$ .

We are told that triangle $S$ has area $9\sqrt{3}$ greater than triangle $T$ , so

A_S = A_T + 9\sqrt{3}.

Substitute the expressions for $A_S$ and $A_T$ :

\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}x^2 + 9\sqrt{3}.

Solve for the side length of triangle S in terms of x

We now solve

\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}x^2 + 9\sqrt{3}

for $s$ .

Multiply both sides by $\dfrac{4}{\sqrt{3}}$ to clear the fraction and the $\sqrt{3}$ :

\begin{aligned} \left(\frac{4}{\sqrt{3}}\right)\cdot \frac{\sqrt{3}}{4}s^2 &= \left(\frac{4}{\sqrt{3}}\right)\cdot \left(\frac{\sqrt{3}}{4}x^2 + 9\sqrt{3}\right) \\ s^2 &= x^2 + 36. \end{aligned}

So the side length of triangle $S$ satisfies

s^2 = x^2 + 36.

Because $s$ is a length and must be positive,

s = \sqrt{x^2 + 36}.

Express the perimeter of triangle S and match to g(x)

The perimeter of an equilateral triangle is three times its side length, so the perimeter of triangle $S$ is

P_S = 3s = 3\sqrt{x^2 + 36}.

Therefore the function that gives the perimeter of triangle $S$ in terms of $x$ is

\boxed{g(x) = 3\sqrt{x^2 + 36}},

which corresponds to choice C.

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation