Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

Example Question #2 : How To Graph A Function

The chord of a \(\displaystyle 120^{\circ }\) central angle of a circle with area \(\displaystyle 32 \pi\) has what length?

Possible Answers:

\(\displaystyle 4\sqrt{3}\)

\(\displaystyle 4 \sqrt{6}\)

\(\displaystyle 4\)

\(\displaystyle 4 \sqrt{2}\)

\(\displaystyle 8\)

Correct answer:

\(\displaystyle 4 \sqrt{6}\)

Explanation:

The radius \(\displaystyle r\) of a circle with area \(\displaystyle 32 \pi\) can be found as follows:

\(\displaystyle \pi r^{2} = A\)

\(\displaystyle \pi r^{2} = 32 \pi\)

\(\displaystyle r^{2} = 32\)

\(\displaystyle r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4 \sqrt{2}\)

The circle, the central angle, and the chord are shown below, along with \(\displaystyle \overline{AM}\), which bisects isosceles \(\displaystyle \Delta AOB\)

Chord

We concentrate on \(\displaystyle \Delta AOM\), a 30-60-90 triangle. By the 30-60-90 Theorem,

\(\displaystyle OM = \frac{1}{2}AO = \frac{1}{2} \cdot 4 \sqrt{2} = 2\sqrt{2}\)

and 

\(\displaystyle AM = OM \sqrt{3} = 2 \sqrt{2} \cdot \sqrt{3} = 2 \sqrt{6}\)

The chord \(\displaystyle \overline{AB}\) has length twice this, or

\(\displaystyle 2 \cdot 2 \sqrt{6} = 4 \sqrt{6}\)

Example Question #211 : Advanced Geometry

The chord of a \(\displaystyle 60^{\circ }\) central angle of a circle with circumference \(\displaystyle 90 \pi\) has what length?

Possible Answers:

\(\displaystyle 90\sqrt{2}\)

\(\displaystyle 45\)

\(\displaystyle 90\)

\(\displaystyle 45\sqrt{3}\)

\(\displaystyle 45\sqrt{2}\)

Correct answer:

\(\displaystyle 45\)

Explanation:

A circle with circumference \(\displaystyle 90 \pi\) has as its radius

\(\displaystyle 90\pi \div 2\pi = 45\).

The circle, the central angle, and the chord are shown below:

Chord

By way of the Isosceles Triangle Theorem, \(\displaystyle \Delta AOB\) can be proved equilateral, so \(\displaystyle AB = AO= 45\), the correct response.

Example Question #1 : How To Graph A Function

What is the domain of y = 4 - x^{2}\(\displaystyle y = 4 - x^{2}\)?

Possible Answers:

x \leq 0\(\displaystyle x \leq 0\)

x \leq 4\(\displaystyle x \leq 4\)

x \geq 4\(\displaystyle x \geq 4\)

all real numbers

\(\displaystyle x=0\)

Correct answer:

all real numbers

Explanation:

The domain of the function specifies the values that \(\displaystyle x\) can take.  Here, 4-x^{2}\(\displaystyle 4-x^{2}\) is defined for every value of \(\displaystyle x\), so the domain is all real numbers. 

Example Question #212 : Advanced Geometry

What is the domain of y=-2\sqrt{x}\(\displaystyle y=-2\sqrt{x}\)?

Possible Answers:

\(\displaystyle x>0\)

\(\displaystyle x=0\)

x\leq 0\(\displaystyle x\leq 0\)

\(\displaystyle x< -2\)

x\geq 0\(\displaystyle x\geq 0\)

Correct answer:

x\geq 0\(\displaystyle x\geq 0\)

Explanation:

To find the domain, we need to decide which values \(\displaystyle x\) can take.  The \(\displaystyle x\) is under a square root sign, so \(\displaystyle x\) cannot be negative.  \(\displaystyle x\) can, however, be 0, because we can take the square root of zero.  Therefore the domain is x\geq 0\(\displaystyle x\geq 0\).

Example Question #3 : Graphing A Function

What is the domain of the function y=\sqrt{4-x^{2}}\(\displaystyle y=\sqrt{4-x^{2}}\)?

Possible Answers:

\(\displaystyle \left ( -2,2 \right )\)

\(\displaystyle \left [ -2,2 \right ]\)

x\leq -2\(\displaystyle x\leq -2\)

\(\displaystyle x< -2\)

\(\displaystyle x>-2\)

Correct answer:

\(\displaystyle \left [ -2,2 \right ]\)

Explanation:

To find the domain, we must find the interval on which \sqrt{4-x^{2}}\(\displaystyle \sqrt{4-x^{2}}\) is defined.  We know that the expression under the radical must be positive or 0, so \sqrt{4-x^{2}}\(\displaystyle \sqrt{4-x^{2}}\) is defined when x^{2}\leq 4\(\displaystyle x^{2}\leq 4\).  This occurs when x \geq -2\(\displaystyle x \geq -2\) and x \leq 2\(\displaystyle x \leq 2\).  In interval notation, the domain is \(\displaystyle \left [ -2,2 \right ]\).

Example Question #4 : Graphing A Function

Define the functions \(\displaystyle f\) and \(\displaystyle g\) as follows:

\(\displaystyle \small \small f (x) = 2x + \sqrt{x - 1}\) 

\(\displaystyle \small g (x) = 6x - \sqrt{x-1}\)

What is the domain of the function \(\displaystyle (f+g)(x)\) ?   

Possible Answers:

\(\displaystyle \small \small [1, \infty )\)

\(\displaystyle \small \small (-\infty , -1] \cup [1, \infty )\)

\(\displaystyle \small (-\infty ,\infty )\)

\(\displaystyle \small (-\infty , 1) \cup (1, \infty )\)

\(\displaystyle \small (1, \infty )\)

Correct answer:

\(\displaystyle \small \small [1, \infty )\)

Explanation:

The domain of \(\displaystyle (f+g)\) is the intersection of the domains of \(\displaystyle f\) and \(\displaystyle g\)\(\displaystyle f\) and \(\displaystyle g\) are each restricted to all values of \(\displaystyle x\) that allow the radicand \(\displaystyle x-1\) to be nonnegative - that is, 

\(\displaystyle \small \small x - 1\geq 0\), or 

\(\displaystyle \small \small \small x \geq 1\)

Since the domains of \(\displaystyle f\) and \(\displaystyle g\) are the same, the domain of \(\displaystyle (f+g)\) is also the same. In interval form the domain of \(\displaystyle (f+g)\) is \(\displaystyle \small \small [1, \infty )\)

Example Question #4 : How To Graph A Function

Define \(\displaystyle g (x) = \frac{1}{\sqrt[3]{x -27}}\)

What is the natural domain of \(\displaystyle g\)?

Possible Answers:

\(\displaystyle (-\infty , - 27)\cup (27, \infty )\)

\(\displaystyle (-\infty , - 3)\cup (3, \infty )\)

\(\displaystyle (-\infty , 27)\cup (27, \infty )\)

\(\displaystyle (3, \infty )\)

\(\displaystyle (27, \infty )\)

Correct answer:

\(\displaystyle (-\infty , 27)\cup (27, \infty )\)

Explanation:

The radical in and of itself does not restrict the domain, since every real number has a real cube root. However, since the expression \(\displaystyle \sqrt[3]{x -27}\) is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which 

\(\displaystyle \sqrt[3]{x -27} = 0\)

\(\displaystyle \left (\sqrt[3]{x -27} \right )^{3}= 0^{3}\)

\(\displaystyle x -27 = 0\)

\(\displaystyle x = 27\)

27 is the only number excluded from the domain.

Example Question #5 : Graphing A Function

Define \(\displaystyle g(x) = \frac{x}{x^{2}+3x-4 }\)

What is the natural domain of \(\displaystyle g\) ?

Possible Answers:

\(\displaystyle ( - \infty ,-1 ) \cup (-1,4) \cup (4, \infty )\)

\(\displaystyle ( - \infty ,-1 ) \cup (-1,0) \cup (0,4) \cup (4, \infty )\)

\(\displaystyle ( - \infty ,-4 ) \cup (-4,0) \cup (0,1) \cup (1, \infty )\)

\(\displaystyle ( - \infty ,-4 ) \cup (-4,1) \cup (1, \infty )\)

\(\displaystyle ( - \infty ,0) \cup (0, \infty )\)

Correct answer:

\(\displaystyle ( - \infty ,-4 ) \cup (-4,1) \cup (1, \infty )\)

Explanation:

Since the expression \(\displaystyle x^{2}+3x-4\) is in a denominator, it cannot be equal to zero, so the domain excludes the value(s) for which \(\displaystyle x^{2}+3x-4 =0\). We solve for \(\displaystyle x\) by factoring the polynomial, which we can do as follows:

\(\displaystyle x^{2}+3x-4 =0\)

\(\displaystyle (x+?)(x+?)=0\)

Replacing the question marks with integers whose product is \(\displaystyle -4\) and whose sum is 3:

\(\displaystyle (x+4)(x-1)=0\)

\(\displaystyle x + 4 = 0 \Rightarrow x = -4\)

\(\displaystyle x -1 = 0 \Rightarrow x = 1\)

Therefore, the domain excludes these two values of \(\displaystyle x\).

Example Question #1 : How To Graph A Function

Define \(\displaystyle f (x) = \frac{1}{x ^{2}- 25 }\).

What is the natural domain of \(\displaystyle f\)?

 

Possible Answers:

\(\displaystyle (-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)\)

\(\displaystyle (-\infty , -5 ) \cup \left ( -5, \infty)\)

\(\displaystyle (-5 , 5)\)

\(\displaystyle (-\infty , -5 ) \cup (5, \infty)\)

\(\displaystyle (-\infty , 5 ) \cup \left ( 5, \infty)\)

Correct answer:

\(\displaystyle (-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)\)

Explanation:

The only restriction on the domain of \(\displaystyle f\) is that the denominator cannot be 0. We set the denominator to 0 and solve for \(\displaystyle x\) to find the excluded values:

\(\displaystyle x ^{2}-25 =0\)

\(\displaystyle x ^{2}= 25\)

\(\displaystyle x = -5 \textrm{ or }x = 5\)

The domain is the set of all real numbers except those two - that is, 

\(\displaystyle (-\infty , -5 ) \cup \left ( -5,5\right ) \cup (5, \infty)\).

Example Question #121 : Coordinate Geometry

2

The figure above shows the graph of y = f(x). Which of the following is the graph of y = |f(x)|?

Possible Answers:

6

4

2

5

3

Correct answer:

2

Explanation:

One of the properties of taking an absolute value of a function is that the values are all made positive. The values themselves do not change; only their signs do. In this graph, none of the y-values are negative, so none of them would change. Thus the two graphs should be identical.

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