Question

1. a segment of a circle with a circumference of 20π miles is bounded by two radii and an arc that measures 4π miles. what is the area bounded by these lines?

a) 20π miles²
b) 25 miles²
c) 2π miles²
d) 80π miles²

1. jeffrey believes he has discovered a pattern in the pythagorean triples (three positive integers that satisfy the pythagorean theorem relationship).

in every pythagorean triple, at least two of the members differ by 1 unit.
which of the following is a counter - example that disproves the above statement?
a) (3,4,5)
b) (5,7,11)
c) (5,12,13)
d) (8,15,17)

1. a cube is inscribed in a sphere. if the surface area of the sphere is 96π square units, what is the surface area of the cube (in square units)?

a) 36
b) 128
c) 224
d) 512

1. if the sine of angle β of a right triangle is 1/3, and the side adjacent to that angle measures 5√2 units, which of the following must be true?

a) cosβ = 2√2/3
b) cosβ = √6/3
c) tanβ = 1/√6
d) tanβ = 1/2√2

Explanation:

Step1: Solve the first - circle sector area problem

The circumference of a circle is $C = 2\pi r$. Given $C=20\pi$, then $2\pi r=20\pi$, so $r = 10$ miles. The area of a circle is $A=\pi r^{2}=\pi\times10^{2}=100\pi$ square - miles. If the arc measures $4\pi$ miles, and the formula for the length of an arc is $l = r\theta$ (where $\theta$ is the central angle in radians), then $4\pi=10\theta$, so $\theta=\frac{2\pi}{5}$ radians. The area of a sector of a circle is $A_{s}=\frac{1}{2}r^{2}\theta$. Substituting $r = 10$ and $\theta=\frac{2\pi}{5}$, we get $A_{s}=\frac{1}{2}\times10^{2}\times\frac{2\pi}{5}=20\pi$ square - miles.

Step2: Solve the Pythagorean triples counter - example problem

For a Pythagorean triple $(a,b,c)$ where $a^{2}+b^{2}=c^{2}$.

• For option A: $(3,4,5)$ since $3^{2}+4^{2}=9 + 16=25=5^{2}$, and $4 - 3=1$.
• For option B: $(5,7,11)$ since $5^{2}+7^{2}=25 + 49 = 74

eq11^{2}=121$, it is not a Pythagorean triple.

• For option C: $(5,12,13)$ since $5^{2}+12^{2}=25+144 = 169=13^{2}$, and $13 - 12 = 1$.
• For option D: $(8,15,17)$ since $8^{2}+15^{2}=64 + 225=289=17^{2}$, and $17-15 = 2$, $15 - 8=7$, so $(8,15,17)$ is a counter - example.

Step3: Solve the cube - inscribed in a sphere problem

If a cube is inscribed in a sphere of radius $R$, the relationship between the side length $a$ of the cube and the radius of the sphere is $a=\frac{2R}{\sqrt{3}}$. The surface area of a sphere is $S_{s}=4\pi R^{2}=96\pi$, then $4\pi R^{2}=96\pi$, so $R^{2}=24$, $R = 2\sqrt{6}$. The side length of the cube $a=\frac{2R}{\sqrt{3}}=\frac{2\times2\sqrt{6}}{\sqrt{3}}=4\sqrt{2}$. The surface area of a cube is $S_{c}=6a^{2}$, substituting $a = 4\sqrt{2}$, we get $S_{c}=6\times(4\sqrt{2})^{2}=6\times32 = 192$.

Step4: Solve the right - triangle trigonometry problem

Given $\sin\beta=\frac{1}{3}$ and the adjacent side $x = 5\sqrt{2}$. Using the Pythagorean identity $\sin^{2}\beta+\cos^{2}\beta = 1$, $\cos\beta=\sqrt{1-\sin^{2}\beta}=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}$. Also, $\tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}}=\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$.

Answer:

1. A. $20\pi$ miles$^{2}$
2. D. $(8,15,17)$
3. (There is an error in the options, the correct surface area of the cube is 192)
4. None of the given options are correct.