Question

1. jeffrey believes he has discovered a pattern in the pythagorean triples (three - positive integers that satisfy the pythagorean theorem). in every pythagorean triple, at least two of the numbers 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 64π square units, what is the surface area of the cube (in square units)?

○a) 96
○b) 128
○c) 384
○d) 512

Explanation:

Step1: Check Pythagorean triples and difference condition for 118

A Pythagorean triple satisfies $a^{2}+b^{2}=c^{2}$.

• 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 $13 - 12 = 1$, $12-5

eq1$.

• For option D: $(8,15,17)$ since $8^{2}+15^{2}=64 + 225=289=17^{2}$, and $17-15 = 2$, $15 - 8=7$, it is a counter - example as no two numbers differ by 1.

Step2: Solve for the surface area of the cube in 119

The surface area of a sphere is $S = 4\pi r^{2}$. Given $S=64\pi$, then $4\pi r^{2}=64\pi$.
Dividing both sides by $4\pi$ gives $r^{2}=16$, so $r = 4$.
If a cube is inscribed in a sphere, the diameter of the sphere is the space - diagonal of the cube. The diameter $d = 8$, and if the side length of the cube is $a$, the space - diagonal of the cube $D=\sqrt{3}a$. So $\sqrt{3}a=8$, then $a=\frac{8}{\sqrt{3}}$.
The surface area of a cube $A = 6a^{2}=6\times(\frac{8}{\sqrt{3}})^{2}=6\times\frac{64}{3}=128$.

Answer:

1. D. $(8,15,17)$
2. B. 128