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Luden [163]
3 years ago
6

The diagonal of the rectangle is 13 inches, and the side lengths of the triangles are Pythagorean triples. To the nearest tenth,

what is the area of the shaded part of the figure?
Mathematics
1 answer:
frosja888 [35]3 years ago
3 0
1) A Pythagorean triples is a set of three integers, a, b, c,  that follow the rule a^2 + b^2 = c^2

2) Assumin c is the diagonal of your rectangle and the hypotenuse of the triangles, you are looking this:

13^2 = a^2 + b^2

=> 169 = a^2 + b^2

Two integer numbers that meet that equality are 5 and 12:

5^2 = 25
12^2 = 144
=> 5^2 + 12^2 = 25 + 144 = 169

 => 13^2 = 5^2 + 12^2

Now that you have the three numbers you know the dimensions of the figure and can calculate the area of the shaded part. I cannot do that for you becasue you did not include the figure.
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Using the Babylonian method find √600 ???????????? √235.<br><br> how?
podryga [215]

Answer:

With the Babylonian method \sqrt{600} \approx 24.494897 and \sqrt{235} \approx 15.329709.

Step-by-step explanation:

  • To find the square root of 600, do the following:

1. Make an initial guess:  Because \sqrt{576}=24 and \sqrt{625} =25 you can start with x_{0}=24 as your initial guess.

2. Apply the formula:

x_{1}=\frac{(x_{0}+\frac{x}{x_{0}})}{2} where x=600

x_{1}=\frac{(24+\frac{600}{24})}{2}\\x_{1}=24.5

The number x_{1} is a better approximation to \sqrt{600}

3. Iterate until convergence:

For this apply the formula:

x_{n+1}=\frac{(x_{n}+\frac{x}{x_{n}})}{2}

Convergence is achieved when the digits of x_{n+1} and x_{n} agree to as many decimal places as you desire.

x_{2}=\frac{(x_{1}+\frac{x}{x_{1}})}{2}\\x_{2}=\frac{(24.5+\frac{600}{24.5})}{2}\\x_{2}=24.494897

x_{3}=\frac{(x_{2}+\frac{x}{x_{2}})}{2}\\x_{3}=\frac{(24.494897+\frac{600}{24.494897})}{2}\\x_{3}=24.494897\\

Because x_{2} and x_{3} agree to six decimal places. we can say that an estimate for \sqrt{600} \approx 24.494897.

You can compare with the value that WolframAlpha gives you which is \sqrt{600} \approx 24.49489742 you can see that it agrees to six decimal places.

  • To find the square root of 235, do the following:

1. Make an initial guess:  Because \sqrt{225}=15 and \sqrt{256} =16 you can start with x_{0}=15 as your initial guess.

2. Apply the formula:

x_{1}=\frac{(x_{0}+\frac{x}{x_{0}})}{2} where x=235

x_{1}=\frac{(15+\frac{235}{15})}{2}\\x_{1}=\frac{46}{3}

3. Iterate until convergence:

Apply the formula:

x_{n+1}=\frac{(x_{n}+\frac{x}{x_{n}})}{2}

x_{2}=\frac{(x_{1}+\frac{235}{x_{1}})}{2}\\x_{2}=\frac{(\frac{46}{3}+\frac{235}{\frac{46}{3}})}{2} \\x_{2}=15.329710

x_{3}=\frac{(x_{2}+\frac{235}{x_{2}})}{2}\\x_{3}=\frac{(15.329710+\frac{235}{15.329710})}{2} \\x_{3}=15.329709

x_{4}=\frac{(x_{3}+\frac{235}{x_{3}})}{2}\\x_{4}=\frac{(15.329709+\frac{235}{15.329709})}{2} \\x_{3}=15.329709

Because x_{3} and x_{4} agree to six decimal places. We can say that an estimate for \sqrt{235} \approx 15.329709.

WolframAlpha gives you \sqrt{235} \approx 15.329709716

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