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3 years ago

7

What is the circumference of the circle that circumscribes a triangle with side lengths 3, 4, and 5?

1 answer:

djyliett [7]3 years ago

6 0

Answer:

5π, about 15.708 units

Step-by-step explanation:

A 3-4-5 triangle is a right triangle. When a circle circumscribes a right triangle, the hypotenuse is the diameter of the circle. The circumference of a circle is given by ...

C = πd

For a diameter of 5 units, the circumference is ...

C = π(5) = 5π = 15.708 . . . units

_____

<em>Additional comment</em>

The (3, 4, 5) triple is one of the first Pythagorean triples you run across. It is the smallest integer triple, and the only primitive triple with values in an arithmetic sequence. You can show this is a Pythagorean triple by ...

3² + 4² = 9 +16 = 25 = 5²

That is, these numbers satisfy the Pythagorean theorem relation for sides of a right triangle.

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Read 2 more answers

A rectangular box is to have a square base and a volume of 12 ft3. If the material for the base costs $0.17/ft2, the material fo

katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the box $V=x^2z$

Volume of the box $=12 ft^3\\$

$Therefore, x^2z=12\\z=\frac{12}{x^2}$

The material for the base costs $\$0.17/ft^2$ , the material for the sides costs $\$0.10/ft^2$ , and the material for the top costs $\$0.13/ft^2$ .

Area of the base $=x^2$

Cost of the Base $=\$0.17x^2$

Area of the sides $=4xz$

Cost of the sides= $=\$0.10(4xz)$

Area of the Top $=x^2$

Cost of the Base $=\$0.13x^2$

Total Cost, $C(x,z) =0.17x^2+0.13x^2+0.10(4xz)$

Substituting $z=\frac{12}{x^2}$

$C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}$

To minimize C(x), we solve for the derivative and obtain its critical point

$C'(x)=\dfrac{0.6x^3-4.8}{x^2}\\Setting \:C'(x)=0\\0.6x^3-4.8=0\\0.6x^3=4.8\\x^3=4.8\div 0.6\\x^3=8\\x=\sqrt[3]{8}=2$

Recall: $z=\frac{12}{x^2}=\frac{12}{2^2}=3\\$

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

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