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zhuklara [117]
2 years ago
12

In a track field competition, A discus has a radius of 11 centimeters. What is the circumference, I centimeters, of the discus?

Mathematics
1 answer:
Morgarella [4.7K]2 years ago
3 0

Given the radius of the discus used in the track field competition, the circumference of the discus is 69.08cm.

<h3>What is the circumference of the discus?</h3>

Note that; a circumference is the perimeter of a circle or ellipse.

It is expressed as C = 2πr

Where r is the radius and π is the constant pi ( π = 3.14 )

Given that;

  • Radius of the discus r = 11cm
  • Circumference C = ?

C = 2πr

C = 2 × 3.14 × 11cm

C = 69.08cm

Therefore, given the radius of the discus used in the track field competition, the circumference of the discus is 69.08cm.

Learn more about circumference and area circles here: brainly.com/question/11952845

#SPJ1

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52.9 inches .

Step-by-step explanation:

Given that the triangular indentation has an area of 100 in.² and the base and height of this traingle are represented by expressions 3x and x+3 respectively . We need to find out the <u>perimeter </u>to the nearest tenth.

As we know that the area of triangle is ,

\rm\longrightarrow Area_{\triangle}=\dfrac{1}{2}(base)(height)

Substituting the respective values,

\rm\longrightarrow 100\ =\dfrac{1}{2}(x+3)(3x)\\\\\rm\longrightarrow 2(100) = 3x(x+3)\\\\\rm\longrightarrow200 = 3x^2+9x\\\\\rm\longrightarrow 3x^2+9x-200=0

On using the <u>Quadratic</u><u> formula</u> , we have;

\rm\longrightarrow x =\dfrac{-9\pm\sqrt{9^2-4(3)(-200)}}{2(3)} \\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{81+1200}}{6}\\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{1281}}{6}

On simplifying above , we will get ,

\rm\longrightarrow x = 6.802,-9.802

Since sides can't be negative, therefore,

\rm\longrightarrow\underline{\underline{ x =6.802}}

Therefore ,

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\rm\longrightarrow h=\sqrt{ b^2+p^2}\\

\rm\longrightarrow h =\sqrt{ (20.406)^2+(9.802)^2}\\

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Now we may find perimeter as ,

\rm\longrightarrow P = p + b + h \\

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\\\rm\longrightarrow P = 52.89 \ in.\\

\rm\longrightarrow \underline{\underline{ Perimeter = 52.9\ in .}}

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