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oksian1 [2.3K]
3 years ago
6

Perimeter of a quarter circle is 17.85 what is the quarter circles radius

Mathematics
1 answer:
givi [52]3 years ago
4 0

The radius of the given circle  is 2.84 units

Step-by-step explanation:

Step 1 :

Given

The perimeter of a quarter table is 17.85

We need to find the circle's radius.

Step 2 :

The circle's perimeter is given by the formula, 2 \pi r where r represents the circle's radius

So from this we have ,

2 \pi r = 17.85

We can determine the radius by dividing the perimeter value by 2 \pi

= >  r = 17.85 divided by 2 \pi r

        = 2.84 [ taking  \pi  = \frac{22}{7}]

Step 3 :

Answer :

The radius of the given circle  is 2.84 units

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Where |a| and |b| are the norms of vectors <em>a</em> and b, and cosФ is the cosine of the angle between either vector <em>a</em> and <em>b</em>.

If a = [ \\ a_{1}, a_{2}, a_{3} ] and b = [\\ b_{1}, b_{2}, b_{3}], then the dot product is simply a number (not a vector) obtained from:

a . b = \\ a_{1}*b_{1} + a_{2}*b_{2} + a_{3}*b_{3}

The norm of a vector (its length) is, for instance, |a| = \\ \sqrt{{a_{1}^2 + {a_{2}}^2 + {a_{3}}^2}, for a vector in \\ R^{3}.

Having all that into account, we can determine the angles of the triangle for each vertex using equation [1] and solving it for Ф.

<h3>Angle of the triangle for vertex in (1, 1, 1)</h3>

The vectors which form an angle from this vertex are the result of subtracting the vertex (1, 1, 1) to any of the remaining points (1, -3, 2) and (-3, 2, 5):

v(1, 1, 1) - v(1, -3, 2) = \\ (1 - 1, 1 - (-3), 1 - 2) = (0, 1 + 3, -1) = (0, 4, -1)

v(1, 1, 1) - v(-3, 2, 5) = \\ (1 - (-3), 1 - 2, 1 - 5) = (1 + 3, -1, -4) = (4, -1, -4)

The <em>dot product</em> for these vectors is:

[0, 4, -1] . [4, -1, -4] = [0 * 4 + 4 * -1 + -1 * -4] = 0 - 4 + 4 = 0

The norm for each vector is:

|(0, 4, -1)| = \\ \sqrt{0^2 + 4^2 + -1^2} = \sqrt{0 + 16 + 1} = \sqrt{17}

|(4, -1, -4)| = \\ \sqrt{4^2 + -1^2 + -4^2} = \sqrt{16 + 1 + 16} = \sqrt{33}

So

a . b = |a| |b| cosФ

\\ 0 = \sqrt{17} * \sqrt{33} * cos{\theta}

\\ \frac{0}{\sqrt{17} * \sqrt{33}} = cos{\theta}

\\ cos^{-1}{0}} = cos^{-1}(cos{\theta})

\\ 90 = \theta

In vertex (1, 1, 1) the angle of the triangle is 90 degrees. We have here a right triangle.

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Therefore, the other angles are 35.67 degrees and 180 - (90 + 35.67) = 180 - 125.67 = 54.33 degrees.

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