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# PLS HELP NOW PLS DUE TMROW

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1) Find the area of a regular 12-gon inscribed in a unit circle.

2) A regular hexagon has side length 6. If the perimeter and area of the hexagon are p and A, respectively, what is the value of (p^4)/(a^2)?

3)Isosceles triangle OPQ has legs OP = OQ, base PQ = 2, and and angle POQ = 45 degrees. Find the distance from O to PQ.

4) A, B, C, D and E are points on a circle of radius 2 in counterclockwise order. We know AB = BC = DE = 2 and CD = EA Find [ABCDE].

5)Two unit squares share the same center. The overlapping region of the two squares is an octagon with perimeter 3.5. What is the area of the octagon?

Jan 21, 2018

#1
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1) Find the area of a regular 12-gon inscribed in a unit circle.

We have 12 identical triangles  identical isosceles triangles with equal sides of 1 and whose apex angle  between these sides =  360/12  = 30”

The  area will  be given  by    (1/2) (1)^2*sin (360/12°)  =

(1/2 sin (30°)  =  1/2  *  1/2  =   (1/4 ) units^2

2) A regular hexagon has side length 6. If the perimeter and area of the hexagon are p and A, respectively, what is the value of (p^4)/(a^2)?

The perimeter, p, is 36  ⇒   p^4  =   36^4

The area, a,  is   (1/2)6^2sin (60)  =  18*√3/2  =  9√2  ⇒ a^2  = (√182) ^2   = 162

So p^4 / a^2  =    36^4  / 162   =   10368

3)Isosceles triangle OPQ has legs OP = OQ, base PQ = 2, and and angle POQ = 45 degrees. Find the distance from O to PQ.

The distance from O to PQ  is the altitude....call the point where the altitude intersects the base, R

And this altitude bisects POQ....so angle POR  =  22.5°

And the altitude also bisects the base.... so PR  =  1

Using the tangent function....we have that

tan (22.5)  =  1 / altitude        rearrange as

altitude  =   1  / tan (22.5)  ≈ 2.4142  =    1 + √2

4) A, B, C, D and E are points on a circle of radius 2 in counterclockwise order. We know AB = BC = DE = 2 and CD = EA Find [ABCDE].

This is a regular pentagon inscribed in a circle....I assume you want the area of ABCDE??

If so...... the area is   5 (1/2)(2)^2sin (72)  = 10sin (72)  = 10 √  [  5/8  + √5/8 ]  units^2

If you want the perimeter...we  have that  the half side lengh  =

2sin(36)

And we have 10 half side lengths comprising the perimeter....so....the perimeter  =

10 * 2 *  sin(36)   = 20√ [ 5/8  - √5/8 ] units   Jan 21, 2018
#2
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If the perimeter is  3.5....then one side  = 3.5/ 8  = .4375

This serves as the base of one of the 8 congruent triangles comprising the octagon

And we  can  find the side, s, of one of these triangles thusly

(.4375)^2  =  2s^2  - (s^2)√2

(/4375)^2  = s^2 ( 2 - √2)

s  =  .4375 / √ (2 - √2)

So....the area of the octagon is given by

8 *  (1/2) s^2 * sin (45°)

8 *  (1/2) [ (.4375)^2 / (2 - √2) ] (√2) / 2  =

2√2 * (.4375)^2 / ( 2 - √2)    ≈   .9242  units^2   Jan 22, 2018