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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].
Enter your answer in the form x + y√z in simplest radical form.

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
+101746
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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].
Enter your answer in the form x + y√z in simplest radical form.

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
+101746
+1

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