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?
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
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