CPhill

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 #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].
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

 

 

cool cool cool

CPhill vor 7 Stunden
 #2
avatar+81063 
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CPhill vor 10 Stunden
 #1
avatar+81063 
+2

 

Probably a way easier way to do this with Geometry...but....I didn't see it...so.....

 

By SAS,   triangles BCF  and ABE  are congruent

 

Therefore angle EAB  = angle FBC

And angle DAE   = angle BEA

But angle DAE + angle EAB  = 90

But angle FBC +  angle FBA  = 90

So  angle BEA =  angle FBA

So...by  AA congruence....triangles BEG   and ABG are similar

But angle BGA  = angle EGB...so....each must = 90

 

So triangles BEG an ABG are right triangles

And AB  = 2BE

So  AG  = 2 BG

So AB^2  =  AG^2  + BG^2

AB^2  =  (2BG)^2  + BG^2

AB^2  =  5BG^2

So AB   = sqrt(5)BG    =  BC

And  (1/2)AB   =  [sqrt(5)/2] BG

cos angle ABG  = BG/ AB  =  BG / sqrt(5)BG   =  1/ sqrt(5)

sin angle ABG =2BG/sqrt (5)BG  =  2/sqrt(5)

And angle ABG  = angle CFB

So sin angle ABG  = sin angle CFB

But  angle CFB  and angle GFD are supplementary so their sines are equal

So....sin ABG  = sin GFD     

And GFD is obtuse...so   

So... cos GFD  =  -sqrt [  1  - sin^2(ABG) ]  =  -sqrt [ 1 -  sin^2(GFD) ]  = 

 - sqrt [ 1 - (2/sqrt(5))^2 ]  = -sqrt [ 1 - 4/5]    = -sqrt (1 /5)  =  -1/sqrt (5)

 

And

Triangles ABE  and BCF  are congruent by SAS

EA  = FB     BC  = AB     and BE  = CF =  (1/2)AB  = FD

EA^2  =  BC^2 + BE^2    .... so.....

FB^2  =  AB^2  +  (AB/2)^2

FB^2  =  5BG^2 + AB^2/4

FB^2  =  5BG^2 +  (5/4)BG^2

FB   =   BGsqrt  (5 + 5/4)

FB =  BGsqrt [ 25/4]  =  (5/2)BG

 

And  

FG  =  FB - BG  =     (5/2)BG - BG  =  (3/2)BG

 

So  using the Law of Cosines

 

GD^2  =  FG^2  + FD^2 - 2(FG)(FD)cosGFD

GD^2  =  FG^2  + (AB/2)^2 - 2(FG)(AB/2)cosGFD

GD^2  = (9/4)BG^2 +   (5/4)BG^2 - 2 (3/2)BG * ( sqrt(5)BG/2) * [ - (1/sqrt(5) ) ]

 

GD^2  =  (14/4)BG^2 + (3/2)BG^2

GD^2  =  (7/2)BG^2 +  (3/2) BG^2

GD^2  =  (10/2)BG^2

GD^2 =  5BG^2

GD  =  sqrt(5)BG

DG  =  sqrt(5)BG  =  AB

 

 

cool cool cool

CPhill 21.01.2018 01:11