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1)

In triangle DEF, DE=5, DF=6, and EF=7. Circles are drawn centered at D, E, and F, so that the circles are pairwise externally tangent. Find the sum of the areas of the three circles.

2)

Chords AB and CD of a circle are perpendicular, and meet at P. If AP = 3, BP = 4, CP = 6, and DP = 2, then find the diameter of the circle.

3)

Peter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit. He has also built a pathway around the gazebo, of constant width 1 unit, as shown below. (Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.) Find the area of the pathway.

4)

Trapezoid ABCD is inscribed in a semicircle, as shown below. The length of base CD is 7 and the lengths of legs AD and BC are 3. Find the radius of the semicircle.

 Dec 16, 2019
 #1
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-1

2) Let O be the center of the circle, and draw OA.  Drop the perpendicular from O to AB, and you'll hit the midpoint of chord AB.  You can build right triangles for the other chord, which will eventually give you an answer of sqrt(51).

 Dec 16, 2019
 #6
avatar+111321 
+1

sqrt (51)  ≈    7.14    

 

This would mean that the diameter would be shorter than chord  CD......but....this is impossible since the diameter is the longest chord in any  circle 

 

 

cool cool cool

CPhill  Dec 18, 2019
 #2
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4) Drop pependiculars from C and D to AB, and you can work with the right triangles.   By Pythagoras, r^2 - (3/2)^2 = 7^2 - (r - 7/2)^2.  Therefore, r = 13/2.

 Dec 16, 2019
 #3
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0

Are these homework problems?

 Dec 16, 2019
 #4
avatar+692 
+1

1)    3 circles area = 91.10618695 n2

 

2)     Circle's diameter =  8.062257748 

 

3)     The pathway area = 10.14159265 u2

 

4)      Radius  r = 4.5 units    indecision

 Dec 17, 2019
edited by Dragan  Dec 17, 2019
edited by Dragan  Dec 18, 2019
edited by Dragan  Mar 5, 2020
 #5
avatar+111321 
+1

1.   Let the  radius of te circle with center D  = x

    Let the radius  of the circle with center E  = y

    Let the radius  of the circle with the center F  = z

 

So we have these equations

x + y  = 5

x + z = 6

y + z  = 7

 

Subtract thesecond equation from the  first  and we have that  y - z  =  -1

Add this to equation to the third equation  and we have that

 

2y  = 6

y = 3  

x = 2

z = 4

 

So...the total area of the circles  are   pi ( 2^2 + 3^2 +4^2)  =  pi  ( 29)  =   29 pi units^2  ≈

 

91.1 units^2

 

 

cool cool cool

 Dec 17, 2019
 #7
avatar+111321 
0

Last one

 

Draw DE  and  CF  perpendicular to AB

 

So   AE  =  BF   =   sqrt (3^2 - H^2)       where H is the height of the trapezoid

 

And AB  = 2R....so....

 

2R  =AE + BF + 7

 

2R = 2*AE + 7

 

2R  = 2 sqrt ( 9 - H^2)  + 7

 

R  =   sqrt (9- H^2)  + 3/2

 

R - 3/2  =  sqrt (9 - H^2)       square both sides 

 

R^2 - 3R + 9/4  =  9 - H^2      rearrange as 

 

H^2  =   3R - R^2 + 6.75

 

Let the midpoint of AB  =  O ......  connect OD   and   draw OM  perpendicular to   DC

 

And  OM  =   H     and DM =  3.5    and OD  is the  radius of the  semi-circle = R

 

So....triangle  ODM  is right    and we have that

 

OD^2  -  OM^2  =  DM^2     ....so..... 

 

R^2 - H^2  =  3.5^2

 

R^2  -  (3R - R^2 + 6.75) =  3.5^2        rearrange as

 

R^2 + R^2 - 3R -6.75  = 12.25      subtract   12.25 from each side

 

2R^2 - 3R - 19  =  0

 

Using  the quad formula  we have that

 

R =        3  ± √ [ 9  + 4* 2 * 19 ]            3 ± √ [ 161]

           ___________________   =    ___________ 

                         4                                        4

 

Take the positive answer   and we get that

 

R =     3 + √161

          _________  ≈    3.92

                 4

 

And the diameter ≈  3 * 3.92  =   7.84

 

 

 

cool cool cool

 Dec 18, 2019

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