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1) The centers of three non-intersecting circles lie on the same straight line. Prove that if a fourth circle is tangent to all three of the given circles, then its radius is greater than that of at least one of the given three.

 

2) Two parallelograms ABCD and DCEF are put together along their common side CD. Prove that the quadrilateral ABEF is also a parallelogram.

3) Circle O has diameter AE with AE = 8 cm. Point C is on the circumference of the circle such that AC and CE are congruent. Also, AC is a diameter of semicircle ABC, and CE is a diameter of semicircle CDE, as shown in the figure. In square centimeters, what is the total combined area of the shaded regions?

4) Three coplanar squares with sides of lengths two, four and six units, respectively, are arranged side-by-side, as shown so that one side of each square lies on line AB and a segment connects the bottom left corner of the smallest square to the upper right corner of the largest square. What is the area of the shaded quadrilateral?

 

5) In the following diagram, two sides of a square are tangent to a circle with diameter 8. One corner of the square lies on the circle. Find the area of the square.

 Mar 23, 2020
 #1
avatar+111329 
+2

4) Three coplanar squares with sides of lengths two, four and six units, respectively, are arranged side-by-side, as shown so that one side of each square lies on line AB and a segment connects the bottom left corner of the smallest square to the upper right corner of the largest square. What is the area of the shaded quadrilateral?

 

 

Let h be  the height of the shaded  region

 

By similar triangles, we  have  that

 

6/12  =  h/ 6

 

6 * 6 /12 = h 

 

36/12  = h

 

3  = h

 

So......the  shaded area  plus  the smaller triangular area at the bottom of the left-most square has the area

 

(1/2)(6)(3)  =  9  units^2

 

And using similar triangles again....the height of this smaller triangular area is

 

h / 2  = 6 / 12

 

h = 2*6 / 12  = 1

 

So....the area of this triangular region  is  (1/2) (2)(1)  = 1  units^2

 

So....the shaded area =  9 - 1 =   8 units^2

 

 

cool cool cool

 Mar 23, 2020
 #2
avatar
0

Thanks, helps a ton 

 Mar 23, 2020
 #3
avatar+111329 
+2

(5)

 

Let  the  center of the  circle  be  O

 

Let  the  sides of the square  be tanglent to the  circle at  M  and N

 

Let  the bottom right vertex of the square  be   P

 

So  OMPN  forms a smaller square with sides  of 4  and  a diagonal of  4√2

 

So....the diagonal of the  larger square  =    4√2 + 4  =  4 ( √2 + 1)

 

So....the side of the  larger square is  =   (4/√2) ( √2 + 1) 

 

So....the area of this square is

 

[ (4/√2)   ( √2 + 1 )  ]^2  =

 

( 8)  ( 2  + 2√2 + 1 ]   =

 

(8) [ 3 + 2√2]  =

 

[24 + 16√2]  units^2

 

cool cool cool

 Mar 23, 2020
 #4
avatar+111329 
+2

3) Circle O has diameter AE with AE = 8 cm. Point C is on the circumference of the circle such that AC and CE are congruent. Also, AC is a diameter of semicircle ABC, and CE is a diameter of semicircle CDE, as shown in the figure. In square centimeters, what is the total combined area of the shaded regions?

 

Believe it or not, the two shaded regions  have  the same combined area  as  that of right triangle ACE

 

Proof  :

 

Radius  of semi-circle  CDE  = (1/2)CE

 

So...area  of   semi-circle CDE  = (1/2)pi [(1/2)CE]*2 =  (pi / 8) CE^2

 

And similarly.....the area of semi-circle  ABC =  (pi/8)AC^2

 

Area  of  right triangle  ACE  =  (1/2)(AC)(CE)

 

Area  of    two  unshaded   regions  between the right triangle and  the circle  =

 

(1/2)pi[(1/2)AE]^2  - (1/2)(AC)(CE)  =  

 

(1/2) pi (1/4) [ AE]^2  - (1/2)(AC)(CE)  =    { note that AE^2  = CE^2 + AC^2 }

 

(1/8) pi [ CE^2 + AC^2 ]  - (1/2)(AC)(CE)

 

So....the  area  of  the two shaded regions  =

 

(pi/8) (CE^2 + AC^2]  -  [ (1/8)pi (CE^2 + AC^2 ) - (1/2)(AC)(CE) ]  =

 

(1/2)(AC)(CE).....which is the same  area as right triangle ACE

 

And  the area of  right triangle  ACE = (1/2) (CE)(AC) =   (1/2)(8/√2)(8/√2)  = 64 / 4  =  16 cm^2 = area of two  shaded regions

 

 

cool cool cool

 Mar 23, 2020
edited by CPhill  Mar 23, 2020
edited by CPhill  Mar 23, 2020
 #5
avatar+2854 
+1

you should  get paid for this

CalculatorUser  Mar 23, 2020
 #6
avatar+111329 
+1

HAHAHAHA!!!!!

 

I once suggested that to the site manager.....

 

He wouldn't go for it .......

 

 

 

 

cool cool cool

CPhill  Mar 23, 2020
edited by CPhill  Mar 23, 2020
 #11
avatar
0

wait isnt \((1/2)(8 sqrt2)(8sqrt2) = 1/2(128)= 64\)

Guest Mar 25, 2020
 #7
avatar+109524 
+1

Chris should get paid (by this cheater not by the site manager) AND, for giving all his/her assignment to Chris to answer, the asker should get shot, or at least failed.

 

The asker was lucky Chris saw the questions before I did. I would have deleted them all.

 Mar 23, 2020
edited by Melody  Mar 23, 2020
 #8
avatar+278 
+1

Shot? I mean, thats a bit harsh, isn't it? Chris does deserve to get paid, though.

 Mar 23, 2020
edited by DragonLord  Mar 23, 2020
 #9
avatar+109524 
0

Yeah, failed and shamed would be adequate punishment.

 

I don't think that any of us should answer for obvious cheats though.

I know there are a lot of cheats here but this one was blatant.

Melody  Mar 24, 2020
 #10
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0

True. I don't know if there is a possible way to identify cheats, though. Some of them are extremely obvious, and others are hidden. People sometimes rephrase a problem differently to make it seem like its not a test problem.

 Mar 24, 2020
 #12
avatar+337 
0

Cheats? How do you cheat on this?

 Mar 26, 2020
 #13
avatar+13 
+1

Can someone answer the other equations?

 Apr 1, 2020

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