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.

Guest Mar 23, 2020

#1**+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

CPhill Mar 23, 2020

#3**+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

CPhill Mar 23, 2020

#4**+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

CPhill Mar 23, 2020

#7**+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.

Melody Mar 23, 2020

#8**+1 **

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

DragonLord Mar 23, 2020