# Help Me With This Problem.

1. Dec 14, 2004

### Chikawakajones

The Area Of These 2 Adjacent Squares Are 4 cm^2 and 196 cm^2.

Find The Length Of The Segment Joining The Centers Of Their Inscribed Circled

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2. Dec 14, 2004

### theriddler876

it's the square root of four plus the square root of 196 and all that over two

3. Dec 14, 2004

### Chikawakajones

does anyone agree/disagree?

4. Dec 14, 2004

### daster

I disagree.

Hint: Construct a right-angled triangle with the connecting segment as its hyp

5. Dec 15, 2004

### Chikawakajones

tell me step by step........

6. Dec 15, 2004

### Tide

It will be the hypotenuse of a triangle whose sides are (14-2)/2 and (14+2)/2.

Last edited: Dec 16, 2004
7. Dec 16, 2004

### futb0l

make a triangle using the line as the hypotenuse

side 1:

Get the radius of the big circle and subtract the radius of the small circle. Giving you 7 - 1 = 6

side 2:

Get the radius of the big and the small circle.
Giving you 7 + 1 = 8

Hopefully you can solve the problem now.

Last edited by a moderator: Dec 16, 2004
8. Dec 16, 2004

### mahesh_2961

let me see ..
draw two lines, one from each center perpendicular to the bottom side (as in the figure ) Then draw a line parallel to the bottom line through the center of the smaller circle.. Then u will see a
right angled triangle formed with hypotenuse being the line connecting the two centers ...
u can now very easiy find the length of thwo sides and hence that of the hypotenuse ..

answer = sqrt (6^2+8^2) = 10

9. Dec 21, 2004

### theriddler876

youre making it all more complicated, the center of an inscribed circle is righ in the center of the square so the distance from the center to one side is the half of what the whole length of the square is so you need to square root the areas given to find the lengths divide each length by two, and add them together, the answer should be nine

10. Dec 21, 2004

### HallsofIvy

Staff Emeritus
That's simple alright! It just suffers from the minor problem of being wrong.

It would be correct IF the line through the denters were parallel to the sides- but that's not true.

11. Dec 30, 2004

### theriddler876

hmnn so the squares make more of a stairstep and not a pyramid.

12. Dec 30, 2004

### dextercioby

What pyramid??This is plane geometry...
$$d=\sqrt{106}$$
The question is:"how did I get that??"

Daniel.

13. Dec 30, 2004

### dextercioby

2.There's more common the expression "perimeter of a square".Similar,the"circumference of a circle".

Daniel.

14. Dec 30, 2004

### apchemstudent

Here's where you went wrong:

106 = 9^2 + 5^2

wrong numbers. You forgot that the radius of the smaller circle is only 1 not 2.
Thus 7-1 = 6 not 7-2.

As well 7+1 = 8 not 7+2.