Calculating the Length of a Segment Joining Centers of Inscribed Circles

[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

 

[theriddler876]

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

 

[Chikawakajones]

does anyone agree/disagree?

 

[daster]

I disagree.

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

 

[Chikawakajones]

tell me step by step...

 

[Tide]

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

 

[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.

 

[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

 

[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

 

[HallsofIvy]

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

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.

 

[theriddler876]

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

 

[dextercioby]

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

Daniel.

 

[dextercioby]

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

Good method,bad answer.

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

1.Wrong answer,wrong method.
2.There's more common the expression "perimeter of a square".Similar,the"circumference of a circle".

Daniel.

 

[apchemstudent]

Good method,bad answer...

Daniel.

No your answer was wrong... Mahesh's was the right one.

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.

I agree with Mahesh's answer...

 

[dextercioby]

Yeah,you're right,sorry,it must be because it's almost 2 am.
Advice:don't do math at 2 am. :tongue2:

Daniel.