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**joojoo****Guest**

Can anyone help me with this problem and show me the steps it takes to do it?

There are three tangent congruent circles with their centers on the diagonal of a square. The radius of each circle is 1 inch. The circles on the ends are tangent to two sides of the square. What is the area of the square? Give an exact answer (don't round).

If anyone could help that would be AWESOME thanks!!!

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

I don't think that anyone can tell you precisely how to do it. You must have a decent imagination and a good understanding of geometry. With that said:

The top circle is in the corner of the square, therfore the radius of the circle is one inch away from each side. Using the Pythagorean Theorem (x^2 + y^2 = r^2) tells you that the center of the first circle is 2^(1/2) away from the corner. This relationship will hold in the opposite corner as well.

So the total diagonal distance will be those two distances from the opposite corners plus the entire diameter of the center circle plus the two half circle or radius remaiders from the corner circles.

The diagonal distance would then be: 2(2)^(1/2) + 2 + 2

The sides of the square will also be solved using the Pythagorean Theorem with the diagonal distance being the hypotenuse.

x^2 + y^2 = r^2, since x = y in this case, 2x^2 = r^2

Our radius from earlier was 4 + 2(2)^(1/2) = 4 + 8^(1/2)

2x^2 = 4 + 8^(1/2)

x^2 = (4 + 8^(1/2)) / 2

x = (2 + 2^(1/2))^(1/2)

Since the area of a square is simply a side squared the area is:

A = 2 + 2^(1/2)

That's the exact answer. It is approximately 3.41421356237.................

*Last edited by irspow (2005-12-07 11:13:13)*

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

(((1+1+1+1)cos45°)+1+1)² inches

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,585

**igloo** **myrtilles** **fourmis**

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Good catch John, I missed the 2x^2 term equaling the r^2 and not the radius. Do I get partial credit? It should have read:

2x^2 = 16 + 8 + 8^(1/2)

x^2 = 12 + 4*8^(1/2)

x^2 = 3 + 8^(1/2)

Since x^2 = Area, this is your answer.

Sorry for any inconvenience joojoo.

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**joojoo****Guest**

wait whats the answer??

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

irspow and John E. Franklin together solved your problem, giving a final answer of 3 + √8 square inches.

Why did the vector cross the road?

It wanted to be normal.

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