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What is the area of a semicircular region tangent to two sides of a unit square, with endpoints of its diameter on the other two sides?

My teacher told me the answer is pi(3-2sr2)
how do you do this?

2006-12-17 06:06:39 · 4 answers · asked by aleena 1 in Science & Mathematics Mathematics

there was no diagram

2006-12-17 06:34:38 · update #1

sr means square root

2006-12-17 07:36:36 · update #2

4 answers

The area is pie diameter

2006-12-17 06:10:33 · answer #1 · answered by Sammy N 2 · 0 1

Circle area(A) = πr²

A units square is a square with sides equal to 1 (generic units). So, the circle has diameter = 1, and radius r = 1/2 (generic units).

So, r² = 1/4 (generic units)²

So the area of this circle is: A = π/4 (generic units)²

A semi circle is half a circle, and has half the area. So its area (a) is:

a = A/2 = π/8 (generic units)²

I can't say I know what your expression means......no one wants to use standard math symbols, so I will not try and 'guess' what you mean by what you mean by: pi(3-2sr2)....particularly the 2sr2 part....I have no clue what you mean.

A power is expressed as r^2

Multiplication can be expressed as 2r......I don't know what "s" is.....so if you intend to submit questions, why take the time to type out a question and forgoe adding a *little* more info so that everyone knows exactly what you mean? Doesn't make any sense at all to include ambiguity in a question and expect to get correct, quality answers.

The answers you have thus far definitely are testiment to this fact.

2006-12-17 14:36:31 · answer #2 · answered by Anonymous · 0 0

the area of a circle is pi*r^2
the area of a semicircle is, therefore, (pi*r^2)/2
i have no idea on how to address the part of your question which relates to being inscribed in a unit square

comment to fizixx:
i agree with you concerning the clear statement of this question. however, if the circle is a unit circle, it could not satisfy the description of its location inside the unit square.

2006-12-17 14:42:07 · answer #3 · answered by michaell 6 · 0 0

I'm not sure of the tangent part, a diagram would help?

2006-12-17 14:14:48 · answer #4 · answered by teekshi33 4 · 0 0

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