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A frame 2 inches wide is placed around a rectangular picture

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A frame 2 inches wide is placed around a rectangular picture [#permalink] New post 06 Jul 2003, 18:44
A frame 2 inches wide is placed around a rectangular picture with dimensions 8 inches by 12 inches. What is the area of the frame in square inches?

A line segment joining two points on the circumference of a circle is 1 inch from the center of the circle at its closest point. If the circle has a 2-inch radius, what is the length of the line segment?
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 [#permalink] New post 06 Jul 2003, 19:05
Q1 : Ans: 44
Q2: Ans: 2*SQRT(3)
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Explanations [#permalink] New post 06 Jul 2003, 19:27
:lol:

Do I look like a math Stolyar, need explanations, thanks!
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Need Explanations, please [#permalink] New post 07 Jul 2003, 10:55
:lol:

Very difficult to understand without explanations, guys
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 [#permalink] New post 07 Jul 2003, 11:17
please post explanations!!!!!!!!!!!! :cry:
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 [#permalink] New post 07 Jul 2003, 17:32
96.

You have a picture which has an area of 96 (12*8).

Then we have a border. with the border the total area becomes 16*12 (since we added 2 inch border to each side 12+4 and 8+4).

As the result, the area of the frame is total new area - area of the picture.


16*12-12*8 = 12*8(2-1) = 12*8= 96


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Re: Geometry Duo [#permalink] New post 07 Jul 2003, 20:20
Curly05 wrote:
A frame 2 inches wide is placed around a rectangular picture with dimensions 8 inches by 12 inches. What is the area of the frame in square inches?

A line segment joining two points on the circumference of a circle is 1 inch from the center of the circle at its closest point. If the circle has a 2-inch radius, what is the length of the line segment?


(1) seems to be sufficiently detailed

(2) draw a picture in which you have a circle of r=2. from the radius upward (or downward), draw a line of length=1. this is the bisector of the line segment mentioned in the question (let's refer to as L).

Thus you have a right triangle with hypotenuse=2 and leg A=1, so leg B=sqrt(3).

But sqrt(3) is only 1/2 the distance of L. So L is 2sqrt(3).
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Draw it out on MSFT Word [#permalink] New post 07 Jul 2003, 21:50
Hey can you draw out the pic for Me, Jesse?

Victor
Draw it out on MSFT Word   [#permalink] 07 Jul 2003, 21:50
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