Question 1046550
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I presume you mean that the perimeter is 17.5 cm.  Yes, spelling is important.  Yes, using the correct terminology is important.  No, I don't "know" what you mean, I can only make plausible assumptions.


Let *[tex \Large w] represent the width of the rectangle.  Then the length must be *[tex \Large w\ +\ 1.75].


The perimeter of a rectangle is given by 2 times the length plus 2 time the width, so in this case:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  2(w\ +\ 1.75)\ +\ 2w\ =\ 17.5]


Solve the equation for *[tex \Large w], then calculate *[tex \Large w\ +\ 1.75] to determine the length.


The area is given by the length times the width.  Find the product of the two results above.


Given correct arithmetic, you will arrive at an answer for the area that has 3 decimal places.  It is an error to report the answer to this question with 3 decimal place precision since the least precise measurement provided in the stem of the problem is to 1 decimal place.  The correct answers to both parts of this question should be rounded to the nearest one-tenth centimeter.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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