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getzgetzu
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A rectangle is 10 long and 8 wide. The area of a regular 06 May 2006, 03:48
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A rectangle is 10 long and 8 wide. The area of a regular road surrounding the rectangle is 144. What is the width of the road?



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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 08:09
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getzgetzu
A rectangle is 10 long and 8 wide. The area of a regular road surrounding the rectangle is 144. What is the width of the road?
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6? what is regular?
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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 08:21
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getzgetzu
A rectangle is 10 long and 8 wide. The area of a regular road surrounding the rectangle is 144. What is the width of the road?

6? what is regular?
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Let 'x' the be width of the regular road around the rectangle. By regular I think the question means that the width is uniform all across the rectangle.

So (10+2x)(8+2x) - 10.8 = 144 ==>x = -12, 3.

So x = 3.
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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 08:38
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getzgetzu
A rectangle is 10 long and 8 wide. The area of a regular road surrounding the rectangle is 144. What is the width of the road?

6? what is regular?
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one more typo :wink: = 6/2 = 3
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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 09:58
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getzgetzu
A rectangle is 10 long and 8 wide. The area of a regular road surrounding the rectangle is 144. What is the width of the road?

6? what is regular?

one more typo :wink: = 6/2 = 3
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No probs! I am a master of typos! :wink:
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shevy
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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 11:53
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agree with 3 :wink:
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positive soul
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A rectangle is 10 long and 8 wide. The area of a regular 07 May 2006, 13:51
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I got 3 as well ,but I used a different approach.

1) I added the area of the "inner" rectangale and the area of the road to get the total area of the "outter" rectangle. 80 + 144 = 224.

2) If we assumed that the width of the road is consistent, then the ratio of width to length should be the same in both rectangles. So, the ratio should be 8:10, or 4:5, or 2:2.5

3) Therefore, in order to get the an area of 224, the width and the length should be 14 and 16 (14 * 16 = 224).

4) If you substract 16 from the original length (16 -10 =6), and then divide by 2 (for each side), then we get 3. Same applies for 14 (14-8 =6, 6/2 =3)
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A rectangle is 10 long and 8 wide. The area of a regular 08 May 2006, 06:03
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This is a good method. Nice work soul!!
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A rectangle is 10 long and 8 wide. The area of a regular 08 May 2006, 09:26
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It is 3.
(10+2x)(8+2x) – 80=144
x^2+9x-18=0
x=3



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