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If the length and width are denoted by l and w, then

area = lw ( the product of l time w)

and perimeter = 2(l+w) that is, twice the sum of l and w.

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Q: How would you find the area and perimiter of a rectangle?
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Is the area the inside or the outside?

The area is the inside. The perimiter is around. If its a rectangle, to find the area you multiply the lengths of two sides


A rectangle has perimiter 84 ft you also know that the length of the rectangle is six times of the width find the area of this rectangle?

L + W = P/2 = 42 ft. 1/7th of 42 is 6 so L = 36 and W = 6 making the area 216 sqft.


When the perimiter of the rectangle is 38 inches If the rectangle is less than 4 times the width find the area of the rectangle Confused how to do this?

I assume you mean that the length is less than four times the width. Here is the outline. First assume, for simplicity, that the length is EQUAL to 4 times the width. Write two equatios for that - one for the perimeter, one for the area. Solve it, and find the corresponding area. That's basically the minimum area. At the other extreme, make the length equal to the width, and solve again. That would be the maximum area.


How do you find the area and perimiter of a triangle?

To find the area of a triangle find the base and the height of the triangle. Then multiply the base by the height, then divide by 2. To find the perimiter of a triangle add together the outside edge of the triangle. To find the area of a triangle find the base and the height of the triangle. Then multiply the base by the height, then divide by 2. To find the perimiter of a triangle add together the outside edge of the triangle.


How would you find the area of a rectangle if the rectangle was 20cm -9cm?

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