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Solve for rectangle perimeter and area not using A squared B squared

label Algebra
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Find perimeter and area of rectangle given a width of 5" and a diagonal of 13"

Oct 18th, 2017



So in this question we know the width of the rectangle, but in order to find the area and the perimeter of the rectangle (area = length*width and perimeter = 2*(length + width)) we need both the width and the length. So we know that to solve this problem we just need to find the length. One of the key facts that we know about rectangles is that they have right angles. So the the diagonal, the length, and the width form a right triangle, as is shown in the picture above.

Now in this case we don't need to use the Pythagorean theorem that because you can see that the length = 12 as it is a 5, 12, 13 right triangle (one of the Pythagorean triples). So if you know your Pythagorean triples, you can immediately see that 5,12,13 is a Pythagorean triple and the length then must be 12. Now we know that area = length*width = 12*5 = 60 square inches. We can also now solve for the perimeter of the rectangle. Perimeter = 2*(length + width) = 2 * (5+12) = 34 inches

Apr 23rd, 2015

If anything is unclear please let me know and I would be more than happy to further explain it.

Apr 23rd, 2015

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