Q&A for How to Find the Area of a Rectangle Using the Diagonal

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Such a rectangle would have its sides and diagonal in the ratio of 2:1:√5 (per the Pythagorean theorem). The ratio of √5 to the diagonal (20) is equal to the ratio of 2 to the longer side (or the ratio of 1 to the shorter side). Thus √5 : 20 = 2 : x, where x is the longer side. Expressed as fractions, that's √5 / 20 = 2 / x. Cross-multiplying: (√5)(x) = (20)(2). Dividing by √5: x = 40 / √5 = 40 / 2.236 = 17.9 cm. The shorter side is half that, or 8.94 cm.

Let s be the shorter side. Then (by the Pythagorean Theorem) s² + 48² = 50². So s² + 2304 = 2500. Then s² = 196, and s = 14 m. The area is 48 x 14 = 672 square meters.

Let the width be w. Then the length is (w+2). By the Pythagorean theorem, (w²) + (w+2)² = (√20)². So w² + (w² + 4w + 4) = 2w² + 4w + 4 = 20. Divide by 2: w² + 2w + 2 = 10. Subtract 10 from both sides: w² + 2w - 8 = 0. Then (w + 4)(w - 2) = 0. So w equals either -4 or +2. A width cannot be a negative number, so w = 2. The width is 2 inches, the length is 2+2 or 4 inches, and the area of the rectangle is 2x4 or 8 square inches.

Start by defining unknowns; l = length and w = width. We're given two equations; l^2 + w^2 = 19^2 = 361 and lw=120. You can solve by substitution from here -- you'll get a quartic, but it's really a quadratic in l^2 (or w^2 if you substitute the other way) so you can still solve it. But it goes a bit nicer if we don't break symmetry just yet and calculate (l+w)^2 = l^2 + w^2 + 2lw = 361 + 240 = 601, and (l-w)^2 = l^2 + w^2 -2lw = 361 - 240 = 121. So l+w = sqrt(601) and l-w = 11. (Here, I assume l > w > 0.) So l = (sqrt(601) + 11)/2 and w = (sqrt(601)-11)/2.