The width of a rectangle is 7 less than twice its length. If the area of the rectangle is 22 cm², what is the length of the diagonal?
x = length
2x  7 = width {width is 7 less than twice length} Area of a rectangle is length x width x(2x  7) = 22 {area is length x width} 2x²  7x = 22 {used distributive property} 2x²  7x  22 = 0 {subtracted 22 from both sides} (2x  11)(x + 2) = 0 {factored into two binomials} 2x  11 = 0 or x+ 2 = 0 {set each factor equal to 0} 2x = 11 or x =  2 {added 11 and subtracted 2, respectively} x = 11/2 or x = 2 {divided left equation by 2} x = 11/2 {length cannot be 2} 2x  7 = 4 {substituted 11/2, in for x, into 2x  7} length is 11/2 cm width is 4 cm Pythagorean theorem will be used to find the diagonal, since the width, length and diagonal make a right triangle. In it, a and b are the legs {width and length}, while c is the hypotenuse {the diagonal} a² + b² = c² {the pythagorean theorem} (11/2)² + 4² = c² {substituted into pythagorean theorem} 121/4 + 16 = c² {squared 11/2 and 4} 185/4 = c² {added 121/4 and 16} √185  = c {took the square root of both sides} 2 √185  cm = the diagonal ≈ 6.8 cm {in decimal form} 2 Ask Algebra House
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