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 ≈

------- cm

**6.8 cm**{in decimal form}

**2**

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