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Mathematics

The lengths of the parallel sides of a trapezium are (x + 8) cm and (2x + 3) cm, and the distance between them is (x + 4) cm. If its area is 590 cm2, find the value of x.

Quadratic Equations

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Answer

Given,

Length of parallel sides of trapezium = (x + 8) cm and (2x + 3) cm.

The distance between the parallel lines (height) = (x + 4) cm.

Area of trapezium = 590 cm2

By formula,

Area of trapezium = 12\dfrac{1}{2} × (sum of parallel sides) × height

⇒ 590 = 12\dfrac{1}{2} (x + 8 + 2x + 3)(x + 4)

⇒ 590 × 2 = (3x + 11)(x + 4)

⇒ 1180 = 3x2 + 12x + 11x + 44

⇒ 1180 = 3x2 + 23x + 44

⇒ 0 = 3x2 + 23x + 44 - 1180

⇒ 3x2 + 23x - 1136 = 0

⇒ 3x2 - 48x + 71x - 1136 = 0

⇒ 3x(x - 16) + 71(x - 16) = 0

⇒ (3x + 71)(x - 16) = 0

⇒ (x - 16) = 0 or (3x + 71) = 0     [Using zero-product rule]

⇒ x = 16 or x = −713\dfrac{-71}{3}

Since, length cannot be negative.

Thus, x = 16

Hence, value of x = 16.

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