Question

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.

Answer 1

Area of a trapezium $$= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$. 

Parallel sides are $$(x + 8)\text{ cm}$$ and $$(2x + 3)\text{ cm}$$. 

Height $$= (x + 4)\text{ cm}$$. 

Given area $$= 590\text{ cm}^2$$:

$$\frac{1}{2} \times [(x + 8) + (2x + 3)] \times (x + 4) = 590$$ 

$$\frac{1}{2} \times (3x + 11) \times (x + 4) = 590$$

$$(3x + 11)(x + 4) = 1180$$

$$3x^2 + 12x + 11x + 44 = 1180$$

$$3x^2 + 23x - 1136 = 0$$ 

Using the quadratic formula: $$x = \frac{-23 \pm \sqrt{23^2 - 4(3)(-1136)}}{2(3)}$$ 

$$x = \frac{-23 \pm \sqrt{529 + 13632}}{6}$$

$$= \frac{-23 \pm \sqrt{14161}}{6}$$

$$= \frac{-23 \pm 119}{6}$$ 

Since $$x$$ must be positive: $$x = \frac{-23 + 119}{6}$$

$$= \frac{96}{6}$$

$$= 16$$ 

Hence, $$x = 16$$.