The parallel sides of a trapezium and its height are in an arithmetic progression with a common difference of 4 . If the beight is the highest term and the area of the trapezium is 160sq. units, find the ratio of length of greatest parallel side to that of the smallest parallel side.

Correct option is D

Solution:

Let the terms of the arithmetic progression (AP) be:

- Smallest parallel side = a

- Second term = a + 4

- Height (highest term) = a + 8

The area of a trapezium is given by:

​$$\text{Area} = \frac{1}{2} \times \text{Height} \times (\text{Sum of parallel sides})$$​​

​$$160 = \frac{1}{2} \times (a + 8) \times (a + (a + 4)) \\\ \\160 = \frac{1}{2} \times (a + 8) \times (2a + 4)\\\ \\160 = \frac{1}{2} \times (2a^2 + 4a + 16a + 32)\\\ \\160 = a^2 + 10a + 16\\\ \\a^2 + 10a - 144 = 0\\\\ \\a = \frac{-10 \pm \sqrt{10^2 - 4(1)(-144)}}{2(1)}\\\\ \\a = \frac{-10 \pm \sqrt{676}}{2}\\\\ \\a = \frac{-10 \pm 26}{2}$$​​

a = 8  (since length cannot be negative)

The smallest parallel side = a = 8

- The greatest parallel side = a + 4 = 12

- The ratio of the greatest parallel side to the smallest parallel side = $$\frac{12}{8} = \frac{3}{2}$$​​

The ratio of the greatest parallel side to the smallest parallel side is 3:2 .