The figure (ii) given below is a trapezium. Find

(i) AB

(ii) area of trapezium ABCD.

(i) Construct a perpendicular from C to AD parallel to AB.

So, ABCM is a rectangle. Since, opposite sides of a rectangle are equal.

∴ AM = CB = 2 units.

From figure,

⇒ AD = AM + MD

⇒ MD = AD - AM = 8 - 2 = 6 units.

In right angle triangle MDC,

⇒ CD² = MD² + CM²

⇒ 10² = 6² + CM²

⇒ 100 = 36 + CM²

⇒ CM² = 64

⇒ CM = $\sqrt{64}$ = 8 units.

Since, ABCM is a rectangle.

∴ AB = CM = 8 units.

Hence, AB = 8 units.

(ii) Area of trapezium ABCD = $\dfrac{1}{2}$ × (sum of || sides) × distance between them

= $\dfrac{1}{2}$ × (AD + BC) × AB

= $\dfrac{1}{2}$ × (8 + 2) × 8

= 40 sq. units.

Hence, area of trapezium ABCD = 40 sq. units.