In the given figure, ABCD is a trapezium in which AD = 13 cm, BC = 5 cm, CD = 17 cm and ∠A = ∠B = 90°. Calculate :

(i) AB

(ii) Area of trap. ABCD

(i) Draw CE ⊥ AD.

∴ DE = 13 - 5 = 8 cm.

In triangle DEC,

By pythagoras theorem,

⇒ CD² = EC² + DE²

⇒ 17² = EC² + 8²

⇒ EC² = 17² - 8²

⇒ EC² = 289 - 64

⇒ EC² = 225

⇒ EC = $\sqrt{225}$

⇒ EC = 15 cm

From figure,

AB = EC = 15 cm.

Hence, AB = 15 cm.

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

= $\dfrac{1}{2}$ × (AD + BC) × EC

= $\dfrac{1}{2}$ × (13 + 5) × 15

= $\dfrac{1}{2}$ × 18 × 15

= 9 × 15 = 135 cm².

Hence, area of trapezium = 135 cm².