In the given figure, ∠ABC = 90° = ∠DEC, AC = 15 cm and AB = 9 cm. If the area of the quadrilateral ABCD is 72 cm²; find the length of DE.

ABC is a right angle triangle. Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ BC² = AC² - AB²

= 15² - 9²

= 225 - 81

= 144

⇒ BC = $\sqrt{144}$

= 12

As we know that area of triangle = $\dfrac{1}{2}$ x base x height

Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ BCD

⇒ 72 = $\dfrac{1}{2}$ x AB x BC + $\dfrac{1}{2}$ x BC x DE

⇒ 72 = $\dfrac{1}{2}$ x 9 x 12 + $\dfrac{1}{2}$ x 12 x DE

⇒ 72 = 9 x 6 + 6 x DE

⇒ 72 = 54 + 6 x DE

⇒ 6DE = 72 - 54

⇒ 6DE = 18

⇒ DE = $\dfrac{18}{6}$

⇒ DE = 3 cm

Hence, the value of DE = 3 cm.