In the adjoining figure, △ABC is right angled at A, BC = 7.5 cm and AB = 4.5 cm. If the area of quad. ABCD is 30 cm² and DL is the altitude of △DAC, calculate the length DL.

Given,

△ABC is right-angled at A.

BC = 7.5 cm (hypotenuse)

AB = 4.5 cm

Area of quadrilateral ABCD = 30 cm²

In △ABC,

By using pythagoras theorem,

⇒ BC² = AB² + AC²

⇒ (7.5)² = (4.5)² + AC²

⇒ 56.25 = 20.25 + AC²

⇒ AC² = 56.25 - 20.25

⇒ AC² = 36

⇒ AC = $\sqrt{36}$ = 6 cm.

$\text{Area of right angle △ABC} = \dfrac{1}{2} \times \text{Base} \times \text{Height} \\[1em] = \dfrac{1}{2} \times AC \times AB \\[1em] = \dfrac{1}{2} \times 6 \times 4.5 \\[1em] = 3 \times 4.5 \\[1em] = 13.5 \text{ cm}^2.$

From figure,

Area of △DAC = Area of quad. ABCD − Area of △ABC

= 30 - 13.5

= 16.5 cm².

$\text{Area of △DAC} = \dfrac{1}{2} \times AC \times DL \\[1em] \Rightarrow 16.5 = \dfrac{1}{2} × 6 × DL \\[1em] \Rightarrow 33 = 6 \times DL \\[1em] \Rightarrow DL = \dfrac{33}{6} \\[1em] \Rightarrow DL = 5.5 \text{ cm}.$

Hence, DL = 5.5 cm.