In triangle ABC, ∠BAC = 90°, AB = 6 cm and BC = 10 cm. A circle is drawn inside the triangle which touches all the sides of the triangle (i.e. an incircle of △ABC is drawn). Find the area of the triangle excluding the circle.

△ABC is shown in the figure below:

In △ABC,

By pythagoras theorem,

⇒ BC² = AB² + AC²

⇒ 10² = 6² + AC²

⇒ AC² = 100 - 36

⇒ AC² = 64

⇒ AC = $\sqrt{64}$ = 8 cm.

A tangent line is perpendicular to the radius line from the center to the point of contact

∴ DF ⊥ BC, DG ⊥ AC and DH ⊥ AB.

Let radius of circle be r.

∴ DF = DG = DH = r

By formula,

Area of triangle = $\dfrac{1}{2} \times$ base × height

From figure,

Area of △ABC = Area of △ADC + Area of △CDB + Area of △ADB

$\Rightarrow \dfrac{1}{2} \times AC \times AB = \dfrac{1}{2} \times DG \times AC + \dfrac{1}{2} \times DF \times BC + \dfrac{1}{2} \times DH \times AB \\[1em] \Rightarrow \dfrac{1}{2} \times 8 \times 6 = \dfrac{1}{2} \times r \times 8 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 6 \\[1em] \Rightarrow \dfrac{1}{2} \times 48 = \dfrac{1}{2}\Big(8r + 10r + 6r\Big) \\[1em] \Rightarrow 48 = 8r + 10r + 6r \\[1em] \Rightarrow 24r = 48 \\[1em] \Rightarrow r = \dfrac{48}{24} = 2\text{ cm}.$

Area of triangle (excluding circle) = Area of triangle - Area of circle

= $\dfrac{1}{2} \times AC \times AB$ - πr²

= $\dfrac{1}{2} \times 8 \times 6 - \dfrac{22}{7}$ × (2)²

= 24 - $\dfrac{88}{7}$

= $\dfrac{168 - 88}{7}$

= $\dfrac{80}{7}$

= $11\dfrac{3}{7}$ cm².

Hence, area of triangle excluding circle = $11\dfrac{3}{7}$ cm².