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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.

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Answer

△ABC is shown in the figure below:

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. Mixed Practice, Concise Mathematics Solutions ICSE Class 10.

In △ABC,

By pythagoras theorem,

⇒ BC2 = AB2 + AC2

⇒ 102 = 62 + AC2

⇒ AC2 = 100 - 36

⇒ AC2 = 64

⇒ AC = 64\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 = 12×\dfrac{1}{2} \times base × height

From figure,

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

12×AC×AB=12×DG×AC+12×DF×BC+12×DH×AB12×8×6=12×r×8+12×r×10+12×r×612×48=12(8r+10r+6r)48=8r+10r+6r24r=48r=4824=2 cm.\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

= 12×AC×AB\dfrac{1}{2} \times AC \times AB - πr2

= 12×8×6227\dfrac{1}{2} \times 8 \times 6 - \dfrac{22}{7} × (2)2

= 24 - 887\dfrac{88}{7}

= 168887\dfrac{168 - 88}{7}

= 807\dfrac{80}{7}

= 113711\dfrac{3}{7} cm2.

Hence, area of triangle excluding circle = 113711\dfrac{3}{7} cm2.

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