# ABC is a right angle triangle,right angle at A.Acircle is inscribed in it.The length of two sides containing angle A is 12cm and 5cm find the radius

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from which chapter

Similar triangles

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from which chapter

Similar triangles

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The length of two sides containing angle A is 12cm and 5cm

So let AB = 12cm and AC = 5cm. BC is opposite to angle A, so BC is hypotenuse.

Using Pythagoras theorem,

BC² = AC² + AB²

⇒ BC² = 5² + 12²

⇒ BC² = 25+144

⇒ BC² = 169

⇒ BC = √169 = 13cm

Now from the centre of the circle O, draw perpendiculars to the three sides of the triangle. Note that OP=ON=OM = radius of circle = r. Now we can write that

Area of ΔABC = Area of ΔOAC + Area of ΔOAB + Area of ΔOBC

⇒ 0.5×AB×AC = 0.5×AC×ON + 0.5×AB×OP + 0.5×BC×OM

⇒ 0.5×12×5 = 0.5×( 5×r + 12×r + 13×r )

⇒ 12×5 = 5r + 12r + 13r

⇒ 60 = 30r

⇒ r = 60/30

⇒ r = 2cm

**Thus the radius of the inscribed circle is 2cm.**

So let AB = 12cm and AC = 5cm. BC is opposite to angle A, so BC is hypotenuse.

Using Pythagoras theorem,

BC² = AC² + AB²

⇒ BC² = 5² + 12²

⇒ BC² = 25+144

⇒ BC² = 169

⇒ BC = √169 = 13cm

Now from the centre of the circle O, draw perpendiculars to the three sides of the triangle. Note that OP=ON=OM = radius of circle = r. Now we can write that

Area of ΔABC = Area of ΔOAC + Area of ΔOAB + Area of ΔOBC

⇒ 0.5×AB×AC = 0.5×AC×ON + 0.5×AB×OP + 0.5×BC×OM

⇒ 0.5×12×5 = 0.5×( 5×r + 12×r + 13×r )

⇒ 12×5 = 5r + 12r + 13r

⇒ 60 = 30r

⇒ r = 60/30

⇒ r = 2cm