The radius of the incircle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and
8 cm. Determine the other two sides of the triangle.
(Hint: Equate the areas of the triangle found by using the formula √[s(s-a)(s-b)(s-c)] and also found by dividing it into three triangles.)
Given: The radius of the in circle of a triangle is 4 cm and the segments into which one side is divided by the point of contact are 6 cm and 8 cm.
To find: AC and BC.

Proof:

Area of a triangle
=
=
=
…………(i)
Area of triangle = 2 area of Δ AOP + 2 area of Δ COR + 2 area of Δ PBO
= 2 ×
× 4 × 6 + 2 ×
× 4 × 8 + 2 ×
× 4 × x
= 24 + 32 + 4x
= 56 + 4x ………..(ii)
Equating (i) and (ii)
48x(x + 14) = (56 + 4x)2
48x(x + 14) = 16(x + 14)2
48x = 16(x + 14) or x + 14 = 0
48x = 16x + 224
32x = 224
x = 7 or x = -14(which we ignore)
AC = 6 + 7 = 13 cm
BC = 8 + 7 = 15 cm
The other two sides of the triangles are 13 cm and 15 cm.
To find: AC and BC.

Proof:

Area of a triangle
=
=

=
…………(i)Area of triangle = 2 area of Δ AOP + 2 area of Δ COR + 2 area of Δ PBO
= 2 ×
× 4 × 6 + 2 ×
× 4 × 8 + 2 ×
× 4 × x= 24 + 32 + 4x
= 56 + 4x ………..(ii)
Equating (i) and (ii)
48x(x + 14) = (56 + 4x)2
48x(x + 14) = 16(x + 14)2
48x = 16(x + 14) or x + 14 = 0
48x = 16x + 224
32x = 224
x = 7 or x = -14(which we ignore)
AC = 6 + 7 = 13 cm
BC = 8 + 7 = 15 cm
The other two sides of the triangles are 13 cm and 15 cm.
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