Question

A circle with centre O is inscribed in ∆ABC. Then prove that A(ABC) = r × semi perimeter of ∆ABC where r is a radius of incircle.
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Answer 1

• When a circle is inscribed in a triangle, Area of ∆ = r × semi perimeter of the ∆ .

Given :- A circle with centre O is inscribed in ∆ABC.

To Prove :-

• Area (∆ABC) = r × semi perimeter of ∆ABC (where r is a radius of incircle .)

Formula used :-

• Area of ∆ = (1/2) × Base × Perpendicular height .

Construction :-

• Let the circle intersect the sides of the triangle at D and F .
• Join OD , OE and OF .
• Also join OA, OB and OC .

Solution :-

Let,

→ OD = OE = OF = r { Radius of the circle }

Now,

→ A(∆AOB) = (1/2) × Base × Perpendicular height

→ A(∆AOB) = (1/2) × AB × OD

→ A(∆AOB) = (1/2) × AB × r -------- Equation (1)

similarly,

→ A(∆AOC) = (1/2) × Base × Perpendicular height

→ A(∆AOC) = (1/2) × AC × OE

→ A(∆AOC) = (1/2) × AC × r -------- Equation (2)

similarly,

→ A(∆BOC) = (1/2) × Base × Perpendicular height

→ A(∆BOC) = (1/2) × BC × OF

→ A(∆BOC) = (1/2) × BC × r -------- Equation (3)

adding all three Equation's we get,

→ A(∆AOB) + A(∆AOC) + A(∆BOC) = (1/2) × AB × r + (1/2) × AC × r + (1/2) × BC × r

→ A(ABC) = (1/2)•r•(AB + BC + AC)

→ A(ABC) = r × {(AB + BC + AC) ÷ 2)}

→ A(ABC) = r × (Perimeter of ∆ABC ÷ 2)

→ A(ABC) = r × semi perimeter of ∆ABC (Proved)

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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Answer 2

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