Question

A circle is inscribed in a right angled triangle of perimeter 7pi . Then the ratio of numerical values of casamference of the circle to the are of right angled triangled​

Answer 1

Step-by-step explanation:

we know if a triangle inscribes the circle then its area=perimeter×radius/2

In this question it is given'''

perimeter=7pi

therefore Area =7pi×r/2......1

Now we know..

circumference =2pi×r.......2

Dividing eq2 by eq1

circumference =2pi×r/7pi×r×1÷2

area of triangle

=4/7

Answer 2

Given :- A circle is inscribed in a right angled triangle of perimeter 7π .

To Find :- Circumference of circle : Area of right angled triangle .

Formula used :-

• Inradius (r) = Area of triangle / semi perimeter .
• semi perimeter = (sum of all sides)/2
• circumference of circle = 2 × π × radius .

Solution :-

given that,

→ Perimeter of right angled ∆ = 7π

So,

→ Semi - perimeter of right angled ∆ = 7π/2

now, let us assume that, inradius of circle is equal to r .

then,

→ r = Area of right angled ∆ / Semi - perimeter of right angled ∆ = (Area of right angled ∆)/(7π/2) = (2• Area of right angled ∆)/7π ------- Equation (1)

now,

→ circumference of incircle = 2πr

putting value of Equation (1),

→ circumference of incircle = 2π × {(2• Area of right angled ∆)/7π}

→ circumference of incircle = (4/7)•Area of right angled ∆ ---------- Equation (2)

then,

→ Circumference of incircle : Area of right angled triangle = (4/7)•Area of right angled ∆ : Area of right angled ∆

→ Circumference of incircle : Area of right angled triangle = (4/7) : 1

→ Circumference of incircle : Area of right angled triangle = 4 : 7 (Ans.)

Hence, required ratio is equal to 4 : 7 .

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