Question

B. s_1 and s_2 are the inscribed and circumscribed circles of a triangle with sides 3, 4, 5, then area of s_1/area of s_2 =

Hint: Answer must be 4/25​

Answer 1

Answer:

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Answer 2

Given :-

S₁ and S₂ are the inscribed and circumscribed circles of a triangle.

Length of sides of the triangle are 3, 4 and 5 units.

To find :-

(area of S₁ ) /( area of S₂)

Concept and formula used :-

Let ABC be a triangle of length AB = c units , BC = a units and AC = b units. [ Refer fig-a ]

1. Radius of inscribed circle = (Area of triangle) / (semi perimeter of triangle )

2. Area of triangle = 1/2 × base × height

3. Semi perimeter of triangle = (Sum of all sides of triangle)/2

4. Circumradius = a/2sinA = b/2SinB = c/2SinC

or

Circumradius = (abc)/(4× area of triangle)

5. Area of circle = π × r²

Solution :-

Let's consider a triangle EFG of length 3, 4 and 5 units.

Now, we can observe that :-

=> 3²+ 4² = 5²

=> 9+16 = 25

=> 25 = 25

Triangle EFG is a right angle triangle because it's sides are forming Pythagoras triplet. [ Refer fig-b ]

→ Radius of inscribed circle = (Area of triangle) / (semi perimeter of triangle )

➡️Area of triangle EFG = (1/2) × 3 × 4

➡️Area of triangle EFG = 12/2 square units

⏩semi perimeter of triangle EFG = (3+4+5)/2

⏩semi perimeter of triangle EFG = 12/2 units

→ Radius of inscribed circle = (12/2 ) / (12/2) = 1 unit

Area of inscribed circle (S₁ )= π×1² = π square units ...(1)

→ Circumradius = GF/2sinE

Now, ∠E = 90° and GF = 5 units

→ Circumradius = 5/(2× Sin90°)

→ 5/2 units

Area of circumscribed circle (S₂)= π × (5/2)² = (25π)/(4) square units ...(2)

From (1) and (2) :-

➡️(area of S₁ ) /( area of S₂) = π /[(25π)/(4)]

➡️π × (4/25π)

➡️1 × 4/25

➡️ 4/25

• Answer : 4/25