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

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Answer 1

$\large\underline{\sf{Given- }}$

Let us consider a triangle ABC having sides,

• BC, a = 3 units

• CA, b = 4 units

• AB, c = 5 units

Consider,

$\rm :\longmapsto\: {c}^{2} = 25$

and

$\rm :\longmapsto\: {a}^{2} + {b}^{2} = {3}^{2} + {4}^{2} = 9 + 16 = 25$

$\rm :\implies\: {a}^{2} + {b}^{2} = {c}^{2}$

$\bf \: \therefore \: \triangle \: \: ABC \: is \: right \: angled \: at \: B.$

We know,

If r is radius of in-circle of triangle ABC,

then

$\rm :\longmapsto\:r = \dfrac{ar( \triangle \: ABC)}{semi - perimeter \: of \: \triangle \: ABC}$

$\rm :\longmapsto\:r \: = \: \dfrac{\dfrac{1}{2} \times a \times b }{\dfrac{1}{2}(a + b + c) }$

$\rm :\longmapsto\:r = \dfrac{3 \times 4}{3 + 4 + 5}$

$\rm :\longmapsto\:r = \dfrac{ \cancel{12}}{ \cancel{12}} = 1$

So, Area of incircle of radius r is

$\bf \: \therefore \: Area \: of \: s_1 = \pi \: {r}^{2} = \pi \: {(1)}^{2} = \pi \: sq. \: units - - (1)$

Since,

• ABC is right angled triangle.

So,

• Perpendicular bisectors of sides AB and BC meets at the midpoint of Hypotenuse AB.

So,

• Radius of circumcircle = half the length of Hypotenuse AB.

Let R be the radius of circum-circle.

$\bf\implies \:R = \dfrac{1}{2} \times c = \dfrac{1}{2} \times 5 = \dfrac{5}{2}$

So,

Area of circum-circle of radius R, is

$\sf \therefore \: Area \: of \: s_2 = \pi \: \times {\bigg( \dfrac{5}{2} \bigg) }^{2} = \dfrac{25}{4} \pi \: sq. \: units - - (2)$

Now,

Consider,

$\sf \therefore \: \dfrac{Area \: of \: s_1}{Area \: of \: s_2} = \dfrac{ \cancel\pi}{ \: \: \: \: \dfrac{25}{4} \cancel\pi \: \: \: } = \dfrac{4}{25}$

Additional Information :-

$\boxed{ \sf \: \dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC} = 2R}$

$\boxed{ \sf \: R = \dfrac{abc}{ 4\triangle}}$

$\boxed{ \sf \: r = 4R \: sin\dfrac{A}{2} \: sin\dfrac{B}{2} \: sin\dfrac{C}{2}}$

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