Question

radius of blue circle ?

Answer 1

The radius of blue circle is 0.02414 cm.

Step-by-step explanation:

Area of 3 blue circles = 1/4 (4 - 3.14)

                           3π$r^{2}$ = 1/4(0.86)

                            3π$r^{2}$ = 0.215

                                 $r^{2}$ = 0.0583

                                 r = 0.02414  cm

Answer 2

• When radius of bigger circle is 1 unit, radius of each blue circles is equal to (1/9) unit .

Given :-

• Three small blue circles of equal radius .
• One big circle of radius 1 unit .

To Find :-

• Radius of blue circle ?

Solution :-

Let us assume that, radius of each blue circle is equal to r unit and let centre of small blue circles is A, B and C while centre of big circle is O .

So,

→ GB = BI = BJ = IC = CK = CE = HF = r unit .

and,

→ OF = 1 unit .

then,

→ GC = GB + BI + IC

→ GC = r + r + r = 3r unit .

Now, since GH is parallel to radius of circle . GH is also equal to 1 unit .

→ CH = GH - GC

→ CH = (1 - 3r) unit ------- Equation (1)

also,

→ OH = OF - HF

→ OH = (1 - r) unit ------------ Equation (2)

and,

→ CO = CK + KO

→ CO = (r + 1) unit ---------- Equation (3)

therefore, in right angled triangle OHC we have,

→ OH² + CH² = CO² { By pythagoras theorem }

putting values from Equation (1), Equation (2) and Equation (3) we get,

→ (1 - r)² + (1 - 3r)² = (r + 1)²

using (a - b)² = a² + b² - 2ab in LHS and (a + b)² = a² + b² + 2ab in RHS we get,

→ (1 + r² - 2r) + (1 + 9r² - 6r) = 1 + r² + 2r

cancel (1 + r²) from both sides,

→ 9r² - 2r - 6r + 1 = 2r

→ 9r² - 8r - 2r + 1 = 0

→ 9r² - 10r + 1 = 0

splitting the middle term now,

→ 9r² - 9r - r + 1 = 0

→ 9r(r - 1) - 1(r - 1) = 0

taking (r - 1) common,

→ (r - 1)(9r - 1) = 0

putting both equal to 0 now we get,

→ r = 1 unit or (1/9) unit .

since radius of bigger circle is equal to 1 unit . Radius of smaller blue circle cannot be equal to 1 unit .

Hence, we can conclude that, radius of each blue circles is equal to (1/9) unit .

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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