Question

Two circles touch externally. The sum of their areas is 130π sq cm and the distance between their centres is 14 cm. Find the radii of the circles.​

Answer 1

➟ Given :-

$⟶$The sum of their areas is 130π sq cm

$⟶$ the distance between their centres is 14 cm.

➟ To find :-

$⟶$the radii of the circle

➟ Solution :-

since the given circles touch externally,we have

sum of their radii = distance between their centres = 14 cm

Let the radii of the given circles be x cm and (14 – x) cm.

→ Sum of their areas = [πx² + π(14 – x)²] cm²

$⟹$πx² +π(14 – x)² = 130

$⟹$ x² + (14 – x)² = 130

$⟹$2x² – 28x + 66 = 0

$⟹$ x² – 14x + 33 = 0

$⟹$ (x – 11)(x – 3) = 0

$⟹$ x – 11 = 0 or x – 3 = 0

$⟹$ x = 11 or x = 3.

Now, x = 11 = (14 – x) = (14 - 11) = 3.

And, x = 3 = (14 - x) = (14 - 3) = 11.

Hence, the radii of the given circles are 11 cm and 3 cm.

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Answer 2

Given :-

• Sum of areas of both circular plots = 130π .
• Distance between their centres = 14 m.

To Find :-

Radii of the circles ?

Solution :-

Let us assume that, radius of bigger plot is R m and radius of smaller plot is r m.

so,

→ R + r = Distance between their centre = 14m. -------- Eqn.(1)

and,

→ πR² + πr² = 130π

→ π(R² + r²) = 130π

→ R² + r² = 130 ------------ Eqn.(2)

now, from Eqn.(1) ,

→ (R + r) = 14

squaring both sides,

→ R² + r² + 2Rr = 196

putting value of Eqn.(2),

→ 130 + 2Rr = 196

→ 2Rr = 196 - 130

→ 2Rr = 66

→ Rr = 33 ----------------- Eqn.(3)

now, we know that,

• (a - b)² = (a + b)² - 4ab.

so, using Eqn.(1) and Eqn.(3) now,

→ (R - r)² = (R+r)² - 4*R*r

→ (R - r)² = (14)² - 4*33

→ (R - r)² = 196 - 132

→ (R - r)² = 64

→ (R - r) = 8 .

adding the result in Eqn.(1) , we get,

→ (R + r) + (R - r) = 14 + 8

→ 2R = 22

→ R = 11 m.

therefore,

→ R + r = 14

→ 11 + r = 14

→ r = 14 - 11

→ r = 3 m.

therefore, the radii of the circles is 11m and 3m.

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