Question

the radius of a circle is greater than twice the radius of the other circle by 2cm, sum of their areas is 73 pi sq.cm. Find the radius of the greater circle

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Answer 1

Given :

The radius of a circle is greater than twice the radius of the other circle by 2cm, sum of their areas is 73 pi sq.cm.

To find :

The radius of the greater circle .

Solution :

Let the radius of smaller circle be r cm

∴ Radius of greater circle = (2r + 2) cm

As we know,

Area of circle = πr²

ATQ,

⇒ πr² + π(2r + 2)² = 73π

⇒ π[r² + (2r + 2)²] = 73π

⇒ r² + 4r² + 8r + 4 = 73

⇒ 5r² + 8r + 4 - 73 = 0

⇒ 5r² + 8r - 69 = 0

⇒ 5r² - 15r + 23r - 69 = 0

⇒ 5r(r - 3) + 23(r - 3) = 0

⇒ (5r + 23) (r - 3) = 0

⇒ 5r + 23 = 0          or,           r - 3 = 0

⇒ 5r = - 23              or,            r = 3

∵ Radius can't be negative.

            ∴ r = 3 cm

∴ Radius of smaller circle = r = 3 cm

∴ Radius of greater circle = 2r + 2 = 2(3) + 2 = 6 + 2 = 8 cm

Therefore,

Radius of greater circle = 8 cm.

Answer 2

Answer :-

$\: \: \underline{\boxed{\rm{\green{\mapsto \: \: First \: let \: us \: understand \: concept \: used \: here}}}}$

Here we see that in this question, we are given two unknown values that is radius of circles. Now we can find these values by using Quadratic equation and Linear Equation in Two Variables. By using Linear Equation, we can easily get our answer. Using this concept, let us solve this question.

• Area of circle = πr²

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★ Question :-

The radius of a circle is greater than twice the radius of the other circle by 2cm, sum of their areas is 73 π sq.cm. Find the radius of the greater circle.

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$\: \: \huge{\underline{\sf{\pink{\mapsto \: \: Solution \: :-}}}}$

Given,

» The radius of larger circle = 2(radius of smaller circle) + 2

» Area of larger circle + Area of smaller circle = 73 π cm²

$\: \: \: \: \bf{\blue{\longrightarrow \: \: Let \: the \: radius \: of \: larger \: circle \: be \: \red{y \: cm}}}$

$\: \: \: \: \bf{\blue{\longrightarrow \: \: Let \: the \: radius \: of \: smaller \: circle \: be \: \red{x \: cm}}}$

Then, according to the question, :-

~ Case I :-

⌬ y = 2x + 2 ... (i)

~ Case II :-

⌬ πy² + πx² = 73 π cm²

⌬ π(y² + x²) = 73 × π

$\: \: \huge{\rm{\orange{\Longrightarrow \: \: (y^{2} \: + \: x^{2}) \: = \: \dfrac{73 \: \times \: \pi}{\pi}}}}$

Cancelling π , we get.

⌬ y² + x² = 73

⌬ (2x + 2)² + x² = 73

⌬ 4x² + 4 + 8x + x² = 73

⌬ 5x² + 8x + 4 - 73 = 0

⌬ 5x² + 8x - 69 = 0

⌬ 5x² - 15x + 23x - 69 = 0

⌬ 5x(x - 3) + 23(x - 3) = 0

⌬ (5x + 23)(x - 3) = 0

Here either, (5x + 23) = 0 or (x - 3) = 0

Then,

⌬ 5x + 23 = 0 or x - 3 = 0

$\: \: \: \: \huge{\rm{\Longrightarrow \: \: \: \blue{x \: = \: \dfrac{-23}{5} \: \: \: or \: \: \: x \: = \: 3}}}$

We already know that radius of circle js never a negative value. So here, we are ignoring (-23)/5 .

So, the value of x = 3.

→ Radius of smaller circle = x = 3 cm

→ Radius of larger circle = y = 2(x) + 2 = 2(3) + 2 = 6 + 2 = 8 cm

$\: \: \: \underline{\boxed{\rm{\purple{\mapsto \: \: Hence, \: the \: radius \: of \: the \: smaller \: circle \: is \: \underline{3 \: cm} \: and \: that \: of \: larger \: circle \: is \: \underline{8 \: cm}}}}}$

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$\: \: \underline{\boxed{\red{\rm{Confused?, \: Don't \: worry \: let's \: verify \: it}}}}$

For verifying, we must simply apply the value we got into our equation.

~ Case I :-

=> y = 2x + 2

=> 8 = 2(3) + 2

=> 8 = 8

Clearly, LHS = RHS.

~ Case II :-

=> y² + x² = 73

=> (8)² + (3)² = 73

=> 64 + 9 = 73

=> 73 = 73

Clearly, LHS = RHS

Here both teh conditions satisfy, so our answer is correct. Hence, verified.

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$\: \: \huge{\underline{\boxed{\sf{\pink{\mapsto \: \: More \: to \: know \: :-}}}}}$

• Quadratic Equation is the form of equation which is made of constant and variables but they can be of ang degree.

• Standard form of Linear Equation :-

ax² + bx + c = 0

• Linear Equations in Two Variables are the form of equations where we use two variables to find the solution. Standard form is given as :-

px + qy + d = 0

ax + by + c = 0