Question

two circles touch externally the sum of their area is 130 Pi square cm and the distance between their Centre is 14 cm find the radius of the circle​

Answer 1

$\huge\underline{\overline{\mid{\bold{\red{⇲Question-}}\mid}}}$

• Two circles touch externally the sum of their area is 130π² cm and the distance between their Centre is 14 cm find the radius of the circle .

$\huge\underline{\overline{\mid{\bold{\red{⎆Solútion-}}\mid}}}$

$\green{ \large \underline{ \mathbb{\underline{✰GIVEN : }}}}</p><p>$

➫ The two circles touch externally .

$\red{ \large \underline{ \mathbb{\underline{✰Hère \:: }}}}</p><p>$

➭ Sum of their radii = distance between their centre = 14 cm.

☞ So, the radius of the circle be x cm and (14 - x) cm.

$\sf Sum \: of \: their \: areas = [ \pi x^2 + \pi (14 - x)^2] cm^2$

$\longrightarrow \sf \pi x^2 + \pi (14 - x)^2 = 10\pi$

$\longrightarrow \sf \pi( x^2 + (14 - x)^2) = 130\pi$

$\longrightarrow \sf ( x^2 + (14 - x)^2) = \dfrac{130\pi}{\pi}$

$\longrightarrow \sf ( x^2 + (14 - x)^2) = 130$

$\longrightarrow \sf {2x}^{2} - 28x + 66 = 0$

$\longrightarrow \sf {x}^{2} - 14x + 33 = 0$

$\longrightarrow \sf {x}^{2} - 11x - 3x + 33 = 0$

$\longrightarrow \sf x(x - 11) - 3(x - 11) = 0$

$\longrightarrow \sf (x - 3) (x - 11) = 0$

$\longrightarrow \sf x - 3 = 0 \: and \: x - 11 = 0$

$\longrightarrow \sf x = 3 \: and \: x = 11$

➟ Thus, the radius of the circle are 11cm and 3cm.