Question

Two circles touch each other externally. The sum of their areas is $\left(106\pi \right)c{m}^2$ and the distance between their centers is 14 cm. Find the sum diameters of the circles.

• A. 14 cm
• B. 15 cm
• C. 28 cm
• D. 30 cm

Answer 1

The correct option is C

28 cm

Let the radius of bigger and smaller circle be 'R' and 'r' respectively.

Sum of areas $=\pi R^2+\pi r^2=\pi (R^2+r^2)$

According to question, $\pi (R^2+r^2)=106\pi$

$\Rightarrow R^2+r^2=106---\left(I\right)$

Also, $R+r=14$

$\Rightarrow R=14-r----\left(II\right)$

Putting (II) in (I)

$(14-r{)}^2+r^2=106$

$196+r^2-28r+r^2=106$

$2r^2-28r+90=0$

$r^2-14r+45=0$

$r^2-9r-5r+45=0$

$r(r-9)-5(r-9)=0$

$(r-5)(r-9)=0$

$r=9orr=5$

$\Rightarrow r=9$ cm can't be possible because that means $R=5$ cm which can't be true for our assumption. So, $r=5$ cm and $R=9$ cm. Sum of diameters =$(10+18)cm=28cm$