Question

Area of a circular ring enclosed between two concentric circles is 286cm^2.Find the diameter of the circles if their difference is 14cm.
please answer..​

Answer 1

Solution :-

Let :-

Radius of first circle = $\sf r_1$

Radius of second circle = $\sf r_2$

Diameter of the first circle = $\sf 2r_1$

Diameter of the second circle = $\sf 2r_2$

$\bullet \sf \ Area \ of \ first \ circle = \pi (r_1)^2 \\\\\bullet \ Area \ of \ second \ circle = \pi( r_2)^2$

Diameter of the first circle - Diameter of the second circle = 14

$\longrightarrow \sf 2r_1-2r_2 = 14 \\\\\longrightarrow r_1-r_2 = 7 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ................(1)$

Area of first circle - Area of second circle = 286

$\longrightarrow \sf \pi r_1^2-\pi r_2^2 = 286 \\\\\longrightarrow \pi (r_1^2-r_2^2) = 286 \\\\\longrightarrow\pi \times (r_1+r_2) \times (r_1-r_2)= 286$

Substituting the value of $\sf r_1-r_2$,

$\longrightarrow \sf \pi \times 7 \times (r_1+r_2) = 286 \\\\\longrightarrow \dfrac{22}{7} \times 7 \times (r_1+r_2) = 286 \\\\\longrightarrow 22 \times (r_1+r_2) = 286 \\\\\longrightarrow r_1+r_2 = 13 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(2)$

Adding eq (1) and eq (2) :-

$\longrightarrow \sf 2r_1 = 20 \\\\\longrightarrow \bf r_1 = 10$

Substituting the value of $\sf r_1$ in eq (1) :-

$\longrightarrow \sf 10+r_2 = 13 \\\\\longrightarrow \bf r_2 = 3$

Diameter of the first circle = 10 × 2 = 20 cm

Diameter of the second circle = 3 × 2 = 6 cm