An elastic belt is placed around …

{tex}\cos \theta = \frac { 1 } { 2 } \text { or, } \theta = 60 ^ { \circ }{/tex}
Reflex {tex}\angle A O B = 240 ^ { \circ }{/tex}
{tex}\therefore A D B = \frac { 2 \times 3.14 \times 5 \times 240 } { 360 } = 20.93 \mathrm { cm }{/tex}
Hence length of elastic in contact = 20.93 cm
Now, {tex}A P = 5 \sqrt { 3 } \mathrm { cm }{/tex}
a ({tex}\triangle O A P{/tex}) = {tex}\;\frac12\times\mathrm{base}\times\mathrm{height}\;=\;\frac12\times5\times5\sqrt3\;=\frac{25\sqrt3}2{/tex}
Area {tex}( \Delta O A P + \Delta O B P )= 2\times\;\frac{25\sqrt3}2 = 25 \sqrt { 3 } = 43.25 \mathrm { cm } ^ { 2 }{/tex}
Area of sector OACB = {tex}\;\frac{\mathrm\theta}{360}\times\mathrm{πr}^2\;{/tex}
= {tex}\frac { 25 \times 3.14 \times 120 } { 360 } = 26.16 \mathrm { cm } ^ { 2 }{/tex}
Shaded Area = 43.25 - 26.16 = 17.09 cm²