Question

7 cm Three currencies in diameter are placed in such a way that each currency touches the other two coins, then what is the area of ​​the field between the three coins?​

Answer 1

join the centre of the circles

we get a equilateral triangle with side 7cm

and 3 sector with angle of 60°

area of triangle

$= \frac{ \sqrt{3} }{4} \times {a}^{2} \\ = \frac{ \sqrt{3} }{4} \times 7 \times 7 \\ = \frac{ \sqrt{3} }{4} \times 49 \\ = 21.22 \\ area \: of \: three \: sectors \\ = 3 \times \frac{60}{360} \times \frac{22}{7} \times 3.5 \times 3.5 \\ = 19.25 \\ area \: in \: between \: circles \\ 21.22 - 19.25 = 1.97$

therefore area in between the 3 circle is 1.97 sqcm