Area of the yellow region

find area of yellow triangle .as sum of white angles is 180 and r is 7.area of white is 1/2 $\pi$ 49=77.then find area of 3 circles. subtract 77 from it to get area of yellow region of circles.add all the areas of yellow region .

The area of the yellow region is equal to the the sum of the areas of the three circles plus the area of the right triangle minus twice the sum of the areas of the three circular sectors.

$\tan B=\dfrac{21}{28}$ $\implies$ $B=\tan^{-1} \dfrac{21}{28} \approx 36.87^\circ$

$\tan C=\dfrac{28}{21}$ $\implies$ $C=\tan^{-1} \dfrac{28}{21} \approx 53.13^\circ$

$A=\dfrac{22}{7}(7^2)(3)+\dfrac{1}{2}(21)(28)-2\left[\dfrac{1}{4}\left(\dfrac{22}{7}\right)(7^2)+\dfrac{36.87}{360}\left(\dfrac{22}{7}\right)(7^2)+\dfrac{53.13}{360}\left(\dfrac{22}{7}\right)(7^2)\right] \approx \boxed{602}$

Since A + B + C = 180, the white arcs that have to be subtracted of the triangles form a half circle. The are thus equals: A(Triangle) - 0.5 A(circle) + 3 A(circle) - 0.5 A(circle) = A (Triangle) + 2 A(circle) = 0.5 28 21 + 22/7 * 49 * 2 = 602.