Areas outside the overlapping circles indicated as shaded regions

Let
r₁ = radius of circle O₁ = 15 cm
r₂ = radius of circle O₂ = 20 cm
r₃ = radius of circle O₃ = 25 cm

α = ∠BAC = arctan (40/30) = 53.13°
β = ∠ACB = arctan (30/40) = 36.87°
 

Solving for A1
$\theta_1 = 180^\circ - 2\alpha = 180^\circ - 2(53.13^\circ)$

$\theta_1 = 73.74^\circ$
 

 

A₁ = semicircle of radius r₁ - circular segment of radius r₃ and central angle θ₁
$A_1 = \frac{1}{2}\pi{r_1}^2 - \frac{1}{2}{r_3}^2(\theta_{1rad} - \sin \theta_{1deg})$   →   Calculator if DEG mode

$A_1 = \frac{1}{2}\pi(15^2) - \frac{1}{2}(25^2)~ \left[ 73.74^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 73.74^\circ \right]$

$A_1 = 353.43 - 102.19$

$A_1 = 251.24 \, \text{ cm}^2$           answer
 

Solving for A₂
$\theta_2 = 180^\circ - 2\beta = 180^\circ - 2(36.87^\circ)$

$\theta_2 = 106.26^\circ$
 

 

A₂ = semicircle of radius r₂ – circular segment segment of radius r₃ and central angle θ₂
$A_2 = \frac{1}{2}\pi{r_2}^2 - \frac{1}{2}{r_3}^2(\theta_{2rad} - \sin \theta_{2deg})$   →   Calculator if DEG mode

$A_2 = \frac{1}{2}\pi(20^2) - \frac{1}{2}(25^2)~ \left[ 106.26^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 106.26^\circ \right]$

$A_2 = 628.32 – 279.56$

$A_2 = 348.76 \, \text{ cm}^2$           answer
 

Solving for A₃
ABC and O₁BO₂ are both 3-4-5 triangles, thus,
$\theta_3 = 2\alpha = 2(53.13^\circ)$

$\theta_3 = 106.26^\circ$
 

$\theta_4 = 2\beta = 2(36.87^\circ)$

$\theta_4 = 73.74^\circ$
 

 

A₃ = circular segment of radius r₁ and central angle θ₃ + circular segment of radius r₂ and central angle θ₄
$A_3 = \frac{1}{2}{r_1}^2(\theta_{3rad} - \sin \theta_{3deg}) + \frac{1}{2}{r_2}^2(\theta_{4rad} - \sin \theta_{4deg})$   →   Calculator if DEG mode

$A_3 = \frac{1}{2}(15^2)~ \left[ 106.26^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 106.26^\circ \right] + \frac{1}{2}(20^2)~ \left[ 73.74^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 73.74^\circ \right]$

$A_3 = 100.64 + 65.40$

$A_3 = 166.04 \, \text{ cm}^2$           answer
 

Solving for A₄
$\theta_5 = 180^\circ - 2\alpha = 180^\circ - 2(53.13^\circ)$

$\theta_5 = 73.74^\circ$
 

$\theta_6 = 180^\circ - 2\beta = 180^\circ - 2(36.87^\circ)$

$\theta_6 = 106.26^\circ$
 

 

A₄ = semicircle of radius r₃ – circular segment of radius r₁ and central angle θ₅ – circular segment of radius r₂ and central angle θ₆
$A_4 = \frac{1}{2}\pi{r_3}^2 - \frac{1}{2}{r_1}^2(\theta_{5rad} - \sin \theta_{5deg}) - \frac{1}{2}{r_2}^2(\theta_{6rad} - \sin \theta_{6deg})$   →   Calculator if DEG mode

$A_4 = \frac{1}{2}\pi(25^2) - \frac{1}{2}(15^2)~ \left[ 73.74^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 73.74^\circ \right] - \frac{1}{2}(20^2)~ \left[ 106.26^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 106.26^\circ \right]$

$A_4 = 981.75 – 36.79 – 178.92$

$A_4 = 766.04 \, \text{ cm}^2$           answer
 

Checking
$A_1 + A_2 + A_3 = 251.24 + 348.76 + 166.04$

$A_1 + A_2 + A_3 = 766.04 \, \text{ cm}^2$
 

$A_1 + A_2 + A_3 = A_4 = 766.04 \, \text{ cm}^2$       (okay!)