Sectors in Triangle

As a right isosceles with a leg of $10$ , the triangle has acute angles of $45°$ and a hypotenuse of $10\sqrt{2}$ , and an area of $A_{\text{triangle}} = \frac{1}{2} \cdot 10 \cdot 10 = 50$ .

Just one of the sectors has a radius that is half the triangle's hypotenuse, $5 \sqrt{2}$ , a $45°$ central angle, and an area of $A_{\text{sector}} = \frac{45°}{360°} \cdot \pi (5 \sqrt{2})^2 = \frac{25 \pi}{4}$ .

Therefore, the area of the red shape is $A_{\text{red}} = A_{\text{triangle}} - 2A_{\text{sector}} = 50 - 2 \cdot \frac{25 \pi}{4} \approx \boxed{11}$ .

$\text{hypotenuse = diameter =} 10\sqrt{2}$ $[\text{by pythagorean theorem}]$

$\text{radius of circular sector =} \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$

$\text{area of red part = area of right triangle - area of two circular sectors}$

$\text{area of red part =} \left[\dfrac{1}{2} \times \text{base} \times \text{height}\right] - \left[\dfrac{\theta}{360} \times \pi \times r^2\right]$

$\text{area of red part =} \left[\dfrac{1}{2} \times 10 \times 10 \right] - \left[\dfrac{90}{360} \times \pi \left(5\sqrt{2}\right)^2\right] \approx \color{#D61F06}\boxed{\large{11 \text{square units}}}$

Main: colored region = Area of Triangle - 2 (sector)

AoT: given by b*h/2

sector: requires radius, from below we have radius, area of sector given by (area of circle with given radius)/8. We divide by 8 because isosceles, and right angle implies other two angles pi/4 which is 1/8 of 2pi (full circle)

radius: given by pythagorean with a, b = 10, then division by 2 , we now have radius

Thus we have AoT and sector so colored region follows.

Answer in this link: Area of the red figure

Since it’s a right triangle, h = 10 sqrt(2) so the r of the circle is half of it or 5 sqrt(2). The area of the 2 sectors is pi*r^2/4 and subtracting that from 50 gives 11