Sectors, Bisectors, and a Touch of Trigonometry

Focus on triangle $GIH$ first. The main objective to solving this problem is finding the sum of the areas of the sectors of the involved circles. Since an angle is required to find the areas of the sectors, use trigonometry to find the angles first.

Since the dashed line bisects the line $GH$ , it is perpendicular--thus, we have a right triangle with the hypotenuse equal to $24$ (since it is isosceles, two sides must be equal) and one of its legs equal to $7$ (the radius of the circle). Since we are only solving for half the angle we need, let angle $\angle I = \dfrac{\theta}{2}$

We can solve for theta by using the common trigonometric identities,

$\sin \left(\dfrac{\theta}{2}\right) = \dfrac{7}{24}$

$\dfrac{\theta}{2} = \sin ^{-1} \left(\dfrac{7}{24}\right)$

$\dfrac{\theta}{2} = 16.957 ^\circ$

$\theta_I \approx 33.915 ^\circ$

Next, we find the angle associated with circle H's sector. Focus again on the whole triangle $GIH$ .With the law of sines, we have the following equation:

$\dfrac{14}{\sin 33.915} = \dfrac{24}{\sin \theta}$

$\sin \theta = \dfrac{24\sin 33.915}{14}$

$\theta = \sin^{-1} \left( \dfrac{24\sin 33.915}{14} \right)$

$\theta_H \approx 73.04 ^\circ$

We now have solved for the angles we need, so we can now use the formula for the area of a sector of circle:

$A = \pi r^{2}\left(\dfrac{\theta}{360}\right)$

Area of Sector H,

$A_H = \pi \cdot 7^{2}\left(\dfrac{73.04}{360}\right)$

$A_H \approx 31.23$ sq. units

Area of Sector I,

But first we must find its radius:

If $\overline{\rm HI} = \overline{\rm GI} = 24$ , then

$r_h + d + r_i = \overline{\rm HI} = 24$

where:

$r_h =$ radius of circle H = $7$

$d =$ distance from the two endpoints of circles H and I = $9$

$r_i =$ radius of circle I

Therefore:

$7 + 9 + r_i = 24$

$r_i = 8$ units

Now that we have solved for its radius, we can proceed to the area:

$A_I = \pi \cdot 8^{2}\left(\dfrac{33.915}{360}\right)$

$A_I \approx 18.941$ sq. units

Since we're only looking for half the area (bisected by the line),

$\dfrac{A_I}{2} = \dfrac{18.941}{2}$

$\dfrac{A_I}{2} = 9.47$ sq. units

Finally,

$A_{shaded} = A_H + \dfrac{A_I}{2}$

$A_{shaded} = 31.23 + 9.47$

$A_{shaded} \approx 40.7$ sq. units

For the bonus question:

Go back to the right triangle earlier. The bisector line is one of the legs of a right triangle with hypotenuse $24$ and another leg $7$ . By the Pythagorean Theorem,

let $x$ = the length of the bisector line

$24^2 = x^2 + 7^2$

$x^2 = 24^2 - 7^2$

$x = \sqrt{24^2 - 7^2}$

$x = \sqrt{527}$ or $\approx 22.956$ units