1. **Problem Statement:** Two circles with centers $O_1$ and $O_2$ each have radius $9$ cm and intersect. We need to find: (a) the length of the common chord $AB$. (b) the area common to the two circles. 2. **Given:** Radius $r = 9$ cm for both circles. Let the distance between centers $O_1O_2 = d$. 3. **Step 1: Find the distance $d$ between centers $O_1$ and $O_2$** Since the problem does not provide $d$, we assume it is given or can be measured from the figure. For this problem, let's denote $d$ as the distance between $O_1$ and $O_2$. 4. **Step 2: Length of common chord $AB$** The common chord $AB$ is perpendicular to the line segment $O_1O_2$ and bisected by it at point $M$. Using the right triangle $O_1MB$: - $O_1B = r = 9$ - $O_1M = \frac{d}{2}$ (since $M$ is midpoint of $AB$ and lies on $O_1O_2$) - $MB = \frac{AB}{2}$ By Pythagoras theorem: $$MB = \sqrt{r^2 - \left(\frac{d}{2}\right)^2}$$ Therefore, $$AB = 2MB = 2\sqrt{r^2 - \left(\frac{d}{2}\right)^2}$$ 5. **Step 3: Area common to the two circles (lens area)** The area common to two intersecting circles of equal radius $r$ and center distance $d$ is given by: $$A = 2r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}$$ This formula comes from summing the areas of two circular segments formed by the chord $AB$. 6. **Summary:** - Length of common chord: $$AB = 2\sqrt{r^2 - \left(\frac{d}{2}\right)^2}$$ - Area common to the two circles: $$A = 2r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}$$ 7. **Note:** To find numerical answers, the distance $d$ between centers $O_1$ and $O_2$ must be known or measured from the figure.