1. **Problem statement:** Find the lengths of $DC$ and $BD$ in the given figure where $BD$ is perpendicular to $BA$ and $DC$ is parallel to $BA$. 2. **Given:** - Radius $AB = 5$ cm - Arc length $BC = 4$ cm - $BD \perp BA$ - $DC \parallel BA$ 3. **Step 1: Find angle $\theta = \angle BAC$** The arc length formula is $l = r\theta$ where $\theta$ is in radians. Given $l = 4$ cm and $r = 5$ cm, $$\theta = \frac{l}{r} = \frac{4}{5} = 0.8 \text{ radians}$$ 4. **Step 2: Use triangle $ABC$ to find $BC$ and coordinates** Since $AB = 5$ cm (radius), and $\theta = 0.8$ radians, - $BC$ is the chord subtending angle $\theta$ at center $A$. - Length of chord $BC = 2r \sin(\frac{\theta}{2}) = 2 \times 5 \times \sin(0.4)$ Calculate: $$BC = 10 \times \sin(0.4) \approx 10 \times 0.3894 = 3.894 \text{ cm}$$ 5. **Step 3: Find coordinates of points assuming $A$ at origin and $AB$ along x-axis** - Let $A = (0,0)$ - $B = (5,0)$ since $AB=5$ - $C$ lies at angle $\theta=0.8$ radians from $AB$, so $$C = (5\cos 0.8, 5\sin 0.8) \approx (5 \times 0.6967, 5 \times 0.7174) = (3.4835, 3.587)$$ 6. **Step 4: Find point $D$ such that $BD \perp BA$ and $DC \parallel BA$** - Since $BA$ is along x-axis, $BD$ is vertical line through $B$. - So $D$ has same x-coordinate as $B$, $x=5$. - $D$ lies on line $AC$. 7. **Step 5: Equation of line $AC$** Slope of $AC$: $$m = \frac{3.587 - 0}{3.4835 - 0} = \frac{3.587}{3.4835} \approx 1.03$$ Equation: $$y = 1.03x$$ 8. **Step 6: Find $D$ on $AC$ with $x=5$** $$y_D = 1.03 \times 5 = 5.15$$ So $D = (5, 5.15)$ 9. **Step 7: Calculate lengths $BD$ and $DC$** - $B = (5,0)$, $D = (5,5.15)$ $$BD = |5.15 - 0| = 5.15 \text{ cm}$$ - $C = (3.4835, 3.587)$, $D = (5, 5.15)$ $$DC = \sqrt{(5 - 3.4835)^2 + (5.15 - 3.587)^2} = \sqrt{(1.5165)^2 + (1.563)^2}$$ $$= \sqrt{2.3 + 2.44} = \sqrt{4.74} \approx 2.18 \text{ cm}$$ **Final answers:** $$BD \approx 5.15 \text{ cm}, \quad DC \approx 2.18 \text{ cm}$$