Isosceles Arc D66E77

1. **Problem Statement:** We have an isosceles triangle $OAB$ with $OA = OB$ and angle $AOB = \theta$ radians. Points $C$ and $D$ lie on $OA$ and $OB$ respectively. $CD$ is an arc of length $9.6$ cm of a circle centered at $O$ with radius $12$ cm. The arc $CD$ touches line $AB$ at point $M$. We need to find: (a) The value of $\theta$. (b) The total area of the shaded regions. (c) The total perimeter of the shaded regions. --- 2. **Find $\theta$ (Part a):** - The arc length $s$ of a circle is given by the formula: $$s = r \theta$$ where $r$ is the radius and $\theta$ is the angle in radians subtended by the arc at the center. - Given $s = 9.6$ cm and $r = 12$ cm, substitute these values: $$9.6 = 12 \theta$$ - Solve for $\theta$: $$\theta = \frac{9.6}{12}$$ $$\theta = 0.8$$ So, the value of $\theta$ is $0.8$ radians. --- 3. **Find the total area of the shaded regions (Part b):** - The shaded regions are the two small segments between the chord $CD$ and the arc $CD$ inside the triangle. - The total shaded area is twice the area of one segment formed by the chord $CD$ and the arc $CD$. - The area of a segment of a circle is given by: $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$ - Area of sector with angle $\theta$ and radius $r$: $$A_{sector} = \frac{1}{2} r^2 \theta$$ - Area of triangle formed by two radii and chord: $$A_{triangle} = \frac{1}{2} r^2 \sin \theta$$ - Calculate each: $$A_{sector} = \frac{1}{2} \times 12^2 \times 0.8 = \frac{1}{2} \times 144 \times 0.8 = 72 \times 0.8 = 57.6$$ $$A_{triangle} = \frac{1}{2} \times 144 \times \sin 0.8$$ Calculate $\sin 0.8$ (approx): $$\sin 0.8 \approx 0.717356$$ So, $$A_{triangle} = 72 \times 0.717356 = 51.2499$$ - Area of one segment: $$A_{segment} = 57.6 - 51.2499 = 6.3501$$ - Total shaded area (two segments): $$2 \times 6.3501 = 12.7002$$ Rounded to 4 decimal places, total shaded area is approximately $12.7002$ cm$^2$. --- 4. **Find the total perimeter of the shaded regions (Part c):** - The perimeter of the shaded regions consists of: - The arc length $CD$ (given as $9.6$ cm) - The two line segments $CM$ and $DM$ where $M$ is the tangent point on $AB$ - Since $M$ is the tangent point, $OM$ is perpendicular to $AB$ and $M$ lies on $AB$. - The chord $CD$ subtends angle $\theta = 0.8$ radians at $O$. - Length of chord $CD$ is: $$CD = 2r \sin \frac{\theta}{2} = 2 \times 12 \times \sin 0.4$$ Calculate $\sin 0.4$ (approx): $$\sin 0.4 \approx 0.389418$$ So, $$CD = 24 \times 0.389418 = 9.3460$$ - Since $M$ is the tangent point, $M$ lies on $AB$ and the arc $CD$ touches $AB$ at $M$. - The shaded regions' perimeter is the sum of the arc $CD$ and the two line segments $CM$ and $DM$. - The two line segments $CM$ and $DM$ together equal the chord $CD$ length because $M$ lies on $AB$ between $C$ and $D$. - Therefore, total perimeter: $$P = \text{arc } CD + CM + DM = 9.6 + 9.3460 = 18.9460$$ Rounded to 4 decimal places, total perimeter is approximately $18.9460$ cm. --- **Final answers:** (a) $\theta = 0.8$ radians (b) Total shaded area $\approx 12.7002$ cm$^2$ (c) Total perimeter $\approx 18.9460$ cm