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(C) Let the center of the circle be $O$. Since $AB$ is a tangent at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$	riangle OAB$ is a right-angl

Vedclass · Accepted Answer

$168$

Vedclass · Answer

(C) Let the center of the circle be $O$. Since $AB$ is a tangent at $A$,the radius $OA$ is perpendicular to $AB$. Thus,$	riangle OAB$ is a right-angled triangle with $\angle OAB = 90^{\circ}$.Given $OA = 10 \, cm$ (radius) and $AB = 24 \, cm$. Using the Pythagorean theorem in $	riangle OAB$:$OB^2 = OA^2 + AB^2 = 10^2 + 24^2 = 100 + 576 = 676$.So,$OB = \sqrt{676} = 26 \, cm$.We are given that $BC$ passes through the center $O$. Since $C$ lies on the circle,$OC$ is the radius of the circle,so $OC = 10 \, cm$.Thus,$BC = BO + OC = 26 + 10 = 36 \, cm$,which matches the given information.Now,consider $	riangle ABC$. The base is $AB = 24 \, cm$. The height of $	riangle ABC$ with respect to base $AB$ is the perpendicular distance from $C$ to the line $AB$. Since $OA \perp AB$,the height is $OA = 10 \, cm$.Area of $	riangle ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AB 	imes OA = \frac{1}{2} 	imes 24 	imes 10 = 120 \, cm^2$.Wait,re-evaluating the geometry: The height of $	riangle ABC$ relative to base $AB$ is not $OA$. The angle $\angle OAB = 90^{\circ}$. The area of $	riangle ABC$ is $\frac{1}{2} 	imes AB 	imes h$,where $h$ is the perpendicular distance from $C$ to $AB$. Since $OA \perp AB$,the line $OA$ is parallel to the altitude from $C$ to $AB$. Using similar triangles or coordinates,the area is $120 \, cm^2$. Checking the options,it seems there might be a calculation error in the problem statement or options. Let's re-verify: Area = $120$. None of the options match. Let's re-read: Is $BC$ the base? If $BC$ is the base,height is the perpendicular from $A$ to $BC$. Let $h$ be the altitude from $A$ to $BC$. Area = $\frac{1}{2} 	imes BC 	imes h$. In $	riangle OAB$,$\sin(\angle OBA) = \frac{OA}{OB} = \frac{10}{26} = \frac{5}{13}$. In $	riangle ABH$ (where $AH \perp BC$),$h = AB \sin(\angle OBA) = 24 	imes \frac{5}{13} = \frac{120}{13}$. Area = $\frac{1}{2} 	imes 36 	imes \frac{120}{13} \approx 166.15$. Given the options,$168$ is the closest.

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