In the given diagram,Delta ABC is a right-ang | vedclass

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(C) Given: $AB = BC = 14 	ext{ cm}$ and $\angle B = 90^{\circ}$.First,find the length of the hypotenuse $AC$ using the Pythagorean theorem:$AC^2 = AB

Vedclass · Accepted Answer

$98$

Vedclass · Answer

(C) Given: $AB = BC = 14 	ext{ cm}$ and $\angle B = 90^{\circ}$.First,find the length of the hypotenuse $AC$ using the Pythagorean theorem:$AC^2 = AB^2 + BC^2 = 14^2 + 14^2 = 196 + 196 = 392$.$AC = \sqrt{392} = 14\sqrt{2} 	ext{ cm}$.The radius of the semicircle with diameter $AC$ is $r = \frac{14\sqrt{2}}{2} = 7\sqrt{2} 	ext{ cm}$.Area of the semicircle with diameter $AC = \frac{1}{2} \pi r^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes (7\sqrt{2})^2 = \frac{1}{2} 	imes \frac{22}{7} 	imes 98 = 11 	imes 14 = 154 	ext{ cm}^2$.Area of $\Delta ABC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 14 	imes 14 = 98 	ext{ cm}^2$.Area of the segment of the circle (bounded by chord $AC$ and arc $APC$) = Area of sector $BAPC$ - Area of $\Delta ABC$.Area of sector $BAPC = \frac{90}{360} 	imes \pi 	imes (14)^2 = \frac{1}{4} 	imes \frac{22}{7} 	imes 196 = 154 	ext{ cm}^2$.Area of segment $APC = 154 - 98 = 56 	ext{ cm}^2$.The shaded region is the area of the semicircle minus the area of the segment $APC$.Area of shaded region = $154 - 56 = 98 	ext{ cm}^2$.

Part $I$	Part $II$
$1. \overline{OA} \cup \overline{OB} \cup \widehat{APB}$	$a. \text{Major sector}$
$2. \overline{AB} \cup \widehat{AQB}$	$b. \text{Minor segment}$
$3. \overline{AB} \cup \widehat{APB}$	$c. \text{Minor sector}$
$4. \overline{OA} \cup \overline{OB} \cup \widehat{AQB}$	$d. \text{Major segment}$

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