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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 = $154 - 56 = 98 	ext{ cm}^2$.

Similar Questions

The area of a circle is $38.5\, m^2$,then its circumference will be $\ldots \ldots \ldots \ldots \, m$.

The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36 \, cm$ and $20 \, cm$ is (in $cm$)

The circumference of a circular ground is $220\, m$. Outside it,runs a road of equal width. If the circumference of the ground with the road is $264\, m$,find the width of the road (in $m$).

The length of a square field is $50 \, m$. $A$ cow is tethered at one of the vertices by a $3 \, m$ long rope. Find the area of the region of the field in which the cow can graze. $(\pi = 3.14)$ (in $m^2$)

The diameter of a circular garden is $210 \,m$. Inside it,all along the boundary,there is a path of uniform width $7 \,m$. Then,the area of the path is $\ldots \ldots \ldots \ldots \,m^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module