In odot(O, 5.6),overline{OA} and overline{OB} | vedclass

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

Question

(C) Given that in $\odot(O, 5.6)$,$\overline{OA} \perp \overline{OB}$.Thus,$OA = OB = 5.6 	ext{ cm}$ and $m\angle AOB = 90^\circ$.The area of the min

Vedclass · Accepted Answer

$15.68$

Vedclass · Answer

(C) Given that in $\odot(O, 5.6)$,$\overline{OA} \perp \overline{OB}$.Thus,$OA = OB = 5.6 	ext{ cm}$ and $m\angle AOB = 90^\circ$.The area of the minor sector formed by minor $\widehat{AB}$ is given by $\frac{	heta}{360^\circ} 	imes \pi r^2$.The area of the corresponding minor segment is given by $(	ext{Area of minor sector}) - (	ext{Area of } \Delta OAB)$.Therefore,the difference between the area of the minor sector and the area of the minor segment is equal to the area of $\Delta OAB$.Area of $\Delta OAB = \frac{1}{2} 	imes OA 	imes OB = \frac{1}{2} 	imes 5.6 	imes 5.6 = 15.68 	ext{ cm}^2$.

In $\odot(O, 5.6)$,$\overline{OA}$ and $\overline{OB}$ are radii perpendicular to each other. Then,the difference between the area of the minor sector formed by minor $\widehat{AB}$ and the corresponding minor segment is $\ldots \ldots \ldots \ldots cm^2$.

Similar Questions

In the figure,two circular flower beds are shown on two sides of a square lawn $ABCD$ of side $56 \, m$. If the centre of each circular flower bed is the point of intersection $O$ of the diagonals of the square lawn,find the sum of the areas of the lawn and the flower beds (in $m^2$).

In a circle with radius $11.2 \, cm$,two radii are perpendicular to each other. Find the area of the minor sector,the major sector,and the minor segment corresponding to these radii.

The radii of two circles are $8 \, cm$ and $6 \, cm$ respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles. [use $\pi = \frac{22}{7}$] (in $cm$)

In a circle,the length of a minor arc is $110 \,cm$ and it subtends an angle of measure $150^{\circ}$ at the centre. Then,the radius of the circle is $\ldots \ldots \ldots \ldots \,cm$.

In $\odot(O, r)$,the length of minor $\widehat{ACB}$ is one-eighth of the circumference of the circle. Then,the measure of the angle subtended at the centre by that arc is $\ldots \ldots \ldots \ldots$ (in $^\circ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module