Let AB be a line segment of length 2. Constru | vedclass

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

Question

(B) Let $O$ be the center of the semicircle $S$ with diameter $AB = 2$. Thus,the radius of $S$ is $r_S = 1$.Since $C$ is the mid-point of the arc $AB$

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) Let $O$ be the center of the semicircle $S$ with diameter $AB = 2$. Thus,the radius of $S$ is $r_S = 1$.Since $C$ is the mid-point of the arc $AB$,$	riangle AOC$ is a right-angled isosceles triangle with $OA = OC = 1$ and $\angle AOC = 90^\circ$.The length of the chord $AC$ is $\sqrt{OA^2 + OC^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.Semicircle $T$ is constructed with $AC$ as diameter,so its radius is $r_T = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.The area of the region inside $T$ but outside $S$ is the area of the semicircle $T$ minus the area of the segment of $S$ cut off by chord $AC$.Area of semicircle $T = \frac{1}{2} \pi r_T^2 = \frac{1}{2} \pi \left(\frac{1}{\sqrt{2}}ight)^2 = \frac{\pi}{4}$.The area of the segment of $S$ cut off by chord $AC$ is (Area of sector $OAC$) - (Area of $	riangle OAC$).Area of sector $OAC = \frac{90^\circ}{360^\circ} \pi (1)^2 = \frac{\pi}{4}$.Area of $	riangle OAC = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.Area of segment = $\frac{\pi}{4} - \frac{1}{2}$.Required Area = (Area of semicircle $T$) - (Area of segment) = $\frac{\pi}{4} - (\frac{\pi}{4} - \frac{1}{2}) = \frac{1}{2}$.

Let $AB$ be a line segment of length $2$. Construct a semicircle $S$ with $AB$ as diameter. Let $C$ be the mid-point of the arc $AB$. Construct another semicircle $T$ external to the $\triangle ABC$ with chord $AC$ as diameter. The area of the region inside the semicircle $T$ but outside $S$ is

Similar Questions

If a circle $C$ passing through the point $(4, 0)$ touches the circle $x^2+y^2+4x-6y=12$ externally at the point $(1, -1)$,then the radius of $C$ is

Let $S=0$ be the circle passing through the points $(2,0)$,$(1,-2)$,and $(-1,1)$. Then the point $(1,2)$

If $(1, a)$ and $(b, 2)$ are conjugate points with respect to the circle $x^2+y^2=25$,then $4a+2b=$

Find the equation of the director circle of the circle $x^2 + y^2 = 8$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module