Question

What is the area of the circle whose diametrically opposite points are $(4sin{40}^{\circ },0)and(0,4sin{50}^{\circ })$respectively

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is C

Diameter of the circle is equal to the distance between the given points
Diameter, d $=\sqrt{(0-4sin{40}^{\circ }{)}^2+(4sin{50}^{\circ }-0{)}^2}d=\sqrt{(-4sin{40}^{\circ }{)}^2+(4sin{50}^{\circ }{)}^2}d=\sqrt{(4sin{40}^{\circ }{)}^2+(4sin{50}^{\circ }{)}^2}d=4\sqrt{(sin{40}^{\circ }{)}^2+(cos{40}^{\circ }{)}^2}(sin{50}^{\circ }=cos{40}^{\circ })d=4\times 1=4(si{n}^2\theta +co{s}^2\theta =1)$
Radius of the circle $=\frac{4}{2}=2$
Area of the circle $=A=\pi r^2=\pi \times 2^2=4\pi$