Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
i) minor segment
ii) major sector

Answer 1

Hint: The diagram of the circle can be drawn, either direct formulas can be applied or geometry of subtracting larger area from smaller to get the required area.
Formulas to be used:
Area of sector = $ \dfrac{\theta }{{360}} \times \pi {r^2} $ where, r is the radius of the circle.
Area of triangle = $ \dfrac{1}{2} \times b \times h $ where, b and h are the base and height of the triangle respectively.

Complete step-by-step answer:
i) Minor segment of the circle is formed by XYZ

The area of this segment can be given by the difference of area formed by the sector OXYZ and triangle XOZ.
🡪 Area of sector OXYZ = $ \dfrac{\theta }{{360}} \times \pi {r^2} $
r = 10 cm (given)
 $ \theta $ = 90 °
Substituting these values:
 $
  \Rightarrow \dfrac{{90}}{{360}} \times 3.14 \times {(10)^2} \\
   = \dfrac{{90}}{{360}} \times \dfrac{{314}}{{{{(10)}^2}}} \times {(10)^2} \\
   = \dfrac{1}{4} \times 314 \\
  $
[ $ \pi $ = 3.14]
= 78.5
Thus, Area of sector OXYZ = 78.5 $ c{m^2} $ _______ (1)
🡪 Area of triangle XYZ = $ \dfrac{1}{2} \times b \times h $
b = h = 10 cm (base and height of triangle are equal to radius of the circle)
Substituting these values:
 $ \to \dfrac{1}{2} \times 10 \times 10 $
= 5 X 10
= 50
Thus, Area of triangle XYZ = 50 $ c{m^2} $ _______ (2)
Required area of segment = equation (1) – equation (2)
                                               = (78.5 – 50) $ c{m^2} $
Required area of segment = 28.5 $ c{m^2} $
Therefore, the area of the corresponding minor segment is 28.5 $ c{m^2} $

ii) Major sector of the circle is formed by XPZ

Area of sector XPZ = $ \dfrac{\theta }{{360}} \times \pi {r^2} $
r = 10 cm (radius of the circle)
90° + $ \theta $ = 360°
 $ \theta $ = 360° - 90°
  $ \theta $ = 270°
Because at O, the inner angle and outer angle together make a circle hence there sum is 360° and inner angle is given to be the right angle.
Substituting these values:
 $
\Rightarrow \dfrac{{270}}{{360}} \times 3.14 \times {(10)^2} \\
   = \dfrac{{270}}{{360}} \times \dfrac{{314}}{{{{(10)}^2}}} \times {(10)^2} \\
   = \dfrac{3}{4} \times 314 \\
  $
= 235.5
Thus, area of corresponding major sector is 235.5 $ c{m^2} $

Note: To avoid confusion between segment, sector or chord remember the following image:

If a sector encloses a lesser area it is called minor and if a greater area is enclosed, it is called major sector. Same thing goes with the segment.
The area enclosed by any chord (straight joining two pint on the circle) is the segment.
Right angle always refers to 90°.