Mix Examples - Areas Related to Circles Questions in English

(B) The diameter of the circle $= 2 \times \text{radius}$.
$= 2 \times 35 = 70 \, cm$.
Since the square is inscribed in the circle,the diagonal of the square is equal to the diameter of the circle.
Therefore,the diagonal of the square $= 70 \, cm$.
The area of a square $= \frac{(\text{diagonal})^{2}}{2}$.
$= \frac{70^{2}}{2} = \frac{4900}{2} = 2450 \, cm^{2}$.

(B) In $\Delta ABC$,the base $BC$ is the diameter of the semicircle.
Given radius $r = 10 \, cm$,so the base $BC = 2r = 2 \times 10 = 20 \, cm$.
The maximum height of the triangle inscribed in a semicircle is equal to the radius of the semicircle.
Therefore,the maximum height $OA = 10 \, cm$.
The maximum area of $\Delta ABC = \frac{1}{2} \times \text{base} \times \text{maximum height}$.
Area $= \frac{1}{2} \times 20 \times 10 = 100 \, cm^2$.

(A) When a square is inscribed in a circle,the diagonal of the square passes through the center of the circle.
Therefore,the length of the diagonal of the square is equal to the diameter of the circle.
Given,the radius of the circle $r = 10\, cm$.
Diameter of the circle $= 2 \times r = 2 \times 10\, cm = 20\, cm$.
Thus,the length of the diagonal of the square is $20\, cm$.

(D) The area of a circle is given by the formula $A = \pi r^2$.
For $\odot(O, 6)$,the radius $r_1 = 6$. Thus,Area $A_1 = \pi(6)^2 = 36\pi$.
For $\odot(P, 12)$,the radius $r_2 = 12$. Thus,Area $A_2 = \pi(12)^2 = 144\pi$.
The ratio of the areas is $\frac{A_1}{A_2} = \frac{36\pi}{144\pi} = \frac{36}{144} = \frac{1}{4}$.
Therefore,the ratio is $1:4$.

(C) Actual area of the circle with radius $r_1 = 5 \, cm$ is $A_1 = \pi r_1^2 = \pi(5)^2 = 25\pi \, cm^2$.
Calculated area of the circle with radius $r_2 = 6 \, cm$ is $A_2 = \pi r_2^2 = \pi(6)^2 = 36\pi \, cm^2$.
The increase in area is $A_2 - A_1 = 36\pi - 25\pi = 11\pi \, cm^2$.
The percentage increase in the calculated area is given by $\frac{\text{Increase in area}}{\text{Actual area}} \times 100 = \frac{11\pi}{25\pi} \times 100$.
$= \frac{11}{25} \times 100 = 11 \times 4 = 44 \%$.

(A) Based on the definitions of circle parts:
$1.$ $\overline{OA} \cup \overline{OB} \cup \widehat{APB}$ represents the region bounded by two radii and the minor arc,which is the $Minor \ sector$ $(c)$.
$2.$ $\overline{AB} \cup \widehat{AQB}$ represents the region bounded by a chord and the major arc,which is the $Major \ segment$ $(d)$.
$3.$ $\overline{AB} \cup \widehat{APB}$ represents the region bounded by a chord and the minor arc,which is the $Minor \ segment$ $(b)$.
$4.$ $\overline{OA} \cup \overline{OB} \cup \widehat{AQB}$ represents the region bounded by two radii and the major arc,which is the $Major \ sector$ $(a)$.
Therefore,the correct matching is $(1-c), (2-d), (3-b), (4-a)$.