Solutions: Areas Related to Circles - Test 3

Q1. (d) $9\pi$ cm$^2$

Diameter of inscribed circle = Side of square = 6 cm. Radius $r = 3$ cm.
Area = $\pi r^2 = \pi (3)^2 = 9\pi$ cm$^2$.

Q2. (b) 128 cm$^2$

Diagonal of inscribed square = Diameter of circle = $16$ cm.
Area = $\frac{1}{2} d^2 = \frac{1}{2} \times 16 \times 16 = 128$ cm$^2$.

Q3. (d) 50 cm

$\pi R^2 = \pi r_1^2 + \pi r_2^2 \Rightarrow R^2 = 24^2 + 7^2 = 576 + 49 = 625$.
$R = 25$ cm. Diameter = 50 cm.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2r \times r = r^2$.

Q5. Area of shaded region = Area of Square - 4 $\times$ Area of Quadrant.

Side $a = 14$ cm. Radius $r = 7$ cm.
Area = $14^2 - 4 \times \frac{1}{4} \times \frac{22}{7} \times 7^2 = 196 - 154 = 42$ cm$^2$.

Q6. Area = Area of Circle + Area of Equilateral Triangle - Area of Common Sector.

Radius $r=6$, Triangle side $a=12$, $\theta=60^\circ$.
Area = $\pi(6)^2 + \frac{\sqrt{3}}{4}(12)^2 - \frac{60}{360}\pi(6)^2$
$= 36\pi + 36\sqrt{3} - 6\pi = 30\pi + 36\sqrt{3}$ cm$^2$.

Q7. Radius of quadrant $r = OB = \sqrt{20^2 + 20^2} = 20\sqrt{2}$ cm.

Area = Area of Quadrant - Area of Square.
$= \frac{1}{4} \times 3.14 \times (20\sqrt{2})^2 - 20^2$
$= \frac{1}{4} \times 3.14 \times 800 - 400 = 628 - 400 = 228$ cm$^2$.

Q8. Area = Area of Sector(R) - Area of Sector(r).

$R=21, r=7, \theta=30^\circ$.
Area = $\frac{30}{360} \times \frac{22}{7} \times (21^2 - 7^2)$
$= \frac{1}{12} \times \frac{22}{7} \times (441 - 49) = \frac{1}{12} \times \frac{22}{7} \times 392$
$= \frac{1}{12} \times 22 \times 56 = \frac{1232}{12} = 102.67$ cm$^2$.

Q9. 9 circles of radius 7 cm $\Rightarrow$ Diameter 14 cm.
Side of square = $3 \times 14 = 42$ cm.
Area of Square = $42 \times 42 = 1764$ cm$^2$.
Area of 9 circles = $9 \times \frac{22}{7} \times 7 \times 7 = 9 \times 154 = 1386$ cm$^2$.
Remaining Area = $1764 - 1386 = 378$ cm$^2$.

Q10. Racing Track.

(i) Distance around inner edge = $2 \times 106 + 2 \times \pi r_{\text{inner}}$.
Inner diameter = 60 m $\Rightarrow r = 30$ m.
Distance = $212 + 2 \times \frac{22}{7} \times 30 = 212 + \frac{1320}{7} = 212 + 188.57 = 400.57$ m.

(ii) Area of track = Area of rectangular parts + Area of circular rings.
Rectangular parts = $2 \times (106 \times 10) = 2120$ m$^2$.
Circular rings = $\pi(R^2 - r^2)$ where $R = 30+10=40, r=30$.
Area = $\frac{22}{7} (40^2 - 30^2) = \frac{22}{7} (1600 - 900) = \frac{22}{7} \times 700 = 2200$ m$^2$.
Total Area = $2120 + 2200 = 4320$ m$^2$.

(i) Perimeter = $2 \times \text{Length} + 2 \times \pi r$.
Length = 100 m. Diameter = 50 m $\Rightarrow r = 25$ m.
Perimeter = $200 + 2 \times \frac{22}{7} \times 25 = 200 + 157.14 = 357.14$ m.

(ii) Area = Area of Rectangle + 2 $\times$ Semicircles (1 Circle).
Area = $100 \times 50 + \pi (25)^2 = 5000 + 1964.28 = 6964.28$ m$^2$.

(iii) Cost = Area $\times$ Rate = $6964.28 \times 10 = \text{Rs } 69642.8$.