Solutions: Areas Related to Circles - Test 4

Q1. (a) $\pi r + 2r$

Perimeter of semicircle = Arc length + Diameter = $\pi r + 2r$.

Q2. (b) 70 cm

Distance = $n \times \pi d \Rightarrow 11000 \text{ m} = 5000 \times \frac{22}{7} \times d$.
$11000 = 5000 \times \frac{22}{7} \times d \Rightarrow d = \frac{11000 \times 7}{5000 \times 22} = \frac{77}{110} = 0.7$ m = 70 cm.

Q3. (b) $\pi h(2r + h)$

Outer radius $R = r + h$. Area = $\pi(R^2 - r^2) = \pi((r+h)^2 - r^2) = \pi(r^2 + h^2 + 2rh - r^2) = \pi(h^2 + 2rh) = \pi h(h + 2r)$.

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

$2 \times \frac{22}{7} \times r = 176 \Rightarrow r = \frac{176 \times 7}{44} = 4 \times 7 = 28$ cm.

Q5. Length of arc $l = \frac{\theta}{360} \times 2\pi r$.

$8.8 = \frac{30}{360} \times 2 \times \frac{22}{7} \times r$
$8.8 = \frac{1}{12} \times \frac{44}{7} \times r \Rightarrow r = \frac{8.8 \times 12 \times 7}{44} = 0.2 \times 12 \times 7 = 16.8$ cm.

Q6. Short hand (Hour hand): $r=4$. Moves 4 rounds in 2 days (48 hrs).
Distance = $4 \times 2\pi(4) = 32\pi$.
Long hand (Minute hand): $r=6$. Moves 48 rounds in 2 days.
Distance = $48 \times 2\pi(6) = 576\pi$.
Total = $32\pi + 576\pi = 608\pi = 608 \times \frac{22}{7} = 1910.85$ cm.

Q7. Area of square = $a^2 = 121 \Rightarrow a = 11$ cm.
Perimeter of square = $4 \times 11 = 44$ cm.
Circumference of circle = 44 cm $\Rightarrow 2\pi r = 44 \Rightarrow r = 7$ cm.
Area of circle = $\pi r^2 = \frac{22}{7} \times 7 \times 7 = 154$ cm$^2$.

Q8. 8 ribs $\Rightarrow$ 8 sectors. Angle $\theta = \frac{360}{8} = 45^\circ$.

Area = $\frac{45}{360} \times \pi r^2 = \frac{1}{8} \times \frac{22}{7} \times 45 \times 45$
$= \frac{11 \times 2025}{28} = \frac{22275}{28} \approx 795.5$ cm$^2$.

Q9. Area of shaded region = Area of Square - 4 $\times$ Area of Quadrant.

Side = 12 cm. Radius of quadrant = $12/2 = 6$ cm.
Area = $12^2 - 4 \times \frac{1}{4} \times 3.14 \times 6^2$
$= 144 - 3.14 \times 36 = 144 - 113.04 = 30.96$ cm$^2$.

Q10. Area of 1 triangular tile using Heron's formula.

$a=9, b=28, c=35$. $s = \frac{9+28+35}{2} = \frac{72}{2} = 36$.
Area = $\sqrt{36(36-9)(36-28)(36-35)} = \sqrt{36 \times 27 \times 8 \times 1}$
$= 6 \sqrt{9 \times 3 \times 4 \times 2} = 6 \times 3 \times 2 \sqrt{6} = 36\sqrt{6}$ cm$^2$.
$= 36 \times 2.45 = 88.2$ cm$^2$.
Total Area = $16 \times 88.2 = 1411.2$ cm$^2$.
Cost = $1411.2 \times 0.50 = \text{Rs } 705.60$.

(i) Area watered = Area of quadrant of circle with $r=14$.
Area = $\frac{1}{4} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{4} \times 22 \times 2 \times 14 = 154$ m$^2$.

(ii) Area of dry lawn = Area of Square - Area watered.
Area = $20^2 - 154 = 400 - 154 = 246$ m$^2$.

(iii) If the sprinkler is at the center with a radius of 10 m, it waters a full circle.
Area = $\pi r^2 = 3.14 \times 10 \times 10 = 314$ m$^2$.