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Question 1: Three circles each of radius 7 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles. (Use $\sqrt{3} = 1.732$)
Solution:
Step 1: The centres of the three circles form an equilateral triangle of side $a = 7+7=14$ cm.
Step 2: Area of $\Delta = \frac{\sqrt{3}}{4} (14)^2 = \frac{1.732}{4} \times 196 = 1.732 \times 49 = 84.868$ cm$^2$.
Step 3: Area of 3 sectors (each $60^\circ$) $= 3 \times \frac{60}{360} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{2} \times 154 = 77$ cm$^2$.
Step 4: Enclosed Area $= 84.868 - 77 = 7.868$ cm$^2$. -
Question 2: Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of $120^\circ$. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Solution:
Step 1: Area of Sector $= \frac{120}{360} \times 3.14 \times 12 \times 12 = \frac{1}{3} \times 3.14 \times 144 = 150.72$ cm$^2$.
Step 2: Area of $\Delta = \frac{1}{2} r^2 \sin 120^\circ = \frac{1}{2} (144) (\frac{\sqrt{3}}{2}) = 36\sqrt{3} = 36(1.73) = 62.28$ cm$^2$.
Step 3: Area of Segment $= 150.72 - 62.28 = 88.44$ cm$^2$. -
Question 3: The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of the distances travelled by their tips in 48 hours. (Use $\pi = 22/7$)
Solution:
Step 1: Short hand (Hour hand): In 48 hours, it completes 4 rounds. Distance $= 4 \times 2\pi(4) = 32\pi$.
Step 2: Long hand (Minute hand): In 48 hours, it completes 48 rounds. Distance $= 48 \times 2\pi(6) = 576\pi$.
Step 3: Total distance $= 32\pi + 576\pi = 608\pi = 608 \times \frac{22}{7} = 1910.86$ cm. -
Question 4: A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. If the rope were 10 m long instead of 5 m, find the increase in the grazing area. (Use $\pi = 3.14$)
Solution:
Step 1: Area is a quadrant. Increase $= \frac{1}{4}\pi(R^2 - r^2)$.
Step 2: $R=10, r=5$. Area $= \frac{1}{4} \times 3.14 \times (100 - 25) = \frac{1}{4} \times 3.14 \times 75$.
Step 3: Area $= 0.785 \times 75 = 58.875$ m$^2$. -
Question 5: The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.
Solution:
Step 1: Perimeter $= 2r + L \Rightarrow 16.4 = 2(5.2) + L \Rightarrow 16.4 = 10.4 + L \Rightarrow L = 6$ cm.
Step 2: Area of Sector $= \frac{1}{2} \times L \times r = \frac{1}{2} \times 6 \times 5.2$.
Step 3: Area $= 3 \times 5.2 = 15.6$ cm$^2$. -
Question 6: In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle. Find the area of the design.
Solution:
Step 1: Area of circle $= \pi (32)^2 = \frac{22}{7} \times 1024 = \frac{22528}{7}$ cm$^2$.
Step 2: For equilateral $\Delta$ inscribed in circle, radius $R = \frac{\text{side}}{\sqrt{3}} \Rightarrow \text{side} = 32\sqrt{3}$.
Step 3: Area of $\Delta = \frac{\sqrt{3}}{4} (32\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 1024 \times 3 = 768\sqrt{3}$ cm$^2$.
Step 4: Area of design $= (\frac{22528}{7} - 768\sqrt{3})$ cm$^2$. -
Question 7: Find the area of the shaded region in the figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD. (Side of square = 12 cm).
Solution:
Step 1: Side $= 12$. Radius of each arc $= 6$ cm.
Step 2: Four quadrants at corners form one complete circle of radius 6 cm.
Step 3: Area of Square $= 12^2 = 144$ cm$^2$. Area of 4 quadrants $= \pi (6)^2 = 36\pi$.
Step 4: Shaded Area $= 144 - 36 \times 3.14 = 144 - 113.04 = 30.96$ cm$^2$. -
Question 8: A car travels 1 km distance in which each wheel makes 450 complete revolutions. Find the radius of its wheels.
Solution:
Step 1: Distance $= 1000$ m $= 100000$ cm.
Step 2: Distance in 1 revolution $= \frac{100000}{450} = \frac{2000}{9}$ cm.
Step 3: $2\pi r = \frac{2000}{9} \Rightarrow 2 \times \frac{22}{7} \times r = \frac{2000}{9}$.
Step 4: $r = \frac{2000 \times 7}{9 \times 44} = \frac{500 \times 7}{9 \times 11} = \frac{3500}{99} \approx 35.35$ cm. -
Question 9: In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle. (Use $\sqrt{3} = 1.732$)
Solution:
Step 1: Area of $\Delta = \frac{\sqrt{3}}{4} (24)^2 = \frac{1.732}{4} \times 576 = 1.732 \times 144 = 249.408$ cm$^2$.
Step 2: Radius of incircle $r = \frac{\text{side}}{2\sqrt{3}} = \frac{24}{2\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3}$.
Step 3: Area of circle $= \pi r^2 = 3.14 \times (4\sqrt{3})^2 = 3.14 \times 48 = 150.72$ cm$^2$.
Step 4: Remaining Area $= 249.408 - 150.72 = 98.688$ cm$^2$. -
Question 10: Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm $\times$ 7 cm. Find the area of the remaining card board.
Solution:
Step 1: Diameter of each circle $= 7$ cm (since width is 7 and length 14 fits two 7cm circles). Radius $= 3.5$ cm.
Step 2: Area of Rectangle $= 14 \times 7 = 98$ cm$^2$.
Step 3: Area of 2 circles $= 2 \times \pi (3.5)^2 = 2 \times \frac{22}{7} \times 12.25 = 77$ cm$^2$.
Step 4: Remaining Area $= 98 - 77 = 21$ cm$^2$.