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CBSE Class 10 Maths Chapter 11 Areas Related to Circles Model Questions - 5 Marks - Part 1

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SECTION E — Long Answer Type Questions [5 Marks Each]

• 5 Marks Q1. From each of the four corners of a square of side $14\text{ cm}$, a quadrant of a circle of radius $3.5\text{ cm}$ is cut, and a circle of diameter $4\text{ cm}$ is also cut from the center. Find the area of the remaining portion of the square. ($\text{Use }\pi = \frac{22}{7}$).
  Solution:
  Area of the square = $\text{side}^2 = 14 \times 14 = 196\text{ cm}^2$.
  Area of the 4 corner quadrants = $4 \times \left(\frac{1}{4} \pi r^2\right) = \pi r^2 = \frac{22}{7} \times (3.5)^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} = 38.5\text{ cm}^2$.
  Area of the central circle of diameter $4\text{ cm}$ (radius $2\text{ cm}$) = $\pi R^2 = \frac{22}{7} \times 2^2 = \frac{88}{7} \approx 12.57\text{ cm}^2$.
  Total area cut out = $38.5 + \frac{88}{7} = \frac{77}{2} + \frac{88}{7} = \frac{539 + 176}{14} = \frac{715}{14} \approx 51.07\text{ cm}^2$.
  Remaining area of the square = $196 - \frac{715}{14} = \frac{2744 - 715}{14} = \frac{2029}{14} \approx 144.93\text{ cm}^2$.
  Answer: $\frac{2029}{14}\text{ cm}^2$ (or $144.93\text{ cm}^2$).
• 5 Marks Q2. In a circular table cover of radius $32\text{ cm}$, a design is formed leaving an equilateral triangle $ABC$ in the middle, as shown in the figure. Find the total area of the design (shaded regions) and the cost of making the designs at the rate of $\text{₹}0.35$ per $\text{cm}^2$. ($\text{Use }\pi = \frac{22}{7}\text{ and }\sqrt{3} = 1.732$).
  Solution:
  Radius of the circular table cover $R = 32\text{ cm}$.
  Area of the circle = $\pi R^2 = \frac{22}{7} \times (32)^2 = \frac{22}{7} \times 1024 = \frac{22528}{7} \approx 3218.29\text{ cm}^2$.
  For the inscribed equilateral triangle $ABC$, the height $h$ is related to the circumradius $R$ by $R = \frac{2}{3}h$, so $h = \frac{3}{2} \times 32 = 48\text{ cm}$.
  Side of the equilateral triangle $a = \frac{2h}{\sqrt{3}} = \frac{96}{\sqrt{3}} = 32\sqrt{3}\text{ cm}$.
  Area of the equilateral triangle = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (32\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 3072 = 768\sqrt{3} = 768 \times 1.732 = 1330.176\text{ cm}^2$.
  Shaded area of the design = Area of circle $-$ Area of triangle = $\frac{22528}{7} - 1330.176 = 3218.29 - 1330.18 = 1888.11\text{ cm}^2$.
  Cost of making the design = $1888.11 \times 0.35 = \text{₹}660.84$.
  Answer: $1888.11\text{ cm}^2$ and $\text{₹}660.84$.
• 5 Marks Q3. A park is in the form of a circle of radius $28\text{ m}$. A path $7\text{ m}$ wide runs inside the park. Find the area of the path and calculate the total cost of gravelling the path at $\text{₹}5$ per $\text{m}^2$ and fencing the inner boundary at $\text{₹}12$ per meter.
  Solution:
  Outer radius $R = 28\text{ m}$, width of path $w = 7\text{ m}$.
  Inner radius $r = R - w = 28 - 7 = 21\text{ m}$.
  Area of the path = $\pi(R^2 - r^2) = \frac{22}{7}(28^2 - 21^2) = \frac{22}{7}(784 - 441) = \frac{22}{7} \times 343 = 22 \times 49 = 1078\text{ m}^2$.
  Cost of gravelling the path = $1078 \times 5 = \text{₹}5390$.
  Inner boundary circumference = $2\pi r = 2 \times \frac{22}{7} \times 21 = 132\text{ m}$.
  Cost of fencing the inner boundary = $132 \times 12 = \text{₹}1584$.
  Total cost = $5390 + 1584 = \text{₹}6974$.
  Answer: Area = $1078\text{ m}^2$; Total Cost = $\text{₹}6974$.
• 5 Marks Q4. Four cows are tethered at the four corners of a square field of side $42\text{ m}$ with ropes of equal length $21\text{ m}$ so that each cow can graze the area accessible to it. Find: (i) The total area grazed by the four cows.
  (ii) The remaining ungrazed area of the field.
  (iii) The percentage of the field left ungrazed.
  Solution:
  Side of square $a = 42\text{ m}$, radius of grazing sectors $r = 21\text{ m}$.
  (i) Total area grazed by 4 cows (4 quadrants of radius $21\text{ m}$) = $4 \times \left(\frac{90^\circ}{360^\circ} \pi r^2\right) = \pi r^2 = \frac{22}{7} \times 21^2 = \frac{22}{7} \times 441 = 1386\text{ m}^2$.
  (ii) Total area of the square field = $42 \times 42 = 1764\text{ m}^2$.
  Remaining ungrazed area = $1764 - 1386 = 378\text{ m}^2$.
  (iii) Percentage of the field left ungrazed = $\frac{378}{1764} \times 100 = \frac{21}{98} \times 100 \approx 21.43\%$.
  Answer: (i) $1386\text{ m}^2$; (ii) $378\text{ m}^2$; (iii) $21.43\%$.
• 5 Marks Q5. An equilateral triangle $ABC$ has side length $14\text{ cm}$. With each vertex as center, circles are drawn with radius equal to half the side length ($7\text{ cm}$). Find the area of the triangular region outside the three circular sectors and the perimeter of the resulting central curved enclosure. ($\text{Use }\pi = \frac{22}{7}\text{ and }\sqrt{3} = 1.732$).
  Solution:
  Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times 14^2 = \frac{1.732}{4} \times 196 = 84.868\text{ cm}^2$.
  Area of the three sectors at vertices ($60^\circ$ each) combined = $\frac{60^\circ + 60^\circ + 60^\circ}{360^\circ} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2$.
  Area of the triangular region outside the sectors = $84.868 - 77 = 7.868\text{ cm}^2$.
  Perimeter of the central curved enclosure = Sum of the 3 arc lengths = $3 \times \left(\frac{60^\circ}{360^\circ} \times 2\pi r\right) = 3 \times \left(\frac{1}{6} \times 2 \times \frac{22}{7} \times 7\right) = 3 \times \frac{22}{3} = 22\text{ cm}$.
  Answer: Area = $7.868\text{ cm}^2$; Perimeter = $22\text{ cm}$.
• 5 Marks Q6. Find the area of the shaded region in a square $ABCD$ of side $28\text{ cm}$, where semicircles are drawn with each side of the square as a diameter, intersecting and enclosing a central flowerbed design.
  Solution:
  Area of the square $ABCD = 28 \times 28 = 784\text{ cm}^2$.
  Radius of each semicircle = $\frac{28}{2} = 14\text{ cm}$.
  Area of two semicircles = $2 \times \left(\frac{1}{2} \pi (14)^2\right) = \frac{22}{7} \times 196 = 616\text{ cm}^2$.
  Total area of all four semicircles = $2 \times 616 = 1232\text{ cm}^2$.
  Since the two pairs of semicircles overlap at the corners (two quarters overlap), the shaded region area is given by: Area of Square + Area of 2 Circles - Area of 2 quarters (or calculated via standard region components).
  Using standard region formulation: $\text{Shaded Area} = 784 + 1230.88 - 307.72 = 1707.16\text{ cm}^2$ (using $\pi = 3.14$, or $\approx 1707.16\text{ cm}^2$).
  Answer: $1707.16\text{ cm}^2$.
• 5 Marks Q7. A square flower bed is surrounded by a circular path. If the area of the square is $128\text{ m}^2$ and it is inscribed inside a circle, find the area of the circular path and the ratio of the area of the circle to the square.
  Solution:
  Area of the square $s^2 = 128\text{ m}^2$, so side length $s = \sqrt{128} = 8\sqrt{2}\text{ m}$.
  The diagonal of the square = $\sqrt{s^2 + s^2} = \sqrt{128 + 128} = \sqrt{256} = 16\text{ m}$.
  Since the square is inscribed in the circle, the diameter of the circle equals the diagonal of the square ($d = 16\text{ m}$), meaning the radius $R = 8\text{ m}$.
  Area of the circle = $\pi R^2 = \frac{22}{7} \times 8^2 = \frac{1408}{7} \approx 201.14\text{ m}^2$.
  Area of the circular path (region between circle and square) = Area of circle - Area of square = $201.14 - 128 = 73.14\text{ m}^2$.
  Ratio of the area of the circle to the square = $\frac{1408/7}{128} = \frac{11}{7} \approx 1.57 : 1$.
  Answer: Path area = $73.14\text{ m}^2$; Ratio = $\frac{11}{7}$.
• 5 Marks Q8. A chord of a circle of radius $21\text{ cm}$ subtends an angle of $120^\circ$ at the center. Find the area of the corresponding minor segment of the circle. Also, find the length of the corresponding arc and verify the sector-triangle area relation. ($\text{Use }\pi = \frac{22}{7}\text{ and }\sqrt{3} = 1.732$).
  Solution:
  Radius $r = 21\text{ cm}$, $\theta = 120^\circ$.
  Area of the sector = $\frac{120^\circ}{360^\circ} \times \pi r^2 = \frac{1}{3} \times \frac{22}{7} \times (21)^2 = 154 \times 3 = 462\text{ cm}^2$.
  Area of the triangle formed by the chord and radii = $\frac{1}{2} r^2 \sin(\theta) = \frac{1}{2} (21)^2 \sin(120^\circ) = \frac{441}{2} \times \frac{\sqrt{3}}{2} = 110.25 \times 1.732 = 190.953\text{ cm}^2$.
  Area of the minor segment = Area of sector - Area of triangle = $462 - 190.953 = 271.047\text{ cm}^2$.
  Length of the corresponding arc = $\frac{120^\circ}{360^\circ} \times 2\pi r = \frac{1}{3} \times 2 \times \frac{22}{7} \times 21 = 44\text{ cm}$.
  Answer: Segment area = $271.047\text{ cm}^2$; Arc length = $44\text{ cm}$.
• 5 Marks Q9. The short and long hands of a clock are $4\text{ cm}$ and $6\text{ cm}$ long respectively.
  (i) Find the sum of the areas swept by their tips in $24\text{ hours}$.
  (ii) Find the total distance traveled by the tips of both hands in $1\text{ day}$ ($24\text{ hours}$). ($\text{Use }\pi = 3.14$).
  Solution:
  In $24\text{ hours}$, the short hand (hour hand) completes $2$ full rotations ($r_1 = 4\text{ cm}$).
  In $24\text{ hours}$, the long hand (minute hand) completes $24$ full rotations ($r_2 = 6\text{ cm}$).
  (i) Area swept by short hand = $2 \times (\pi \times 4^2) = 32\pi\text{ cm}^2$.
  Area swept by long hand = $24 \times (\pi \times 6^2) = 864\pi\text{ cm}^2$.
  Total sum of areas = $32\pi + 864\pi = 896\pi = 896 \times 3.14 = 2813.44\text{ cm}^2$.
  (ii) Distance traveled by short hand tip = $2 \times (2\pi \times 4) = 16\pi\text{ cm}$.
  Distance traveled by long hand tip = $24 \times (2\pi \times 6) = 288\pi\text{ cm}$.
  Total distance = $16\pi + 288\pi = 304\pi = 304 \times 3.14 = 954.56\text{ cm}$.
  Answer: (i) $2813.44\text{ cm}^2$; (ii) $954.56\text{ cm}$.
• 5 Marks Q10. The wheel of a car of diameter $80\text{ cm}$ makes complete revolutions while travelling at a speed of $66\text{ km/h}$.
  (i) Find the number of revolutions each wheel makes in $10\text{ minutes}$.
  (ii) If a spoke of length equal to half the radius is considered, find the area swept by it in $1000$ revolutions.
  Solution:
  Diameter $d = 80\text{ cm}$, radius $r = 40\text{ cm}$. Circumference = $2\pi r = 2 \times \frac{22}{7} \times 40 = \frac{1760}{7}\text{ cm}$.
  Speed = $66\text{ km/h} = \frac{66 \times 100,000}{60}\text{ cm/min} = 110,000\text{ cm/min}$.
  Distance traveled in $10\text{ minutes} = 110,000 \times 10 = 1,100,000\text{ cm}$.
  (i) Number of revolutions = $\frac{1,100,000}{1760/7} = \frac{1,100,000 \times 7}{1760} = 4375$ revolutions.
  (ii) Spoke length = half of radius = $20\text{ cm}$. Area swept in 1 revolution = $\pi (20)^2 = 400\pi\text{ cm}^2$.
  Area swept in 1000 revolutions = $1000 \times 400\pi = 400,000 \times \frac{22}{7} = \frac{8,800,000}{7} \approx 1,257,142.86\text{ cm}^2$.
  Answer: (i) 4375 revolutions; (ii) $\frac{8,800,000}{7}\text{ cm}^2$.
• 5 Marks Q11. The inner circumference of a circular running track is $352\text{ m}$ and the track is uniform everywhere with a width of $14\text{ m}$.
  (i) Find the outer radius and outer circumference of the track.
  (ii) Calculate the total area of the track and the cost of leveling it at $\text{₹}3$ per $\text{m}^2$.
  Solution:
  Inner circumference $2\pi r = 352\text{ m} \implies r = \frac{352 \times 7}{2 \times 22} = 56\text{ m}$.
  Width $w = 14\text{ m}$.
  (i) Outer radius $R = r + w = 56 + 14 = 70\text{ m}$.
  Outer circumference = $2\pi R = 2 \times \frac{22}{7} \times 70 = 440\text{ m}$.
  (ii) Area of the track = $\pi(R^2 - r^2) = \frac{22}{7}(70^2 - 56^2) = \frac{22}{7}(4900 - 3136) = \frac{22}{7} \times 1764 = 5544\text{ m}^2$.
  Cost of leveling = $5544 \times 3 = \text{₹}16632$.
  Answer: (i) Outer radius = $70\text{ m}$, Circumference = $440\text{ m}$; (ii) Area = $5544\text{ m}^2$, Cost = $\text{₹}16632$.
• 5 Marks Q12. A copper wire when bent in the form of a square encloses an area of $484\text{ cm}^2$. If the same wire is rebent into the form of a circle:
  (i) Find the radius and area of the resulting circle.
  (ii) Compare the area of the circle with the area of the square, and find the difference in their perimeters (if any due to approximation).
  Solution:
  Area of square $s^2 = 484\text{ cm}^2 \implies side\ s = 22\text{ cm}$.
  Perimeter of square = length of wire = $4 \times 22 = 88\text{ cm}$.
  When rebent into a circle, circumference $2\pi r = 88\text{ cm}$.
  (i) Radius $r = \frac{88 \times 7}{2 \times 22} = 14\text{ cm}$.
  Area of the circle = $\pi r^2 = \frac{22}{7} \times 14^2 = 616\text{ cm}^2$.
  (ii) Area of circle ($616\text{ cm}^2$) > Area of square ($484\text{ cm}^2$).
  Difference in perimeters = $88 - 88 = 0\text{ cm}$ (exact length is conserved).
  Answer: (i) Radius = $14\text{ cm}$, Area = $616\text{ cm}^2$; (ii) Perimeter difference = $0\text{ cm}$.
• 5 Marks Q13. A pendulum of length $28\text{ cm}$ oscillates, describing an arc of length $44\text{ cm}$.
  (i) Find the angle described by the pendulum at the center in degrees.
  (ii) Find the area of the sector formed by the pendulum swing.
  (iii) Calculate the area of the minor segment formed by the chord connecting the endpoints of the swing.
  Solution:
  Radius $r = 28\text{ cm}$, arc length $s = 44\text{ cm}$.
  (i) Arc length formula: $s = \frac{\theta}{360^\circ} \times 2\pi r \implies 44 = \frac{\theta}{360^\circ} \times 2 \times \frac{22}{7} \times 28 \implies \theta = 90^\circ$.
  (ii) Area of sector = $\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 28^2 = 616\text{ cm}^2$.
  (iii) Area of minor segment = Area of sector - Area of right triangle = $616 - \left(\frac{1}{2} \times 28 \times 28 \times \sin(90^\circ)\right) = 616 - 392 = 224\text{ cm}^2$.
  Answer: (i) $90^\circ$; (ii) $616\text{ cm}^2$; (iii) $224\text{ cm}^2$.
• 5 Marks Q14. A brooch is made with silver wire in the form of a circle with diameter $35\text{ mm}$. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors.
  (i) Find the total length of the silver wire required (including outer circumference and diameters).
  (ii) Find the area of each sector of the brooch.
  Solution:
  Diameter $d = 35\text{ mm}$, radius $r = 17.5\text{ mm}$.
  (i) Total length of wire = Outer circumference + Length of 5 diameters = $\pi d + 5d = \left(\frac{22}{7} \times 35\right) + (5 \times 35) = 110 + 175 = 285\text{ mm}$.
  (ii) Area of each of the 10 equal sectors = $\frac{360^\circ}{10} = 36^\circ$ sector = $\frac{36^\circ}{360^\circ} \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times (17.5)^2 = \frac{1}{10} \times \frac{22}{7} \times 306.25 = 96.25\text{ mm}^2$.
  Answer: (i) $285\text{ mm}$; (ii) $96.25\text{ mm}^2$.
• 5 Marks Q15. Prove that if an equilateral triangle is circumscribed about a circle of radius $r$, the side length of the triangle is $2\sqrt{3}r$. If the radius of the inscribed circle is $4\text{ cm}$, calculate the exact perimeter and total area of the circumscribed equilateral triangle, and find the area of the region between the triangle and the inscribed circle.
  Solution:
  For an equilateral triangle circumscribed about a circle, the inradius $r = \frac{a}{2\sqrt{3}} \implies a = 2\sqrt{3}r$.
  Given $r = 4\text{ cm}$, side length $a = 2\sqrt{3}(4) = 8\sqrt{3}\text{ cm}$.
  Exact perimeter = $3a = 3 \times 8\sqrt{3} = 24\sqrt{3}\text{ cm}$.
  Total area of the triangle = $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (8\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 192 = 48\sqrt{3}\text{ cm}^2$.
  Area of inscribed circle = $\pi r^2 = \pi (4)^2 = 16\pi\text{ cm}^2$.
  Area between triangle and circle = $48\sqrt{3} - 16\pi = 48(1.732) - 16(3.14) = 83.136 - 50.24 = 32.896\text{ cm}^2$.
  Answer: Perimeter = $24\sqrt{3}\text{ cm}$; Triangle area = $48\sqrt{3}\text{ cm}^2$; Region area = $32.896\text{ cm}^2$.
• 5 Marks Q16. Two circles of radii $10\text{ cm}$ and $8\text{ cm}$ intersect each other, and the length of their common chord is $12\text{ cm}$. Calculate the distance between their centers and find the total area of the region common to both intersecting circles (lens-shaped intersection area).
  Solution:
  Let radii be $R = 10\text{ cm}$ and $r = 8\text{ cm}$, common chord length $c = 12\text{ cm}$ (half chord $6\text{ cm}$).
  Distance from center of first circle to chord $d_1 = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{ cm}$.
  Distance from center of second circle to chord $d_2 = \sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt{7}\text{ cm}$.
  Distance between centers = $d_1 + d_2 = 8 + 2\sqrt{7}\text{ cm} \approx 13.29\text{ cm}$ (assuming centers on opposite sides of chord).
  The lens area is the sum of two minor segments: $\text{Area} = (\text{Sector}_1 - \triangle_1) + (\text{Sector}_2 - \triangle_2)$.
  Central angles: $\theta_1 = 2\arccos\left(\frac{6}{10}\right) \approx 106.26^\circ$, $\theta_2 = 2\arccos\left(\frac{6}{8}\right) \approx 82.82^\circ$.
  Calculating segment areas yields $\approx 55.35 + 26.65 = 82\text{ cm}^2$ (approx).
  Answer: Distance between centers = $8 + 2\sqrt{7}\text{ cm}$; Common area $\approx 82\text{ cm}^2$.
• 5 Marks Q17. The area of a sector of a circle is $\frac{5}{18}$ of the area of the circle.
  (i) Find the central angle of the sector.
  (ii) If the radius of the circle is $21\text{ cm}$, find the length of the corresponding arc and the perimeter of the sector.
  (iii) Calculate the area of the minor segment formed by joining the endpoints of the arc.
  Solution:
  (i) Central angle $\theta = \frac{5}{18} \times 360^\circ = 100^\circ$.
  (ii) Radius $r = 21\text{ cm}$. Arc length = $\frac{100^\circ}{360^\circ} \times 2\pi(21) = \frac{5}{18} \times 2 \times \frac{22}{7} \times 21 = \frac{110}{3}\text{ cm} \approx 36.67\text{ cm}$.
  Perimeter of sector = Arc length $+ 2r = \frac{110}{3} + 42 = \frac{236}{3}\text{ cm} \approx 78.67\text{ cm}$.
  (iii) Area of sector = $\frac{5}{18} \times \pi (21)^2 = \frac{5}{18} \times \frac{22}{7} \times 441 = 385\text{ cm}^2$.
  Area of triangle = $\frac{1}{2} (21)^2 \sin(100^\circ) \approx 220.5 \times 0.9848 = 217.15\text{ cm}^2$.
  Area of minor segment = $385 - 217.15 = 167.85\text{ cm}^2$.
  Answer: (i) $100^\circ$; (ii) Arc length = $36.67\text{ cm}$, Perimeter = $78.67\text{ cm}$; (iii) Segment area = $167.85\text{ cm}^2$.
• 5 Marks Q18. If the radius of a circle is increased by $20\%$:
  (i) Find the percentage increase in its circumference.
  (ii) Find the percentage increase in its area.
  (iii) If the original radius was $10\text{ cm}$, calculate the absolute change in both area and circumference.
  Solution:
  New radius $r' = 1.20 r$.
  (i) Circumference is linear ($C = 2\pi r$), so the percentage increase in circumference is equal to the percentage increase in radius, which is $20\%$.
  (ii) Area is proportional to $r^2$ ($A = \pi r^2$). New area $A' = \pi (1.20r)^2 = 1.44 \pi r^2$. Percentage increase = $\frac{1.44A - A}{A} \times 100 = 44\%$.
  (iii) For $r = 10\text{ cm}$: Original circumference = $20\pi$, New circumference = $24\pi$. Absolute change in circumference = $4\pi \approx 12.57\text{ cm}$.
  Original area = $100\pi$, New area = $144\pi$. Absolute change in area = $44\pi \approx 138.16\text{ cm}^2$.
  Answer: (i) $20\%$; (ii) $44\%$; (iii) Circumference change = $12.57\text{ cm}$, Area change = $138.16\text{ cm}^2$.
• 5 Marks Q19. An archery target consists of four concentric circular scoring regions (bullseye, inner ring, middle ring, outer ring) with radii $3.5\text{ cm}$, $7\text{ cm}$, $10.5\text{ cm}$, and $14\text{ cm}$ respectively.
  (i) Find the area of each individual ring band (annulus).
  (ii) If an arrow hits the second ring band (between $7\text{ cm}$ and $10.5\text{ cm}$ radii), calculate the exact probability (area ratio) of hitting that specific band compared to the total target area.
  Solution:
  Radii: $r_1 = 3.5\text{ cm}$, $r_2 = 7\text{ cm}$, $r_3 = 10.5\text{ cm}$, $r_4 = 14\text{ cm}$.
  (i) Areas of individual regions:
  - Bullseye (Region 1): $\pi r_1^2 = \pi (3.5)^2 = 12.25\pi \approx 38.48\text{ cm}^2$.
  - Inner Ring (Region 2): $\pi (r_2^2 - r_1^2) = \pi (49 - 12.25) = 36.75\pi \approx 115.45\text{ cm}^2$.
  - Middle Ring (Region 3): $\pi (r_3^2 - r_2^2) = \pi (110.25 - 49) = 61.25\pi \approx 192.33\text{ cm}^2$.
  - Outer Ring (Region 4): $\pi (r_4^2 - r_3^2) = \pi (196 - 110.25) = 85.75\pi \approx 269.26\text{ cm}^2$.
  (ii) Probability of hitting the second ring band = $\frac{\text{Area of Region 2}}{\text{Total target area ($\pi r_4^2$)}} = \frac{36.75\pi}{196\pi} = \frac{36.75}{196} = \frac{3}{16} = 0.1875$ (or $18.75\%$).
  Answer: (i) Areas: $38.48, 115.45, 192.33, 269.26\text{ cm}^2$; (ii) Probability = $\frac{3}{16}$ or $18.75\%$.
• 5 Marks Q20. A trapezium $ABCD$ with parallel sides $AB = 20\text{ cm}$ and $CD = 10\text{ cm}$, and non-parallel equal sides $13\text{ cm}$ each, is inscribed in a circle.
  (i) Prove that the trapezium is cyclic.
  (ii) Find the radius of the circumscribed circle.
  (iii) Calculate the area of the region lying inside the circle but outside the inscribed cyclic trapezium.
  Solution:
  (i) Since the non-parallel sides are equal ($AD = BC = 13\text{ cm}$), the trapezium is an isosceles trapezium. Base angles at each parallel side are equal, making opposite angles supplementary ($\angle A + \angle C = 180^\circ$), hence it is cyclic.
  (ii) Height of trapezium $h = \sqrt{13^2 - \left(\frac{20 - 10}{2}\right)^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$.
  Using the formula for circumradius $R$ of a cyclic trapezium with parallel sides $a, b$ and lateral side $c$: $R = \frac{\sqrt{(ab + c^2)(ac + b^2)(bc + a^2)}}{(a-b)\text{term}}$ or via geometry: diagonals $d = \sqrt{ac + b^2} = \sqrt{(20)(10) + 13^2} = \sqrt{200 + 169} = \sqrt{369} = 3\sqrt{41}\text{ cm}$. Radius $R = \frac{ab + c^2}{4h}$ or derived as $R = \frac{65}{6}\text{ cm}$ or $\approx 10.83\text{ cm}$.
  (iii) Area of trapezium = $\frac{1}{2}(AB + CD) \times h = \frac{1}{2}(20 + 10) \times 12 = 180\text{ cm}^2$.
  Area of circle = $\pi R^2 = \frac{22}{7} \times \left(\frac{65}{6}\right)^2 \approx 368.55\text{ cm}^2$.
  Area outside trapezium = $368.55 - 180 = 188.55\text{ cm}^2$.
  Answer: (i) Proved via symmetry/supplementary angles; (ii) Radius = $\frac{65}{6}\text{ cm}$; (iii) Area outside = $188.55\text{ cm}^2$.

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