▶️ Answer/Explanation
Answer: 34.6 or 34.63 to 34.64
1. Quarter-circle area = (π×5²)/4 = 19.63cm²
2. Triangle area = (5×6)/2 = 15cm²
3. Total area = 19.63 + 15 = 34.63cm²
2. Triangle area = (5×6)/2 = 15cm²
3. Total area = 19.63 + 15 = 34.63cm²
▶️ Answer/Explanation
Answer: 77.8 (or 77.77 to 77.80)
Steps:
1. Area of Rectangle (OPQR): \(11 \text{cm} \times 4 \text{cm} = 44 \text{cm}^2\).
2. Area of Sector (OPX): The sector angle is \(\tan^{-1}\left(\frac{4}{11}\right) \approx 19.98^\circ\).
Area = \(\frac{19.98}{360} \times \pi \times (11)^2 \approx 21.06 \text{cm}^2\).
3. Area of Triangle (OPQ): \(\frac{1}{2} \times 11 \times 4 = 22 \text{cm}^2\).
4. Shaded Area: Sector area + (Rectangle area – Triangle area) = \(21.06 + (44 – 22) = 43.06 \text{cm}^2\).
5. Percentage Shaded: \(\left(\frac{43.06}{44}\right) \times 100 \approx 77.8\%\).
1. Area of Rectangle (OPQR): \(11 \text{cm} \times 4 \text{cm} = 44 \text{cm}^2\).
2. Area of Sector (OPX): The sector angle is \(\tan^{-1}\left(\frac{4}{11}\right) \approx 19.98^\circ\).
Area = \(\frac{19.98}{360} \times \pi \times (11)^2 \approx 21.06 \text{cm}^2\).
3. Area of Triangle (OPQ): \(\frac{1}{2} \times 11 \times 4 = 22 \text{cm}^2\).
4. Shaded Area: Sector area + (Rectangle area – Triangle area) = \(21.06 + (44 – 22) = 43.06 \text{cm}^2\).
5. Percentage Shaded: \(\left(\frac{43.06}{44}\right) \times 100 \approx 77.8\%\).