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Divide the quarter circle into a triangle and a geometric figure, the triangle is 10cm & # 178;, ask the area of the geometric figure
Divide the quarter circle into a triangle and a geometric figure, the triangle is 10cm & # 178;, ask the area of the geometric figure
3.14×（10×2）×1/4-10
=3.14×5-10
=15.7-10
=5.7 square centimeters
RELATED INFORMATIONS
- 1. a. BC is three positive integers, B & # 178; = 2Ac + 1, a square with B as the side and a rectangle with a and C as the length and width respectively. What is the area of the figure
- 2. If the lengths of three sides of a triangle are a, B, C, and satisfy a ^ 2 + 2B ^ 2-2ab-2ac + C ^ 2 = 0, try to judge what triangle the triangle is and explain the reason
- 3. It is known that a = ½ m + 1, B = ½ m + 2, C = ½ m + 3, find the value of a & #178; + 2Ab + B & #178; - 2Ac + C & #178; - 2BC
- 4. A = half m + 1, B = half m + 2, C = half m + 3, find a ^ 2 + 2Ab + B ^ 2-2ac + C ^ 2-2bc
- 5. Given a = m / 2 + 1, B = m / 2 + 2, C = m / 2 + 3, find the value of a ^ 2 + 2Ab + B ^ 2-2ac-abc + C ^ 2
- 6. When m is the value, the equations {y ^ 2 = 12x, y = 3x + m have two identical real solutions, and the solutions of the equations are obtained
- 7. When m is the value, the equations x + 2y-6 = 0 and y = MX + 3 have two identical real solutions
- 8. When m is the value, the system of equations y = x + m x square + y square = 2 has two equal real solutions
- 9. B ^ 2 + C ^ 2 + 2Ab + 2Ac + 2BC factorization
- 10. B * 2 + C * 2 + 2Ab + 2Ac + 2BC factorization
- 11. Known: a = B & # 178; + C & # 178; - A & # 178; / 2BC, B = A & # 178; + C & # 178; - B & # 178; / 2Ac, C = A & # 178; + B & # 178; - C & # 178; / 2Ab And a + B = C, find the value of a to the power of 2013 + B to the power of 2013 + C to the power of 2013
- 12. Why 2Ab + 2BC + 2Ac = - (A & # 178; + B & # 178; + C & # 178;) = - 100 why a & # 178; + B & # 178; + C & # 178; + 2Ab + 2B + 2BC = 0
- 13. If x is a real number and | x-3 | - | X-1 | > m is constant, then the value range of M is () A. m＞2B. m＜2C. m＞-2D. m＜-2
- 14. Given that a and B belong to positive real numbers and a + b-ab + 3 = 0, then the value range of ab Hurry!
- 15. Solution set of inequality LX + 3L ≤ 1
- 16. Solve the following inequality lx-2l is greater than or equal to l2x + 1L
- 17. The solution set of inequality LX + 1L + lx-1l < LXL + LX + 2L is
- 18. Inequality LX + 1L + lx-2l
- 19. Solving inequality lx-1l + LX + 2L ≤ 5
- 20. Given that a and B are real numbers, and the solution set of inequality | (a + 2) x-2a + 1 | < B about X is - 1 < x < 3, then the value of a + B is () A. 3 or 7b. 13C. 7 or 8D. 8 or 13