📅  最后修改于: 2023-12-03 14:59:06.631000             🧑  作者: Mango
本篇文章介绍的是针对NCERT第3章练习3.1的解决方案。该练习主要是关于四边形的基础知识和性质。
本解决方案旨在:
在开始开发解决方案之前,我们需要分析要解决的问题和需要满足的需求。对于NCERT第3章练习3.1,我们需要完成以下任务:
我们可以通过Python代码实现以上需求。我们将使用以下库:
我们将创建以下函数:
以下是解决方案的代码实现:
import math
import numpy as np
# 获取两点距离
def get_distance(p1, p2):
return math.sqrt((p1[0]-p2[0])**2 + (p1[1]-p2[1])**2)
# 输入四边形的四个顶点
def input_vertices():
vertices = []
for i in ["first", "second", "third", "fourth"]:
x = float(input(f"Enter {i} vertex x-coordinate: "))
y = float(input(f"Enter {i} vertex y-coordinate: "))
vertices.append([x, y])
return vertices
# 计算四边形的四条边的长度
def get_sides(vertices):
sides = []
sides.append(get_distance(vertices[0], vertices[1]))
sides.append(get_distance(vertices[1], vertices[2]))
sides.append(get_distance(vertices[2], vertices[3]))
sides.append(get_distance(vertices[3], vertices[0]))
return sides
# 计算四边形的角度
def get_angles(vertices):
angles = []
v1 = np.array(vertices[1]) - np.array(vertices[0])
v2 = np.array(vertices[2]) - np.array(vertices[1])
angles.append(math.degrees(math.acos(np.dot(v1, v2)/(get_distance(vertices[0],vertices[1])*get_distance(vertices[1],vertices[2])))))
v1 = np.array(vertices[2]) - np.array(vertices[1])
v2 = np.array(vertices[3]) - np.array(vertices[2])
angles.append(math.degrees(math.acos(np.dot(v1, v2)/(get_distance(vertices[1],vertices[2])*get_distance(vertices[2],vertices[3])))))
v1 = np.array(vertices[3]) - np.array(vertices[2])
v2 = np.array(vertices[0]) - np.array(vertices[3])
angles.append(math.degrees(math.acos(np.dot(v1, v2)/(get_distance(vertices[2],vertices[3])*get_distance(vertices[3],vertices[0])))))
v1 = np.array(vertices[0]) - np.array(vertices[3])
v2 = np.array(vertices[1]) - np.array(vertices[0])
angles.append(math.degrees(math.acos(np.dot(v1, v2)/(get_distance(vertices[3],vertices[0])*get_distance(vertices[0],vertices[1])))))
return angles
# 计算四边形的对角线的长度
def get_diagonals(vertices):
d1 = get_distance(vertices[0], vertices[2])
d2 = get_distance(vertices[1], vertices[3])
return d1, d2
# 判断四边形是否是矩形
def is_rectangle(angles):
return all(angle == 90 for angle in angles)
# 判断四边形是否是正方形
def is_square(sides):
return all(side == sides[0] for side in sides)
# 判断四边形是否是菱形
def is_rhombus(sides):
return all(side == sides[0] for side in sides)
# 判断四边形是否是平行四边形
def is_parallelogram(vertices):
return abs(vertices[0][1]-vertices[1][1]) == abs(vertices[2][1]-vertices[3][1]) and abs(vertices[1][1]-vertices[2][1]) == abs(vertices[3][1]-vertices[0][1])
# 判断四边形是否是风筝形
def is_kite(sides, diagonals):
return diagonals[0]*diagonals[1] / 2 == sides[0]*sides[1] / 2 + sides[2]*sides[3] / 2
# 计算四边形的周长
def get_perimeter(sides):
return sum(sides)
# 计算四边形的面积
def get_area(sides, angles, diagonals):
if is_rhombus(sides):
area = (diagonals[0] * diagonals[1]) / 2
else:
area = 0.5 * sides[0] * sides[1] * math.sin(math.radians(angles[0]))
return area
# 主函数,用于调用以上函数
def main():
vertices = input_vertices()
sides = get_sides(vertices)
angles = get_angles(vertices)
diagonals = get_diagonals(vertices)
if is_square(sides):
print("The given quadrilateral is a square.")
elif is_rhombus(sides):
print("The given quadrilateral is a rhombus.")
elif is_rectangle(angles):
print("The given quadrilateral is a rectangle.")
elif is_parallelogram(vertices):
print("The given quadrilateral is a parallelogram.")
elif is_kite(sides, diagonals):
print("The given quadrilateral is a kite.")
else:
print("The given quadrilateral is a general quadrilateral.")
print(f"The perimeter of the quadrilateral is {get_perimeter(sides)}.")
print(f"The area of the quadrilateral is {get_area(sides, angles, diagonals)}.")
if __name__ == "__main__":
main()
本篇文章中,我们介绍了通过Python代码解决NCERT第3章练习3.1的解决方案,并提供了详细的注释和解释,以帮助初学者更好地理解四边形的概念和性质。