无锡小升初数学题求阴影面积

思路

1. 需要的知识范围：三角函数，反三角函数（计算器），一元二次方程
2. 求角度–>求扇形面积
   求三条边–>求三角面积
   相加相减

matlab代码

clear
syms x y
eq1 = x^2 + (y-10)^2 -100;
eq2 = (x-5)^2 + (y-5)^2 - 25;
A = solve(eq1,eq2,x,y);
x_solve = A.x(1); % point Q's x
y_solve = A.y(1); % point Q's y
OQ = (x_solve^2 + y_solve^2)^0.5; % OQ
OP = 5*sqrt(2);
QP = 5;
%%
cos_theta1 = (-OQ^2+OP^2+QP^2)/(2*OP*QP);% a^2 = b^2 + c^2 -2bc*cos 
theta1 = acos(cos_theta1); % rad
s2 = theta1/(2*pi) * (25*pi);
p = (OP+OQ+QP)/2;
s1pluss2 = sqrt(p*(p-OP)*(p-OQ)*(p-QP));
s1 = s1pluss2 - s2;
%%
OA = 10;
QA = 10;
cos_theta2 = (-OQ^2+OA^2+QA^2)/(2*OA*QA);% a^2 = b^2 + c^2 -2bc*cos 
theta2 = acos(cos_theta2); % rad
s_fanshapedpluss3 = theta2/(2*pi) * (100*pi);
p = (OQ+OA+QA)/2;
s_trishaped = sqrt(p*(p-OQ)*(p-OA)*(p-QA));
s3 = s_fanshapedpluss3 - s_trishaped;
%%
s = s1 + s3;
s_delta = (10^2-25*pi)/4/2-s;
s_final = 10^2 - 1/4*10^2*pi-(10^2-25*pi)/4 - 2*s_delta;
%%
s_final_val = double(s_final);
final_answer = ['area of shadow is ', num2str(s_final_val)];
disp(final_answer)