Question

等边三角形ABC是边长为a，求其内切圆的内接正方形DEFG的面积

Answer 1

等腰三角形边长为a 则高为 H=√(a�0�5-(a/2)�0�5)=√(3)/2*a内接园圆心在三角形形心,半径 R=2/3H=√(3)/3*a 直径D=2R=2√(3)/3*a 直径D是元内接正方形对角线,设正方形边长为B 则 B�0�5+B�0�5=D�0�5 2B�0�5=D�0�5=4/3*a�0�5 S=B�0�5=2/3*a�0�5=2a�0�5/3

Answer 2

解:内切圆半径r=2S/C=2×(√3/4)×a^2/3a=√3a/6 S=2r^2=2×(√3a/6)^2=a^2/6