第二题： 11.09

The height of the outer isosceles triagle is 6.

The inscribed right triangle is 30-60-90 degree, so the sides are x (the hypotenuce), x/2 (the shorter leg), sqrt(3)*x/2 (the longer leg). 

The area of the outer triangle is 16*6/2 . It is also 10*(sqrt(3)*x/2)/2=5*sqrt(3)x. 

So x=16*6/(5*sqrt(3)) = 96*sqrt(3)/15 = 11.09. 

所有跟帖： 

• 应该不是30-60-90的三角形。 -shantouchen- ♂ (0 bytes) () 10/29/2022 postreply 23:50:08