Online Course Discussion Forum
hiuwaitong@gmail.com
For Math Challenge I-A geometry, I don't understand the following
6.26; I don't know how to find the height for the 3 smaller equilateral triangles
6.30; I don't understand how they got the triangle with a hypotenuse length x� is x2√�2.
6.24; I don't know how the answer key got 37.5 degrees when the hexagon is outside the octagon when I got 15 degrees instead
6.22; I don't know how to do
p.s Maybe you can have the answer key to list the steps
6.26: For equilateral triangles, as long as the side length is known, the height can be calculated the same way. Remember the height splits the equilateral triangle into two $30$-$60$-$90$ triangles.
6.30: The corner is a $45$-$45$-$90$ triangle. The side length ratio of such a triangle is $1:1:\sqrt{2}$. If teh hypotenuse is $x$, it means $x$ corresponds to the $\sqrt{2}$ part, so the leg length is $\dfrac{x}{\sqrt{2}}$.
6.24: An interior angle of a regular octagon is $135^\circ$.
6.22: You can calculate all the angles of the quadrilateral.
社交网络