You have 2 triangles here making up this single quadrilateral. One of the triangles is a 30-60-90 and the other one is a 45-45-90. These 2 triangles share a hypotenuse. We need the length of that hypotenuse if we are to find the missing side x. The "lower" triangle is the 30-60-90 and the side we are given is across from the 30 degree angle. In the Pythagorean triple for a 30-60-90, the side across from the 30 degree angle is x, the side across from the 60 degree angle is [tex]x \sqrt{3} [/tex], and the hypotenuse is 2x. So we need to find x. If [tex]x=14 \sqrt{3} [/tex] then two times x is equal to [tex]2(14 \sqrt{3}) [/tex] which is equal to [tex]28 \sqrt{3} [/tex]. So the length of the hypotenuse is [tex]28 \sqrt{3} [/tex]. The Pythagorean triple for a 45-45-90 is [tex](x,x,x \sqrt{2}) [/tex] with [tex]x \sqrt{2} [/tex] as the length of the hypotenuse in a 45-45-90. If the length of the hypotenuse is [tex]28 \sqrt{3} [/tex], then we need to solve for x. [tex]x \sqrt{2}=28 \sqrt{3} [/tex]. We solve for x by dividing by the square root of 2, like this: [tex]x= \frac{28 \sqrt{3} }{ \sqrt{2} } [/tex]. The only way to solve this is to multiply by [tex] \frac{ \sqrt{2} }{ \sqrt{2} } [/tex]. Doing this step by step we have [tex] \frac{28 \sqrt{3} }{ \sqrt{2} }* \frac{ \sqrt{2} }{ \sqrt{2} } [/tex]. Multiply straight across the top and straight across the bottom to get [tex] \frac{28 \sqrt{6} }{2} [/tex] which simplifies by reduction to [tex]14 \sqrt{6} [/tex]. First choice above.
