Answer:
1. [tex]x=18, y=6\sqrt{10}[/tex]
2. x=108, y=180
Step-by-step explanation:
The height of the right triangle drawn to the hypotenuse is the geometric mean of two segments of the hypotenuse (two legs' projections)
1. Use this property, so
[tex]CD^2=AD\cdot BD\\ \\6^2=2\cdot x\\ \\36=2x\\ \\x=18[/tex]
Consider right triangle BCD. By the Pythagorean theorem,
[tex]BC^2=BD^2+CD^2\\ \\y^2=18^2+6^2\\ \\y^2=324+36\\ \\y^2=360\\ \\y=\sqrt{360}=6\sqrt{10}[/tex]
2. Consider right triangleGM*. By the Pythagorean theorem,
[tex]G*^2=GM^2+M*^2\\ \\240^2=GM^2+192^2\\ \\GM^2=240^2-192^2=(240-192)(240+192)=48\cdot 432=144^2\\ \\GM=144[/tex]
Use this property to find x:
[tex]GM^2=192\cdot x\\ \\144^2=192\cdot x\\ \\x=\dfrac{144^2}{192}=108[/tex]
Consider right triangle GMCup. By the Pythagorean theorem,
[tex]y^2=x^2+GM^2\\ \\y^2=108^2+144^2\\ \\y^2=32,400\\ \\y=180[/tex]
