Question

Someone please help me find x and y for problems #19 and 20!!! Please explain if you can

Answer 1

Lol did this last year with these ratios. 
19) you have 40 for one leg, 10 for another leg and 50 for a hypotenuse. To find x, the altitude, you need to set up the ratio, 10/x=x/40. You can multiply x by both sides. 10=x^2/40. You can also multiply 40 by both sides. 400=x^2. Square root both sides. 20=x. You now have one variable, let's solve for y now :0. Using the biggest triangle, we have 50 as the hypotenuse. You can use the Pythagorean Theorem to help you out. We first need to find the value of one of the legs, so 40^2+20^2=c^2. This gets me to 2000=c^2. Take the square root of both sides. I get c=44.72 (be aware that I'll always round to the nearest hundredth for every ANSWER! It's a bit more accurate). Now to find the leg of the biggest triangle. y^2+44.72^2=50^2 which is y^2+1999.8784=2500. Subtract 1999.8784 from both sides to get y^2=500.1216. Take the sqrt or square root of both sides. (It's rounded) y=22.36. That's question 19 for ya. 

20) Look at this now! Doesn't it look familiar? Let's start with y. Start with the ratio '21/y=y/4'. Use my ways from before to get (you can pretend the same thing is happening from prob. 19)), '21=(y^2)/4' '84=y^2' 'y=9.17'. That's y, now x. Let's find the leg of the big triangle first. PYTHAGOREAN THEOREM!!! 21^2+9.17^2=c^2, 441+84.0889=c^2,
c^2=525.0889, c=22.91. We can finally figure out x. 22.91^2+x^2=25^2, 524.8681+x^2=625, x^2=100.1319, x=10.01.

Hope you've enjoyed this learning experience about geometric means and whatnot!!!       :3