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# geometry help (again)

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Does anyone know how to do this? thank you

Mar 28, 2018

#1
+7545
+3

Let's call the unlabeled side of the large triangle  " x ".

And let's call the height of that triangle  " h ".

By Pythagorean theorem, we can make the equation

x2 + y2   =   (16 + 4)2

Subtract  y2  from both sides of the equation.

x2   =   (16 + 4)2 - y2

Again by the Pythagorean theorem, we can make the equation

162 + h2   =   x2

Subtract  162  from both sides of the equation.

h2   =   x2 - 162

Substitute   (16 + 4)2 - y2   in for   x2

h2   =   (16 + 4)2 - y2 - 162

Again by the Pythagorean theorem, we can make the equation

42 + h2   =   y2               Substitute   (16 + 4)2 - y2 - 162   in for  h2

42 + (16 + 4)2 - y2 - 162   =   y2      Now we can solve for  y .

Simplify the left side of the equation.

42 + 202 - y2 - 162   =   y2

16 + 400 - y2 - 256   =   y2

160 - y2  =  y2

Add  y2  to both sides of the equation.

160   =   2y2

Divide both sides by  2

80   =   y2

Take the positive square root of both sides.

√80  =  y

y  =  √[2*2 * 2*2 * 5]

y  =  4√5

Mar 28, 2018
#2
+100456
+2

Call the height, h

Since the apex angle  is a right angle... the height is forms a geometric mean, thusly

16/h  = h/4      cross-multiply

h^2  =  64

h  =  8

And by the Pythagoren Theorem

√ [ h^2  + 4^2  ]  =  y

√ [ 8^2 + 4^2 ] = y

√80  =  y

√[16 * 5]   y

4√5  = y

Mar 28, 2018