how do i find the value of x by using the pythagorean theorum on thes 2 triangles?

Question

1) its a right triangle. leg = 1, leg=2, hypotenuse= x

2) also a right triangle. leg= 4, leg= X, hypotenuse= 8

Answer 1

1) x = √5
2) x = 4√3

1) 1² + 2² = x²
x = √(1² + 2² ) = √5
2) 4² + x² = 8²
x = √(8² - 4²) = 4√3

Answer 2

Plug values in :
1² + 2² = c² : 1 + 4 = c² : 5 = c² : c = √5

4² + b² = 8² : 16 + b² = 64 : b² = 48 : b = √48 = 4√3

Answer 3

1) x^2= 1^2 +2^2
x^2= 1+4
x^2=5
x= square root of 5

2) x^2+4^2=8^2
x^2+16=64
x^2= 48
x^2= 16*3
x = sq rt 16 times sq rt3
x= 4times square root of 3

Answer 4

1) 1^2 + 2^2 = x^2
1 + 4 = x^2
x = root 5

2) 4^2 + x^2 = 8^2
16 + x^2 = 64
x^2 = 48
x = root 48 (or 4root 3 after you simplify it)

Answer 5

a2 (squared-times itself) + b2 = c2 (c is always the hypoteneuse)

So
1(1) + 2(2) =x2

Do the multiplication first then the adding. I won't do the math for you, that would be cheating...

Then get the square root of to find x. (Example: the square root of 9 would be 3 b/c when you multiply 3 by itself you get 9)

The other one has a slightly different process because you know the hypoteneuse (c):

a2+b2=c2
4(4)+x2=8(8)

Once you know what a2 and c2 are you need to get x on it's own. You do this by subtracting a2 from both sides.

x2= -answer + answer

So you subtract the answer for a from the answer for c to get one number. This is what x2 is. So you then get the square root and you have x!

Answer 6

1) 1^2 + 2^2 = x^2
1 + 4 = x^2
x = sqrt(5)

2) 4^2 + x^2 = 8^2
16 + x^2 = 64
x^2 = 64 - 16
x = sqrt(48)

Answer 7

Ok so pythagoras' theorem is
a(squared)+b(squared)=c(squared)
where a = leg 1
b = leg 2
and c = hypotenuse

1)1 (squared) + 2(squared) = x(squared)
therefore 1 + 4 = x (squared)
5= x(squared)
hence x= square root of 5

2) 4(squared) + x(sq) = 8 (sq)
therefore 16 + x(sq) = 64
x(squared) = 64 - 18
x(sq) = 46
hence x = square root of 46

No problem =)

Answer 8

A squared plus B squared equals C squared