Question

Please HELP! I DON'T KNOW HOW CAN I SOLVE THIS ​

Answer 1

See below.

Explanation:

These are exercises in the application of the Pythagoras theorem:

The square on the hypotenuse = the sum of the squares on the other 2 sides.

So For No. 42:

x^2 = 121 + 2^2

x^2 = 5

x = √5 .

No. 41.

x^2 = 1^1 + 1^2

x^2 = 2

x = √2.

No 43.

5^2 = 3^2 + x^2

x^2 = 5^2 - 3^2

x^2 = 25 - 9 = 16

x = 4.

Answer 2

41. x = √2

42. x = √5

43. x = 4

44. x = 4√3

Explanation:

The Pythagorean Theorem:

`leg^2+leg^2=hypotenuse^2\to a^2+b^2=c^2`

`41.\\\\a=1,\ b=1,\ c=x\\\\x^2=1^2+1^2\\\\x^2=1+1\\\\x^2=2\to x=\sqrt2`

`42.\\a=1, b=2,\ c=x\\\\x^2=1^2+2^2\\\\x^2=1+4\\\\x^2=5\to x=\sqrt5`

`43.\\a=3,\ b=x,\ c=5\\\\3^2+x^2=5^2\\\\9+x^2=25\qquad\text{subtract 9 from both sides}\\\\x^2=16\to x=√(16)\to x=4`

`44.\\\\a=4,\ b=x,\ c=8\\\\4^2+x^2=8^2\\\\16+x^2=64\qquad\text{subtract 16 from both sides}\\\\x^2=48\to x=√(48)\\\\x=√(16\cdot3)\qquad\text{use}\ √(ab)=√(a)\cdot√(b)\\\\x=√(16)\cdot\sqrt3\\\\x=4\sqrt3`