Question

What is the value of X6?

Show the solution.

Answer 1

Answer : The value of
`x_6` is
`√(7)`.

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of
`x_1`.

Using Pythagoras theorem in triangle 1 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_1)^2=(1)^2+(1)^2`

`x_1=√((1)^2+(1)^2)`

`x_1=√(2)`

Now we have to calculate the value of
`x_2`.

Using Pythagoras theorem in triangle 2 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_2)^2=(1)^2+(X_1)^2`

`(x_2)^2=(1)^2+(√(2))^2`

`x_2=\sqrt{(1)^2+(√(2))^2}`

`x_2=√(3)`

Now we have to calculate the value of
`x_3`.

Using Pythagoras theorem in triangle 3 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_3)^2=(1)^2+(X_2)^2`

`(x_3)^2=(1)^2+(√(3))^2`

`x_3=\sqrt{(1)^2+(√(3))^2}`

`x_3=√(4)`

Now we have to calculate the value of
`x_4`.

Using Pythagoras theorem in triangle 4 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_4)^2=(1)^2+(X_3)^2`

`(x_4)^2=(1)^2+(√(4))^2`

`x_4=\sqrt{(1)^2+(√(4))^2}`

`x_4=√(5)`

Now we have to calculate the value of
`x_5`.

Using Pythagoras theorem in triangle 5 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_5)^2=(1)^2+(X_4)^2`

`(x_5)^2=(1)^2+(√(5))^2`

`x_5=\sqrt{(1)^2+(√(5))^2}`

`x_5=√(6)`

Now we have to calculate the value of
`x_6`.

Using Pythagoras theorem in triangle 6 :

`(Hypotenuse)^2=(Perpendicular)^2+(Base)^2`

`(x_6)^2=(1)^2+(X_5)^2`

`(x_6)^2=(1)^2+(√(6))^2`

`x_6=\sqrt{(1)^2+(√(6))^2}`

`x_6=√(7)`

Therefore, the value of
`x_6` is
`√(7)`.