What is the value of X6?

Show the solution.

Answer : The value of [tex]x_6[/tex] is [tex]\sqrt{7}[/tex].

Explanation :

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

First we have to calculate the value of [tex]x_1[/tex].

Using Pythagoras theorem in triangle 1 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_1)^2=(1)^2+(1)^2[/tex]

[tex]x_1=\sqrt{(1)^2+(1)^2}[/tex]

[tex]x_1=\sqrt{2}[/tex]

Now we have to calculate the value of [tex]x_2[/tex].

Using Pythagoras theorem in triangle 2 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_2)^2=(1)^2+(X_1)^2[/tex]

[tex](x_2)^2=(1)^2+(\sqrt{2})^2[/tex]

[tex]x_2=\sqrt{(1)^2+(\sqrt{2})^2}[/tex]

[tex]x_2=\sqrt{3}[/tex]

Now we have to calculate the value of [tex]x_3[/tex].

Using Pythagoras theorem in triangle 3 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_3)^2=(1)^2+(X_2)^2[/tex]

[tex](x_3)^2=(1)^2+(\sqrt{3})^2[/tex]

[tex]x_3=\sqrt{(1)^2+(\sqrt{3})^2}[/tex]

[tex]x_3=\sqrt{4}[/tex]

Now we have to calculate the value of [tex]x_4[/tex].

Using Pythagoras theorem in triangle 4 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_4)^2=(1)^2+(X_3)^2[/tex]

[tex](x_4)^2=(1)^2+(\sqrt{4})^2[/tex]

[tex]x_4=\sqrt{(1)^2+(\sqrt{4})^2}[/tex]

[tex]x_4=\sqrt{5}[/tex]

Now we have to calculate the value of [tex]x_5[/tex].

Using Pythagoras theorem in triangle 5 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_5)^2=(1)^2+(X_4)^2[/tex]

[tex](x_5)^2=(1)^2+(\sqrt{5})^2[/tex]

[tex]x_5=\sqrt{(1)^2+(\sqrt{5})^2}[/tex]

[tex]x_5=\sqrt{6}[/tex]

Now we have to calculate the value of [tex]x_6[/tex].

Using Pythagoras theorem in triangle 6 :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](x_6)^2=(1)^2+(X_5)^2[/tex]

[tex](x_6)^2=(1)^2+(\sqrt{6})^2[/tex]

[tex]x_6=\sqrt{(1)^2+(\sqrt{6})^2}[/tex]

[tex]x_6=\sqrt{7}[/tex]

Therefore, the value of [tex]x_6[/tex] is [tex]\sqrt{7}[/tex].

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Side = 11 cm, side = 5 cm, side = 6 cm

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