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(All dimensions are in centimeter)

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Let’s analyse the question through the diagram first. Here we have an isosceles triangle $\Delta ABC$, which has an altitude $BE$ on the side $AC$. The altitude divides the $\Delta ABC$ into two right-angled triangles $\Delta BEC$ and $\Delta BEA$. Using all this information we need to find the value of $'x'$

If we notice $\Delta BEC$ in the figure, we can see that $x$ is the hypotenuse of this triangle.

Now we can use Pythagoras’ theorem that states that, “the sum of the squares of the sides of a right triangle is equal to the square on the hypotenuse”

Using Pythagoras’ theorem in $\Delta BEA$, we get:

$ \Rightarrow A{B^2} = B{E^2} + A{E^2}$

Now we can easily substitute the given values in the above equation

$ \Rightarrow {9^2} = B{E^2} + {7^2}$

$ \Rightarrow B{E^2} = 81 - 49 = 32$

$ \Rightarrow BE = \sqrt {32} = 4\sqrt 2 cm$

Similarly, again using Pythagoras’ theorem in $\Delta BEC$, we can see that:

$ \Rightarrow B{C^2} = B{E^2} + E{C^2}$

Let’s substitute the already known values in the above equation

$ \Rightarrow B{C^2} = {\left( {4\sqrt 2 } \right)^2} + {2^2} = 32 + 4 = 36$

$ \Rightarrow BC = x = 6 cm$

Hence, we got the value of $'x'$ as $6 cm$.