Given
\(\triangle\) ABC = 35

I.e \(\frac{1}{2} * BC * 7 = 35\)

or BC = 10

Now in \(\triangle\) BCD,

\(DC^2= \sqrt{(10^2 - 6^2)}\) = 64

or DC = 8
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Another way to solve it is to notice that one leg of BCD is 6 and the hypotenuse is 10, and given that the triangle is a right triangle, we can use the Pitagorean triplet 3:4:5, where each of the sides is doubled. Indeed, in this case we would have 6:x:10 and x = 8.

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