From the given diagram , in smaller triangle BCD , Using Pythagoras theorem we can write
[tex] x^2=p^2+4^2\\\\x^2=p^2+16\\\\ [/tex]
In Triangle ABD , we can write
[tex] q^2=21^2+p^2\\\\ [/tex]
In Triangle ABC
[tex] (21+4)^2=q^2+x^2\\\\q^2+x^2=625\\ [/tex]
Now using the above three equations we can write
[tex] 21^2+p^2+x^2=625\\\\21^2+x^2-16+x^2=625, x^2=p^2+16\\\\2x^2+425=625\\\\2x^2=625-425\\\\2x^2=200\\\\x^2=100\\\\x=10\\ [/tex]
