Answer:
[tex]x=16.4\ y=13.7[/tex]
Step-by-step explanation:
[tex][Kindly\ refer\ the\ attachment.]\\We\ are\ given\ that,\\\triangle ABC\ is\ a\ right-triangle\ and\ has\ it's\ legs\ x\ and\ z.\\It's\ altitude\ CD\ separates\ it\ into\ two\ triangles:\\\triangle BCD\ and\ \triangle ADC,\ both\ of\ which\ are\ right-triangles.\\Now,\\The\ Pythagoras\ Theorem\ tells\ us\ that:\\'The\ sum\ of\ squares\ of\ both\ the\ legs\ is\ equal\ to\ the\ square\ of\ the\\ hypotenuse'.\\Here,\\Lets\ consider\ three\ triangles\ separately.\\[/tex]
[tex]In\ \triangle ADC\ right\ angle\ is\ at\ \angle ADC.\\Hence,\\AC\ is\ the\ hypotenuse, and\ AD\ and\ DC\ are\ it's\ legs.\\Hence,\\AD^2+DC^2=AC^2\\Or,\\9^2+y^2=x^2\\\\Lets\ take\ the\ above\ equation\ as\ E_1.[/tex]
[tex]In\ \triangle BDC\ right\ angle\ is\ at\ \angle BDC.\\Hence,\\BC\ is\ the\ hypotenuse, and\ DC\ and\ DB\ are\ it's\ legs.\\Hence,\\DC^2+DB^2=BC^2\\Or,\\21^2+y^2=z^2\\Lets\ take\ the\ above\ equation\ as\ E_2.[/tex]
[tex]Now,\\In\ whole\ \triangle ACB\ right\ angle\ is\ at\ \angle ACB.\\Hence,\\AB\ is\ the\ hypotenuse, and\ AC\ and\ BC\ are\ it's\ legs.\\Hence,\\AC^2+BC^2=AB^2\\Or,\\x^2+z^2=(21+9)^2=30^2\\Lets\ take\ the\ above\ equation\ as\ E_3.[/tex]
[tex]Lets\ now\ gather\ the\ three\ equations:\\9^2+y^2=x^2\\21^2+y^2=z^2\\x^2+z^2=30^2\\Now,\\Lets\ pair\ and\ add\ E_1\ and\ E_2.\\Hence,\\(9^2+y^2)+(21^2+y^2)=(x^2)+(z^2)\\Hence,\\81+y^2+441+y^2=x^2+z^2\\Hence,\\2y^2+522=x^2+z^2\\Considering\ E_3,\\We\ already\ know\ that,\\x^2+z^2=30^2\\Hence,\\2y^2+522=30^2\\Hence,\\2y^2+522=900\\2y^2=900-522=378\\Hence,\\y^2=\frac{378}{2}=189\\Hence,\\y=\sqrt{189}=13.7\\[/tex]
[tex]Plugging\ in\ y=13.7\ in\ E_1,\\9^2+(13.7)^2=x^2\\81+189=x^2\\x^2=270\\x=\sqrt{270}=16.4[/tex]
