Answer:
Step-by-step explanation:
Q1)
Use Phythogoras theorem:
[tex](AC)^2=(BC)^2+(AB)^2\\BC^2=AC^2-AB^2\\BC^2=6^2-4^2\\BC^2=36-16\\BC^2=20\\\\Apply\ square\ on\ both\ sides\\\\\sqrt{BC^2}=\sqrt{20} \\BC=\sqrt{20} \\BC=\sqrt{10(2)} \\BC=\sqrt{5(2)(2)} \\BC=2\sqrt{5}[/tex]
Q2)
Apply phythogoras theorem:
[tex]CD^2=CE^2+DE^2\\CD^2=7^2+9^2\\CD^2=49+81\\CD^2=130\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{CD^2} = \sqrt{130} \\\\CD=\sqrt{130} \\\\CD=11.4[/tex]
Q3)
Apply phythogoras theorem again:
[tex]BC^2=BD^2+CD^2\\BC^2=7^2+13^2\\BC^2=49+169\\BC^2=218\\\\Apply\ square\ root\ on\ both\ sides\\\\\\sqrt{BC^2} =\sqrt{218} \\\\BC=\sqrt{218} \\\\BC=14.76[/tex]
I have an attached an image for Question 2 for better understanding the length of DE in question equals 15 - 6 = 9
