Respuesta :

gmany
9.
[tex]\text{If QR is tangent to circle then m}\angle PQR=90^o[/tex]

[tex]\text{We can use the converse of Pythagorean theorem}[/tex]

[tex]8\ and\ 4\sqrt3\\\\check:\ 4^2+(4\sqrt3)^2=8^2\\16+16\cdot3=64\\16+48=64\\64=64\ -O.K.\\\\2\ and\ 4\sqrt2\\\\check:\ 4^2+2^2=(4\sqrt2)^2\\16+4=16\cdot2\\20=32\ -FALSE\\\\3\ and\ 5\\\\check:\ 3^2+4^2=5^2\\9+16=25\\25=25\ -O.K.\\\\2\ and\ 2\sqrt5\\\\check:\ 4^2+2^2=(2\sqrt5)^2\\16+4=4\cdot5\\20=20\ -O.K.\\\\6\ and\ 10\\check:\ 4^2+6^2=10^2\\16+36=100\\52=100\ -FALSE[/tex]

[tex]\text{Answer:}\boxed{8\ and\ 4\sqrt3;\ 3\ and\ 5;\ 2\ and\ 2\sqrt5}[/tex]


10.
[tex]\text{Look at the pictures}[/tex]

[tex]\text{Therefore}\\\\C=2\cdot8+2\cdot16+2\cdot9+2\cdot6=16+32+18+12=78cm[/tex]

[tex]Other\ method:\\C=2\cdot(8+16+6+9)=2\cdot39=78cm[/tex]
Ver imagen gmany
Ver imagen gmany

Answer:

\text{If QR is tangent to circle then m}\angle PQR=90^o

\text{We can use the converse of Pythagorean theorem}

8\ and\ 4\sqrt3\\\\check:\ 4^2+(4\sqrt3)^2=8^2\\16+16\cdot3=64\\16+48=64\\64=64\ -O.K.\\\\2\ and\ 4\sqrt2\\\\check:\ 4^2+2^2=(4\sqrt2)^2\\16+4=16\cdot2\\20=32\ -FALSE\\\\3\ and\ 5\\\\check:\ 3^2+4^2=5^2\\9+16=25\\25=25\ -O.K.\\\\2\ and\ 2\sqrt5\\\\check:\ 4^2+2^2=(2\sqrt5)^2\\16+4=4\cdot5\\20=20\ -O.K.\\\\6\ and\ 10\\check:\ 4^2+6^2=10^2\\16+36=100\\52=100\ -FALSE

\text{Answer:}\boxed{8\ and\ 4\sqrt3;\ 3\ and\ 5;\ 2\ and\ 2\sqrt5}

10.

\text{Look at the pictures}

\text{Therefore}\\\\C=2\cdot8+2\cdot16+2\cdot9+2\cdot6=16+32+18+12=78cm

Other\ method:\\C=2\cdot(8+16+6+9)=2\cdot39=78cm

Step-by-step explanation: