MATH SOLVE

4 months ago

Q:
# 15 points!!! GEOMETRY help please!!Questions attached!

Accepted Solution

A:

7.

[tex]\text{If the line segment labeled 6 is tangent to the circle then is rectangle triangle.}\\\\\text{ We can use Pythagorean theorem.}[/tex]

[tex]x^2=2.5^2+6^2\\x^2=6.25+36\\x^2=42.25\\x=\sqrt{42.25}\\x=6.5\\\\Answer:\boxed{6.5}[/tex]

8.

[tex]\text{If BC is tangent to circle then m}\angle OBC=90^o[/tex]

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

[tex]8\ and\ 12\\\\8^2+12^2=20^2\\64+144=400\\208=400\ -FALSE\\\\8;\ 4\sqrt{21}\\\\8^2+(4\sqrt{21})^2=20^2\\64+16\cdot21=400\\64+336=400\\400=400\ -O.K.\\\\12\ and\ 14\\\\12^2+14^2=20^2\\144+196=400\\340=400\ -FALSE\\\\10\ and\ 10\sqrt3\\\\10^2+(10\sqrt3)^2=20^2\\100+100\cdot3=400\100+300=400\\400=400\ -O.K.\\\\14\ and\ 4\sqrt{51};\ 4\sqrt{51} > 20 - FALSE[/tex]

[tex]\text{because the line segment labeled 20 is the longest side of the triangle}[/tex]

[tex]Answer:\ \boxed{8\ and\ 4\sqrt{21};\ 10\ and\ 10\sqrt3}[/tex]

[tex]\text{If the line segment labeled 6 is tangent to the circle then is rectangle triangle.}\\\\\text{ We can use Pythagorean theorem.}[/tex]

[tex]x^2=2.5^2+6^2\\x^2=6.25+36\\x^2=42.25\\x=\sqrt{42.25}\\x=6.5\\\\Answer:\boxed{6.5}[/tex]

8.

[tex]\text{If BC is tangent to circle then m}\angle OBC=90^o[/tex]

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

[tex]8\ and\ 12\\\\8^2+12^2=20^2\\64+144=400\\208=400\ -FALSE\\\\8;\ 4\sqrt{21}\\\\8^2+(4\sqrt{21})^2=20^2\\64+16\cdot21=400\\64+336=400\\400=400\ -O.K.\\\\12\ and\ 14\\\\12^2+14^2=20^2\\144+196=400\\340=400\ -FALSE\\\\10\ and\ 10\sqrt3\\\\10^2+(10\sqrt3)^2=20^2\\100+100\cdot3=400\100+300=400\\400=400\ -O.K.\\\\14\ and\ 4\sqrt{51};\ 4\sqrt{51} > 20 - FALSE[/tex]

[tex]\text{because the line segment labeled 20 is the longest side of the triangle}[/tex]

[tex]Answer:\ \boxed{8\ and\ 4\sqrt{21};\ 10\ and\ 10\sqrt3}[/tex]