Given:
AB=BC=5 inches, and BD=2.5 inches.
BD is the perbenticular bisector of AC.
Required:
We need to find the diameter of the given plate.
Explanation:
Let O be the center of the circle.
The radius of the circle, AO=r.
DO also the radius of the circle, DO=r.
[tex]DO=BD+BO[/tex]
Substitute DO=r and BD=2.5 inches in the equation.
[tex]r=2.5+BO[/tex][tex]BO=r-2.5[/tex]
Consider the right angle triangle ABO.
Use the Pythagorean theorem.
[tex]AO^2=AB^2+BO^2[/tex]
Substitute AO=r, AB=5, and BO=r-2.5 in the equation.
[tex]r^2=5^2+(r-2.5)^2[/tex][tex]Use\text{ }(a-b)^2=a^2-2ab+b^2.\text{ Here a=r and b =2.5.}[/tex]
[tex]r^2=5^2+r^2-2\times2.5r+(2.5)^2[/tex]
[tex]r^2=25+r^2-5r+6.25[/tex]
[tex]r^2=r^2-5r+31.25[/tex]
Solve for r.
[tex]5r=31.25[/tex]
Divide both sides by 5.
[tex]\frac{5r}{5}=\frac{31.25}{5}[/tex][tex]r=6.25[/tex]
We ge radius, r =6.25.
[tex]\text{We know that the diameter =2}\times radius.[/tex]
Substitute radius = 6.25 in the equation.
[tex]Diameter=2\times6.25[/tex]
[tex]Diameter=12.5\text{ inches.}[/tex]
Final answer:
[tex]Diameter=12.5\text{ inches.}[/tex]
