Answer:
No, tyler is wrong, his circle´s area has four times the Lin’s circle area.
Step-by-step explanation:
Hello
Let's remember this about a circle
the area is given by:
[tex]A=\pi r^{2}[/tex]
where r is the radius
and the diameter
[tex]D=2r\\isolating \ x\\r=\frac{D}{2}\\[/tex]
Step 1
according to the question Tyler’s circle has twice the diameter of Lin’s circle,in other terms
Let
[tex]Tylers \ diameter\ (D_{1}) =2\ times\ Lins\ circle(D_{2})\\\\[/tex]
[tex]D_{1}=2D_{2}[/tex]
Step 2
find the areas
Area of Tyler’s circle
[tex]A_{1}=\pi *r_{1} ^{2}[/tex]
replacing
let r_{1}= Tyler’s circle radius
[tex]r_{1}=\frac{D_{1} }{2} \\A_{1}=\pi *r_{1} ^{2}\\A_{1}=\pi *(\frac{D_{1} }{2})^{2}\\A_{1}=\pi *\frac{D_{1}^{2} }{4}[/tex]
[tex]let\ r_{2}= Lin\ circle\ radiusr_{2}=\frac{D_{2} }{2} \\A_{2}=\pi *r_{2} ^{2}\\A_{2}=\pi *(\frac{D_{2} }{2} )^{2}\\A_{2}=\pi *\frac{D_{2}^{2} }{4}\\\\[/tex]
Now, from
[tex]D_{1}=2D_{2}\\so\\\D_{2}=\frac{D_{1} }{2} \\[/tex]
replacing
[tex]A_{2}=\pi *\frac{D_{2}^{2} }{4}\\A_{2}=\pi *\frac{(\frac{D_{1} }{2}) ^{2} }{4}\\\\\\[/tex]
step 3
compare the areas
[tex]A_{1}=\pi *\frac{D_{1}^{2} }{4}\ and\ A_{2}= \pi *\frac{D_{1}^{2}}{16}\\\frac{A_{1}}{A_{2}} =\frac{\pi *\frac{D_{1}^{2} }{4}}{ \pi *\frac{D_{1}^{2}}{16}}\\\frac{A_{1}}{A_{2}} =4\\hence\\A_{1}=4 A_{2}[/tex]
this means that tyler is wrong, his circle´s area has four times the Lin’s circle area.
Have a great day
