Answer:
The radius of the conical cup is 1.95 in.
Step-by-step explanation:
Given:
Height of the conical cup is [tex]h=3[/tex] inches.
Area or curved surface area of the cup is [tex]A=6\rm{in^{2}}[/tex]
Let the radius of the cup is [tex]r[/tex].
Now, the lateral height [tex]l[/tex] of the cone will be,
[tex]l=\sqrt{r^{2}+h^{2}}\\l=\sqrt{r^{2}+3^{2}}[/tex]
The curved surface area of the cone is,
[tex]A=\pi rl\\6=\pi r \sqrt{r^{2}+9}\\r \sqrt{r^{2}+9}=1.91[/tex]
Squaring the both sides,
[tex]r^{2} (r^{2}+9)=3.6[/tex]
Now, let [tex]r^{2}=x[/tex] and solve the resulting quadratic equation.
[tex]x (x+9)=3.6\\x^{2}+9x-3.6=0[/tex]
Solving the above quadratic equation as,
[tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\x=\frac{-9\pm \sqrt{9^{2}-4\times1\times(-3.6)}}{2\times9}\\x=0.38, -9.38[/tex]
Now, the value of [tex]x[/tex] cannot be negative as it is the square of radius.
So, the value of radius of the cone will be,
[tex]r=\sqrt{x}\\r=\sqrt{0.38}\\r=1.95\rm{in}[/tex]
Therefore, the radius of the conical cup is 1.95 in.
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https://brainly.com/question/12267785?referrer=searchResults
