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Solve for d small diameter [tex]A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4}[/tex]
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Solve for d small diameter
[tex]A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4}[/tex]
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Mathematics
1 year
2021-09-04T14:50:34+00:00
2021-09-04T14:50:34+00:00 2 Answers
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Answers ( )
Answer:
[tex] \boxed{ \tt{d = \sqrt{D^{2} -\frac{4A}{\pi}} }}[/tex]
Step-by-step explanation:
[tex]if \: A=\frac{\pi *D^{2} }{4} – \frac{\pi *d^{2} }{4} \\ then \to \\ A = \frac{\pi}{4} (D^{2} – {d}^{2} ) \\ (D^{2} – {d}^{2} ) \pi = 4A \\ D^{2} – {d}^{2} = \frac{4A}{\pi} \\ {d}^{2} = D^{2} -\frac{4A}{\pi} \\ d = \sqrt{D^{2} -\frac{4A}{\pi}} [/tex]
Answer:
[tex]\boxed{\text{\Large \sqrt{-1.27324A+D^2}$}}[/tex]
Step-by-step explanation:
[tex]\displaystyle A=\frac{\pi \times D^2}{4} – \frac{\pi \times d^2}{4}[/tex]
[tex]\displaystyle A=0.785398D^2 – 0.785398d^2[/tex]
Solve for d
[tex]\displaystyle A-0.785398D^2= – 0.785398d^2[/tex]
[tex]\displaystyle \frac{A-0.785398D^2}{- 0.785398} = d^2[/tex]
[tex]-1.27324A+D^2=d^2[/tex]
[tex]\displaystyle \sqrt{-1.27324A+D^2} =d[/tex]