+8581 Mai is putting gold foil around the outside of an elliptical picture frame. The permieter of an ellipse is given by the formula p=3.14/2 Squareroot(2(h^2+w^2), where h is the height and w is the width, as shown in the diagram below. If an elliptical frame has an outside height equal to 4 inches nd an outside width equal to 3 inches. What is its outside perimeter, in inches?
So, What im getting is plug in the number for the variables?
Any help is amazing! Thankss!!
+26393 Mai is putting gold foil around the outside of an elliptical picture frame. The permieter of an ellipse is given by the formula p=3.14/2 Squareroot(2(h^2+w^2), where h is the height and w is the width, as shown in the diagram below. If an elliptical frame has an outside height equal to 4 inches nd an outside width equal to 3 inches. What is its outside perimeter, in inches?
So, What im getting is plug in the number for the variables?
\(\begin{array}{rcl} p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ w^2+h^2 ~\right) } \qquad w = 3\ \text{inch} \qquad h = 4\ \text{inch} \\ p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ 3^2+4^2 ~\right) } \\ p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ 5^2 ~\right) } \qquad 3^2+4^2 = 5^2 \\ p &=& \frac{\pi}{2} \cdot 5\cdot \sqrt{ 2 } \\ p &=& \pi \cdot 2.5 \cdot \sqrt{ 2 } \\ \mathbf{p }& \mathbf{=} & \mathbf{11.1072073454}\ \text{inch} \end{array} \)
+8581 Please DONT comment if you don't know it! Honestly, I could care less on what YOU say to me.
Pretty sure the perimeter of an ellipse is given as approximately 2pi * sqrt((h^2+w^2)/2)
This would give you
2*3.14159*sqrt((4^2+3^2)/2) = 22.214 inches
+26393 Mai is putting gold foil around the outside of an elliptical picture frame. The permieter of an ellipse is given by the formula p=3.14/2 Squareroot(2(h^2+w^2), where h is the height and w is the width, as shown in the diagram below. If an elliptical frame has an outside height equal to 4 inches nd an outside width equal to 3 inches. What is its outside perimeter, in inches?
So, What im getting is plug in the number for the variables?
\(\begin{array}{rcl} p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ w^2+h^2 ~\right) } \qquad w = 3\ \text{inch} \qquad h = 4\ \text{inch} \\ p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ 3^2+4^2 ~\right) } \\ p &=& \frac{\pi}{2} \cdot \sqrt{ 2 \cdot \left(~ 5^2 ~\right) } \qquad 3^2+4^2 = 5^2 \\ p &=& \frac{\pi}{2} \cdot 5\cdot \sqrt{ 2 } \\ p &=& \pi \cdot 2.5 \cdot \sqrt{ 2 } \\ \mathbf{p }& \mathbf{=} & \mathbf{11.1072073454}\ \text{inch} \end{array} \)
From the information you've been given, yes heureka you would be correct however that information is flawed as that is not the correct formula for calculating the perimeter of an ellipse. Not sure who has written that question but they've made an error.
The correct formula is
\(p = 2\pi * \sqrt{0.5*(h^2+w^2)}\)
Couldn't find the symbol but it should be noted that this is approximately equal to, not equal to exactly.
+8581 I jusr copied it down from my worksheet. I did not think that the equation looked right, But was afraid to question it..
If that's the equation you've been given then i'd use Heureka's answer but it may be worth questioning.
+8581 Okay, Thanks everyone! I'll Question it. Thanks everyone for your time, and your help. I greatly appreciated it!
This is why I like to be on here. I get help from wonderful people. ( except from that frog person...)
+26393 \(\begin{array}{lcl} p\approx 2\pi\sqrt{\frac12 (a^2+b^2)} \qquad a \text{ and } b \text{ are the half-axes !!!} \\ \text{ so } a = \frac{w}{2} \text{ and } b = \frac{h}{2} \\ \text{ we have } p\equiv 2\pi\sqrt{\frac12 [ (\frac{w}{2})^2 + (\frac{h}{2})^2 ] } \end{array} \\ \begin{array}{lcl} p &\approx& 2\pi\sqrt{\frac12 [ (\frac{w^2}{4}) + (\frac{h^2}{4}) ] }\\ p &\approx& 2\pi\sqrt{\frac12 (\frac{w^2+h^2}{4}) }\\ p &\approx& 2\pi\sqrt{\frac24 (\frac{w^2+h^2}{4}) }\\ p &\approx& 2\pi \cdot \frac{1}{4} \sqrt{2\cdot (w^2+h^2 ) }\\ p &\approx& \frac{\pi}{2}\cdot \sqrt{2\cdot (w^2+h^2 ) }\\ \end{array} \)
