Semi: Prefix For Half

Geometry Level 2

True or False?

\quad If the perimeter of a circle is π \pi , then the perimeter of a semicircle with the same radius is π 2 \dfrac\pi2 .

True False

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

Relevant wiki: Circles - Radius and Diameter

The semi perimeter will be π 2 + 1 \dfrac π 2 + 1 (you need to add the diameter also)

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Ashish Menon
Aug 3, 2016

2 π r = π r = 1 2 \begin{aligned} 2\pi r & = \pi\\ r & = \dfrac{1}{2} \end{aligned} .

Now, perimeter of semicircle = curved path + diameter = π r + 2 r = π × 1 2 + 2 × 1 2 = π 2 + 1 = \pi r + 2r\\ = \pi×\dfrac{1}{2} + 2×\dfrac{1}{2}\\ = \dfrac{\pi}{2} + \color{#20A900}{1}

So answer is False \color{#3D99F6}{\boxed{\text{False}}} .

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Wonderful! Here's a similar question:

True or False?

\quad If the surface area of a sphere is 4 π 4\pi , then the surface area of a hemisphere of the same radius is 2 π 2\pi .

Pi Han Goh - 4 years, 10 months ago

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False, the answer is 3 π 3\pi .

4 π r 2 = 4 π r = 1 \begin{aligned} 4\pi r^2 & = 4\pi\\ r & = 1 \end{aligned}

Then, surface area of hemisphere = curved surface area of hemisphere + area of cirle formed = 2 π r 2 + π r 2 = 3 π r 2 = 3 π × 1 = 3 π 2\pi r^2 + \pi r^2 = 3\pi r^2 = 3\pi×1 = \color{#3D99F6}{\boxed{3\pi}} .

Ashish Menon - 4 years, 10 months ago

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Wonderful! Now here's a tough one:

If a right circular cylinder is inscribed inside a sphere such that the volume of this cylinder is maximized, prove that the ratio between the radius and the height of this cylinder is r : h = 1 : 2 r : h = 1:\sqrt2 .

Pi Han Goh - 4 years, 10 months ago

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@Pi Han Goh – Hmm. Thats pretty simple.
Get two equations:- Let r and h be the radius and height of the cylinder inscribed. Let R be the radius of the sphere.

First equation:-
Surface area of cylinder(A) = 2 π r 2 + 2 π r h ⟶ 1 2\pi{r}^2 + 2\pi rh \longrightarrow \boxed{1}

Second equation:-
Now, we can imagine that for maximum volume of cylinder, the center of cylinder and center of sphere should coincide. So, there exists a triangle with hypotenuse R, height = h 2 \dfrac{h}{2} and length = r.
Using Pythagoras' theorem:-
r 2 + ( h 4 ) 2 = R 2 r = R 2 − ( h 4 ) 2 ⟶ 2 \begin{aligned} r^2 + {\left(\dfrac{h}{4}\right)}^2 & = R^2\\ \\ r & = \sqrt{R^2 - {\left(\dfrac{h}{4}\right)}^2} \longrightarrow \boxed{2} \end{aligned}

Substitute 2 \boxed{2} in 1 \boxed{1} .
Then get an equation and differentiate it with respect to h. Then put it equal to 0 for maxima. You will get another equation then simple solving of polynomials gives the answer.

Ashish Menon - 4 years, 10 months ago

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@Ashish Menon – That sounds right. You still need to prove that it's a maximum value too.

Now here's another restriction: Solve this same question, but this time, don't use calculus at all.

Pi Han Goh - 4 years, 10 months ago

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@Pi Han Goh – Im sorry, I am not able to find such a way, i think trigonometry would be of some use, but nope I am not getting it.

Ashish Menon - 4 years, 10 months ago

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@Ashish Menon – Hint: AMGM inequality

Pi Han Goh - 4 years, 10 months ago

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@Pi Han Goh – Wow, now that makes sense.

First of all, start of with the same equation used in the above thread.
R 2 = r 2 + ( h 2 ) 2 h = 4 R 2 − 4 r 2 \begin{aligned} R^2 & = \sqrt{r^2 + {\left(\dfrac{h}{2}\right)}^2}\\ \\ h & = \sqrt{4R^2 - 4r^2} \end{aligned}

V = π r 2 h V = π × r 2 × 4 R 2 − 4 r 2 V = π 2 × 2 r 2 × 2 r 2 × ( 4 R 2 − 4 r 2 ) \begin{aligned} V & = \pi r^2 h\\ \\ V & = \pi × r^2 × \sqrt{4R^2 - 4r^2}\\ \\ V & = \dfrac{\pi}{2} × \sqrt{2r^2 × 2r^2 × \left(4R^2 - 4r^2\right)} \end{aligned}

Use the AM-GM inequality,

V ≤ π 2 ( 2 r 2 + 2 r 2 + 4 R 2 − 4 r 2 3 ) 3 V ≤ π 2 ( 4 R 2 3 ) 3 ∴ V m a x = 4 π R 3 3 3 ⟹ h = 2 R 3 ⟹ r = 2 R 3 ∴ r : h = 1 : 2 \begin{aligned} V & \leq \dfrac{\pi}{2}\sqrt{{\left(\dfrac{2r^2 + 2r^2 + 4R^2 - 4r^2}{3}\right)}^3}\\ \\ V & \leq \dfrac{\pi}{2}\sqrt{{\left(\dfrac{4R^2}{3}\right)}^3}\\ \\ \therefore V_{max} & = \dfrac{4\pi R^3}{3\sqrt{3}}\\ \\ \implies h & = \dfrac{2R}{\sqrt{3}}\\ \\ \implies r & = \dfrac{\sqrt{2}R}{\sqrt{3}}\\ \\ \therefore r : h & = \color{#3D99F6}{\boxed{1 : \sqrt{2}}} \end{aligned}

Ashish Menon - 4 years, 10 months ago

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@Ashish Menon – GREATTTTTTTTTTTTTTTT

Pi Han Goh - 4 years, 10 months ago

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@Pi Han Goh – Thanks for the hint!

Ashish Menon - 4 years, 10 months ago

The problem is not well posed. You must mention "disc" instead of "circle. By disque I mean "disque" in french. Maybe in english it doesn t have the same signification. By the way, that s a clever problem, thanks.

Al Far - 4 years, 10 months ago

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Hmm, I don't think there is a distinction. Thanks for your input! =D =D

Pi Han Goh - 4 years, 10 months ago

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So sorry, you were right ! I have to reconsider the perimeter of a demi-circle differently than the lenght of a demi-circle. Best regards

Al Far - 4 years, 10 months ago

It's really not that hard. Half the perimeter of the circle would only be the curved part. You still have the straight line portion. Thus it is false.

Stephanie Stephanie - 4 years, 10 months ago

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Yup, yup! =D =D

Pi Han Goh - 4 years, 10 months ago

oh, come on, it is obviously a trick question cause, duh, come on

Cezar Prodan - 4 years, 10 months ago

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