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Lina20 [59]
1 year ago
8

Andrea has 3 tiles. One is a regular octagon, one is a regular pentagon and one is a regular hexagon. Andrea thinks the 3 tiles

will fit together perfectly, as shown in the diagram. Show calculations to prove that she is wrong.

Mathematics
1 answer:
HACTEHA [7]1 year ago
4 0

From calculations, we can say that the given tiles will not fit together perfectly.

<h3>How to find the sum of interior angles of a Polygon?</h3>

If the tiles join perfectly at a point, sum of all angles around the joining point should be 360°.

Expression for the measure of the interior angle of a polygon,

Interior angle of a polygon = [(n - 2) * 180]/n

Interior angle of a pentagon = [(5 - 2) * 180]/5 = 108°

Interior angle of a hexagon = [(6 - 2) * 180]/6 = 120°

Interior angle of an octagon = [(8 - 2) * 180]/8 = 135°

To prove that the given tiles fit together perfectly → Sum of all the angles around the common point should be 360°

Sum of all interior angles = 108° + 120° + 135° = 363°

Therefore, given tiles will not fit together perfectly.

Read more about Interior angles of a Polygon at; brainly.com/question/224658

#SPJ1

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Answer:

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Step-by-step explanation:

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Given:

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The total surface area is 96 cm cube.

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\begin{gathered} r=\frac{d}{2} \\ r=\frac{2x}{2} \\ r=x\text{ cm} \end{gathered}

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\begin{gathered} S=2\pi rh+2\pi(r)^2 \\ 96=2\pi xh+2\pi(x^2) \\ h=\frac{96-2\pi(x^2)}{2\pi x} \end{gathered}

Volume is,

\begin{gathered} V=\pi(r)^2h \\ =\pi(x^2)\frac{96-2\pi(x^2)}{2\pi x} \\ =\frac{x(96-2\pi(x^2)}{2} \end{gathered}

Now, differentiate with respect to x,

\begin{gathered} \frac{dV}{dx}^{}=\frac{d}{dx}(\frac{x(96-2\pi(x^2)}{2}) \\ =\frac{d}{dx}\mleft(x\mleft(-\pi x^2+48\mright)\mright) \\ =\frac{d}{dx}\mleft(x\mright)\mleft(-\pi x^2+48\mright)+\frac{d}{dx}\mleft(-\pi x^2+48\mright)x \\ =1\cdot\mleft(-\pi x^2+48\mright)+\mleft(-2\pi x\mright)x \\ =84-3\pi(x^2)\ldots\ldots\ldots\ldots\text{.}(1) \end{gathered}

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\begin{gathered} \frac{dV}{dx}=0 \\ 84-3\pi(x^2)=0 \\ x^2=\frac{16}{\pi} \\ x=\sqrt[]{\frac{16}{\pi}} \end{gathered}

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\begin{gathered} \frac{d^2V}{dx^2}=\frac{d}{dx}(84-3\pi(x^2)) \\ =-6\pi x \\ At\text{ x=}\sqrt[]{\frac{16}{\pi}} \\ \frac{d^2V}{dx^2}=-6\pi\sqrt[]{\frac{16}{\pi}}

Since, the double derivative is negative.

So,\text{ the volume is maximum at }\sqrt[]{\frac{16}{\pi}}

So, the volume becomes,

\begin{gathered} V=\pi(x^2)h \\ V=\pi(\sqrt[]{\frac{16}{\pi}})^2h \\ V=\frac{16h}{\pi} \end{gathered}

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Note:
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