What is the length of the apothem of the regular pentagon shown below? Round to one decimal place

Mathematics

1 answer:

DerKrebs [107]2 years ago

7 0

Answer:

Step-by-step explanation:

Draw a diagonal line connecting the centre to a vertex at the bottom. This makes a triangle. It is a regular pentagon so the base of that triangle is 7.6 / 2, which is 3.8. You also know that the angle from the apothem to the base is 90 degrees. You also know that you could split the centre angle into 5 triangles. 360 / 5 = 72 and we have half of it so it is 36 degrees.

Let's see what we have so far:

The base of the triangle is 3.8

The angle at the bottom is 90

The angle at the top is 36

We now have the tan and the opposite and we need the adjacent so rearrange for adjacent.

This gives you adjacent = opposite / tan(angle)

adjacent = 3.8 / tan(36)

adjacent = 5.2m

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correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

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$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$