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Anarel [89]
2 years ago
7

Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base.

The areas of the cross sections are $216\sqrt{3}$ square feet and $486\sqrt{3}$ square feet. The two planes are $8$ feet apart. How far from the apex of the pyramid is the larger cross section?

Mathematics
1 answer:
Tanya [424]2 years ago
8 0

Answer:

The larger cross section is 24 meters away from the apex.

Step-by-step explanation:

The cross section of a right hexagonal pyramid is a hexagon; therefore, let us first get some things clear about a hexagon.

The length of the side of the hexagon is equal to the radius of the circle that inscribes it.

The area is

A=\frac{3\sqrt{3} }{2} r^2

Where r is the radius of the inscribing circle (or the length of side of the hexagon).

Now we are given the areas of the two cross sections of the right hexagonal pyramid:A_1=216\:ft^2\: \:\:\:A_2=486\:ft^2

From these areas we find the radius of the hexagons:

r_1=\sqrt{\frac{2A_1}{3\sqrt{3} } } =\sqrt{\frac{2*216}{3\sqrt{3} } }=\boxed{9.12ft}

r_2=\sqrt{\frac{2A_2}{3\sqrt{3} } } =\sqrt{\frac{2*486}{3\sqrt{3} } }=\boxed{13.68ft}

Now when we look at the right hexagonal pyramid from the sides ( as shown in the figure attached ), we see that r_1 r_2 form similar triangles with length H

Therefore we have:

\frac{H-8}{r_1} =\frac{H}{r_2}

We put in the numerical values of r_1, r_2 and solve for H:

\boxed{H=\frac{8r_2}{r_2-r_1} =\frac{8*13.677}{13.68-9.12} =24\:feet.}

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