Question

Can any one help me with this math problem. I need to find the surface area of this hexagonal pyramid. I got
$48\sqrt{2} + 24 \sqrt{3}$

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Answer 1

Question

What is the surface area of the regular hexagon pyramid?

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Solution

Consider the planar figure in the attachment. It consists of six isosceles triangles and one regular hexagon.

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The base of a triangle is 4 feet. By dropping a perpendicular line, we can find the height $h$.

$\implies h^2+2^2=6^2$

$\implies h^2=32$

$\implies h=4\sqrt{2}$

The area of one isosceles triangle is,

$\dfrac{1}{2} \times 4\times 4\sqrt{2} =\boxed{8\sqrt{2} \ \mathrm{ft^2}}$

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Now, it's time to find the area of the hexagon. A regular hexagon consists of six isosceles triangles. Here the height and base is already given, the area is

$6\times \dfrac{1}{2}\times 4\times 2\sqrt{3} =\boxed{24\sqrt{3} \ \mathrm{ft^2}}$

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Now let's find the surface area.

$\mathrm{Surface\ Area=1\ Hexagon+6\ Isosceles\ Triangles}$

$\implies \mathrm{Surface\ Area}=24\sqrt{3} +6\times 8\sqrt{2}$

$\implies \mathrm{Surface\ Area}=\boxed{48\sqrt{2} +24\sqrt{3}\ \mathrm{ft^2}}$

So, the answer is correct.

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This is the required answer.

Answer 2

Solution :-

We know that

By Pythagoras theorem

$\bf (Hypotenuse)^2 = (Base)^2+ (Perpendicular)^2$

$\sf 6^2 = 2^2+h^2$

$\sf 36 = 4 + h^2$

$\sf 36-4=h^2$

$\sf 32 = h^2$

$\sf\sqrt{32}=h$

$\sf 4\sqrt{2} = h$

Now

$\bf Area_{(Isosceles \; \triangle)} = \dfrac{1}{2} \times Base \times Height$

$\sf Area = \dfrac{1}{2} \times 4\sqrt{2} \times4$

$\sf Area = 4\sqrt{2}\times 2$

$\sf Area = 8\sqrt{2}$

Finding Area of 6 such triangle

$\sf Area = 8\sqrt{2}\times6$

$\sf Area = 48\sqrt{2}$

Finding area of hexagon

$\sf Area_{hexagon} = 6 \times \dfrac{1}{2} \times Base \times Side$

$\sf Area_{Hexagon} = 3 \times 4\times 2\sqrt{3}$

$\sf Area_{Hexagon} = 12\times 2\sqrt{3}$

$\sf Area_{Hexagon} = 24\sqrt{3}$

$\bf Surface \; Area = Area_{Triangle} + Area_{Hexagon}$

$\sf Surface \; Area = 48\sqrt{2} + 24\sqrt{3}$