GMAT Math : Tetrahedrons

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Example Question #9 : Calculating The Surface Area Of A Tetrahedron

Tetra_1

The cube in the above figure has surface area 384. Give the surface area of the tetrahedron with vertices D,E,G,H , shown in red.

Possible Answers:

$96 + 64\sqrt{3}$

$128 \sqrt{3}$

128

$96 + 32 \sqrt{3}$

$128+ 32 \sqrt{3}$

Correct answer:

$96 + 32 \sqrt{3}$

Explanation:

The surface area formula can be used to find the length of each edge of the cube:

$6s^{2}=384$

$s^{2}= 64$

s= 8

Three faces of the tetrahedron - $\bigtriangleup DHE$ , $\bigtriangleup DHG$ , $\bigtriangleup EHG$ - are right triangles with legs of length 8, so the area of each is half the product of the lengths of their legs:

$\frac{1}{2} \cdot 8 \cdot 8 =32$ .

Each triangle is isosceles, so, by the 45-45-90 Theorem, each of their hypotentuses measures $\sqrt{2}$ times a leg, or $8\sqrt{2}$ . is therefore an equilateral triangle with sidelenghth $8\sqrt{2}$ . Its area can be found as follows:

$A = \frac{s^{2}\sqrt{3}}{4}$

$= \frac{(8\sqrt{2})^{2}\sqrt{3}}{4}$

$= \frac{8^{2}\cdot (\sqrt{2})^{2} \cdot \sqrt{3}}{4}$

$= \frac{64\cdot 2 \cdot \sqrt{3}}{4}$

$= 32 \sqrt{3}$

The total surface area is

$32+32+32+ 32 \sqrt{3} = 96 + 32 \sqrt{3}$

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GMAT Math : Tetrahedrons

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Example Question #9 : Calculating The Surface Area Of A Tetrahedron

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