In the right rectangular solid ABCDEFGH, the base ABCD is a rectangle with edge lengths AB… — GRE General Test

In the right rectangular solid $ABCDEFGH$ , the base $ABCD$ is a rectangle with edge lengths $AB = 6$ and $BC = 6\sqrt{3}$ . The vertical edge $CG = 12$ . Point $M$ is the midpoint of edge $CG$ . What is the perimeter of triangle $BDM$ ?

$24 + 6\sqrt{2}$ Answer
B
$12 + 12\sqrt{2}$
C
$24 + 6\sqrt{3}$
D
$18 + 6\sqrt{2} + 6\sqrt{3}$
E
$12 + 6\sqrt{2} + 6\sqrt{5}$

Answer

The perimeter of triangle

BDM

24 + 6\sqrt{2}

The correct answer is obtained by recognizing three right triangles within the 3D figure:

\triangle BCD

has legs

6

and

6\sqrt{3}

giving

BD = 12

;

\triangle BCM

has legs

6\sqrt{3}

and

6

giving

BM = 12

; and

\triangle DCM

has legs

6

and

6

giving

DM = 6\sqrt{2}

. Summing the three sides yields a perimeter of

24 + 6\sqrt{2}

Step-by-Step Solution

Calculate the length of base diagonal

BD

using the right triangle

\triangle BCD

BD = 12

In right triangle

\triangle BCD

, legs are

CD = AB = 6

and

BC = 6\sqrt{3}

. Using the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BD = 2 \times 6 = 12

(or via Pythagorean theorem:

BD = \sqrt{6^2 + (6\sqrt{3})^2} = \sqrt{36 + 108} = \sqrt{144} = 12

Calculate the length of segment

BM

using the right triangle

\triangle BCM

BM = 12

Since

M

is the midpoint of vertical edge

CG = 12

CM = 6

. Vertical edge

CG

is perpendicular to base

ABCD

, so

\triangle BCM

is a right triangle at

C

. The legs are

CM = 6

and

BC = 6\sqrt{3}

. Applying the

30^\circ\text{--}60^\circ\text{--}90^\circ

ratio

(1 : \sqrt{3} : 2)

, hypotenuse

BM = 2 \times 6 = 12

Calculate the length of segment

DM

using the right triangle

\triangle DCM

DM = 6\sqrt{2}

In right triangle

\triangle DCM

, legs are

CD = 6

and

CM = 6

. Since the legs are equal,

\triangle DCM

is a

45^\circ\text{--}45^\circ\text{--}90^\circ

isosceles right triangle with side ratio

1 : 1 : \sqrt{2}

. Thus, hypotenuse

DM = 6\sqrt{2}

Sum the three side lengths to find the perimeter of

\triangle BDM

Perimeter =

12 + 12 + 6\sqrt{2} = 24 + 6\sqrt{2}

The perimeter of

\triangle BDM

BD + BM + DM

Key Concept

Applying special right triangle ratios (

30^\circ\text{--}60^\circ\text{--}90^\circ

and

45^\circ\text{--}45^\circ\text{--}90^\circ

) to 3D rectangular solids.

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