In the figure below, quadrilateral ABCD is divided by diagonal BD into two right triangles… — ACT

In the figure below, quadrilateral $ABCD$ is divided by diagonal $BD$ into two right triangles. In right triangle $ABD$ , the angle at $B$ is a right angle, $\angle ADB = 30^\circ$ , and the hypotenuse $AD = 12$ centimeters. In right triangle $BCD$ , the angle at $C$ is a right angle, and $BC = CD$ . What is the perimeter, in centimeters, of quadrilateral $ABCD$ ?

A
$18 + 6\sqrt{3}$
B
$18 + 12\sqrt{3}$
$18 + 6\sqrt{6}$ Answer
D
$18 + 12\sqrt{6}$
E
$12 + 6\sqrt{3} + 6\sqrt{2}$

Answer

The perimeter of the quadrilateral is

18 + 6\sqrt{6}

centimeters.

The perimeter of quadrilateral

ABCD

is the sum of the lengths of its four outer boundary sides:

AB

BC

CD

, and

DA

. In the

30^\circ-60^\circ-90^\circ

right triangle

ABD

, the shorter leg

AB

is half the hypotenuse

AD

, so

AB = 6\text{ cm}

. The longer leg

BD

AB\sqrt{3} = 6\sqrt{3}\text{ cm}

. In the isosceles right triangle

BCD

, the hypotenuse is

BD = 6\sqrt{3}\text{ cm}

. The legs

BC

and

CD

are congruent, with each length equal to the hypotenuse divided by

\sqrt{2}

, which simplifies to

3\sqrt{6}\text{ cm}

. Adding the four outer side lengths (

6 + 3\sqrt{6} + 3\sqrt{6} + 12

) yields a perimeter of

18 + 6\sqrt{6}\text{ cm}

Step-by-Step Solution

Find the lengths of the legs of right triangle

ABD

using the properties of a

30^\circ-60^\circ-90^\circ

right triangle.

AB = 6\text{ cm}

and

BD = 6\sqrt{3}\text{ cm}

In a

30^\circ-60^\circ-90^\circ

triangle, the leg opposite the

30^\circ

angle is half the length of the hypotenuse (

AB = \frac{12}{2} = 6

), and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the shorter leg (

BD = 6\sqrt{3}

Find the lengths of the legs of the isosceles right triangle

BCD

(

45^\circ-45^\circ-90^\circ

) using the hypotenuse

BD

BC = CD = 3\sqrt{6}\text{ cm}

In a

45^\circ-45^\circ-90^\circ

triangle, the length of each leg is the hypotenuse divided by

\sqrt{2}

. Thus,

BC = CD = \frac{6\sqrt{3}}{\sqrt{2}} = \frac{6\sqrt{3}\cdot\sqrt{2}}{2} = 3\sqrt{6}

Calculate the perimeter of quadrilateral

ABCD

by summing the lengths of its four outer boundary sides:

AB

BC

CD

, and

DA

Perimeter

= 6 + 3\sqrt{6} + 3\sqrt{6} + 12 = 18 + 6\sqrt{6}\text{ cm}

The perimeter is the sum of the outer boundary sides of the quadrilateral, which are

AB

BC

CD

, and

DA

Key Concept

Pythagorean Theorem and Special Right Triangles

Alternative Method

Instead of using special right triangle formulas, the Pythagorean theorem can be used with variables:

AB^2 + BD^2 = AD^2

, where

AB = \frac{1}{2}AD = 6

, so

36 + BD^2 = 144 \implies BD = \sqrt{108} = 6\sqrt{3}

. Then

BC^2 + CD^2 = BD^2 \implies 2BC^2 = 108 \implies BC = \sqrt{54} = 3\sqrt{6}

Estimated Time:1m 30s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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