In quadrilateral ABCD, diagonal AC divides the quadrilateral into two right triangles, \tr… — GRE General Test

In quadrilateral $ABCD$ , diagonal $AC$ divides the quadrilateral into two right triangles, $\triangle ABC$ and $\triangle ADC$ . The measure of angle $ABC$ is $90^\circ$ , the measure of angle $ADC$ is $90^\circ$ , and the measure of angle $BAC$ is $30^\circ$ . If $AB = 12$ and $AD = CD$ , what is the area of quadrilateral $ABCD$ ?

$48 + 24\sqrt{3}$ Answer
B
$36 + 18\sqrt{3}$
C
$192 + 24\sqrt{3}$
D
$96 + 48\sqrt{3}$
E
$24 + 48\sqrt{3}$

Answer

48 + 24\sqrt{3}

The total area of quadrilateral ABCD is the sum of the areas of right triangles ABC and ADC. Using 30-60-90 triangle properties on ABC gives

BC = 4\sqrt{3}

and hypotenuse

AC = 8\sqrt{3}

, resulting in

\text{Area}(\triangle ABC) = 24\sqrt{3}

. Triangle ADC is a 45-45-90 triangle with hypotenuse

8\sqrt{3}

, yielding leg lengths

4\sqrt{6}

and

\text{Area}(\triangle ADC) = 48

. Adding both areas gives

48 + 24\sqrt{3}

Step-by-Step Solution

Analyze right triangle ABC to find side lengths BC and hypotenuse AC.

BC = 4\sqrt{3}

and

AC = 8\sqrt{3}

In 30-60-90 triangle ABC,

\cos(30^\circ) = \frac{AB}{AC} \implies \frac{\sqrt{3}}{2} = \frac{12}{AC} \implies AC = 8\sqrt{3}

. Also,

\tan(30^\circ) = \frac{BC}{AB} \implies \frac{1}{\sqrt{3}} = \frac{BC}{12} \implies BC = 4\sqrt{3}

Calculate the area of right triangle ABC.

\text{Area}(\triangle ABC) = 24\sqrt{3}

\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 12 \times 4\sqrt{3} = 24\sqrt{3}

Analyze isosceles right triangle ADC to find leg length

s = AD = CD

s = 4\sqrt{6}

In 45-45-90 triangle ADC, the hypotenuse

AC = s\sqrt{2} = 8\sqrt{3} \implies s = \frac{8\sqrt{3}}{\sqrt{2}} = 4\sqrt{6}

Calculate the area of right triangle ADC.

\text{Area}(\triangle ADC) = 48

\text{Area}(\triangle ADC) = \frac{1}{2} \times s^2 = \frac{1}{2} \times (4\sqrt{6})^2 = \frac{1}{2} \times 96 = 48

Sum the areas of the two triangles to get total area of quadrilateral ABCD.

\text{Total Area} = 48 + 24\sqrt{3}

\text{Area}(ABCD) = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 24\sqrt{3} + 48 = 48 + 24\sqrt{3}

Key Concept

Special Right Triangle Ratios (30-60-90 and 45-45-90)

Alternative Method

Use trigonometric ratios directly:

\text{Area}(\triangle ABC) = \frac{1}{2} (12)^2 \tan(30^\circ) = 72 \times \frac{1}{\sqrt{3}} = 24\sqrt{3}

, and

\text{Area}(\triangle ADC) = \frac{1}{4} AC^2 = \frac{1}{4} (8\sqrt{3})^2 = 48

Estimated Time:2m 30s

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