In a parallelogram, consecutive angles are supplementary. So, ABC = 180^° - BAD = 180^° - 60^° = 120^°. We can use the Law of Cosines in ABC to find the length of side BC. The Law of Cosines states that c² = a² + b² - 2abcos(C).

In a parallelogram, consecutive angles are supplementary. So, ABC = 180^° - BAD = 180^° - 60^° = 120^°. We can use the Law of Cosines in ABC to find the length of side BC. The Law of Cosines states that c² = a² + b² - 2abcos(C).

Substitute the known values into the Law of Cosines equation. We know AC = (19)/(3), AB = 6, and cos(120^°) = -(1)/(2). This results in a quadratic equation for BC.

Multiply the quadratic equation by 9 to clear the fraction. Then, use the quadratic formula x = (-b ± √(b² - 4ac))/(2a) to solve for BC. Since BC must be a positive length, we take the positive root.

Multiply the quadratic equation by 9 to clear the fraction. Then, use the quadratic formula x = (-b ± √(b² - 4ac))/(2a) to solve for BC. Since BC must be a positive length, we take the positive root.

In parallelogram $ABCD$ , $\angle BAD=60^\circ$ , $AB=6$ , and diagonal $AC=6\tfrac{1}{3}$ . Find the area of $ABCD$ .

Answer: $(-9\sqrt{3} + \sqrt{354}) \text{ square units}$

Step-by-step solution

Step 1: Convert mixed fraction to improper fraction

First, we convert the mixed fraction for the length of diagonal $AC$ into an improper fraction. This makes calculations easier and more precise.

Step 2: Apply the Law of Cosines in triangle ABC

In a parallelogram, consecutive angles are supplementary. So, $\angle ABC = 180^\circ - \angle BAD = 180^\circ - 60^\circ = 120^\circ$ . We can use the Law of Cosines in $\triangle ABC$ to find the length of side $BC$ . The Law of Cosines states that $c^2 = a^2 + b^2 - 2ab\cos(C)$ .

Step 3: Substitute values and solve for BC

Substitute the known values into the Law of Cosines equation. We know $AC = \frac{19}{3}$ , $AB = 6$ , and $\cos(120^\circ) = -\frac{1}{2}$ . This results in a quadratic equation for $BC$ .

Step 4: Solve the quadratic equation for BC

Multiply the quadratic equation by 9 to clear the fraction. Then, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $BC$ . Since $BC$ must be a positive length, we take the positive root.

Step 5: Determine the valid length of BC

Since length cannot be negative, we choose the positive value for $BC$ . The square root of 118 is approximately 10.86, so $\frac{-9 + 10.86}{3} > 0$ .

Step 6: Calculate the area of the parallelogram

The area of a parallelogram is given by the formula $A = ab\sin(\theta)$ , where $a$ and $b$ are the lengths of two adjacent sides and $\theta$ is the angle between them. We use $AB=6$ , $BC=\frac{-9 + \sqrt{118}}{3}$ , and $\angle BAD = 60^\circ$ .

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In parallelogram ABCDABCDABCD, ∠BAD=60∘\angle BAD=60^\circ∠BAD=60∘, AB=6AB=6AB=6, and diagonal AC=613AC=6\tfrac{1}{3}AC=631​. Find the area of ABCDABCDABCD.