[ The Law of Cosines [Isosceles, Inscribed, and Circumscribed Trapezoids]

The bases of the trapezoid are 3 cm and 5 cm. One of the diagonals of the trapezoid is 8 cm, and the angle between the diagonals is $60^{\circ}$. Find the perimeter of the trapezoid.

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Through the vertex of the smaller base of the trapezoid, draw a line parallel to one of the diagonals.

## Solution

Let $A B C D$ be the given trapezoid, with $B C=3$ and $A D=5$ as its bases, and $A C=8$ as the given diagonal. Let $O$ be the point of intersection of the diagonals. Draw a line through vertex $C$ parallel to diagonal $B D$. Let this line intersect the extension of base $A D$ at point $E$. Then $B C E D$ is a parallelogram,

$$
D E=B C=3, A E=A D+D E=A D+B C=3+5=8=A C
$$

Therefore, triangle $C A E$ is isosceles. Hence,

$$
\angle A E C=\angle A C E=\angle A O D
$$

Note that angle $A O D$ cannot be obtuse, as it is equal to angle $A C E$ at the base of the isosceles triangle. Therefore, $\angle A O D=60^{\circ}$. Consequently, triangle $A C E$ is equilateral, so $C E=8$. Since the diagonals $A C$ and $B D$ of trapezoid $A B C D$ are equal, the trapezoid is isosceles. Using the cosine rule in triangle $C E D$, we find that

$$
C D^{2}=C E^{2}+D E^{2}-2 \cdot C E \cdot D E \cos 60^{\circ}=64+9-2 \cdot 8 \cdot 3 \cdot \frac{1}{2}=47
$$

Therefore, $C D=7$. Thus, the perimeter of the trapezoid is $3+5+7+7=22$.

## Answer

22. 

## Hint

## Solution

Answer