Prove that the larger base of the trapezoid is equal to the lateral sides.

## Solution

Let $BC$ and $AD$ be the bases of trapezoid $ABCD$, with $BC=3$, and $CA$ being the bisector of angle $BCD$. Since $\angle CAD = \angle BCA = \angle DCA$, triangle $ACD$ is isosceles. Therefore, $AD = CD = AB = (42 - 3) : 3 = 13$. Drop a perpendicular $CK$ to the base $AD$. Then $DK = 1/2(AD - BC) = 5$, $CK^2 = CD^2 - KD^2 = 12^2$. Therefore, $S_{ABCD} = 1/2(AD + BC) \cdot CK = 96$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-46.jpg?height=391&width=372&top_left_y=-1&top_left_x=846)

## Answer

96.

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$[\quad$ Area of a Trapezoid
Zad [Isosceles, Inscribed, and Circumscribed Trapezoids]

The diagonal of an isosceles trapezoid bisects its obtuse angle. The smaller base of the trapezoid is 3, and the perimeter is 42.

Find the area of the trapezoid.

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