The hypotenuse $AB$ of the right triangle $ABC$ is a chord of a circle with radius 10. The vertex $C$ lies on the diameter of the circle, which is parallel to the hypotenuse, $\angle A=75^{\circ}$. Find the area of triangle $ABC$.
From the center $O$ of the given circle, drop a perpendicular $OM$ to the hypotenuse $AB$. Then $MC = MA = MB$.

Therefore, $\angle MCB = \angle B = 15^{\circ}$, $\angle BCO = \angle B = 15^{\circ}$. Consequently, $\angle MCO = 30^{\circ}$.

Let $OM = x$. From the right triangle $MCO$, we find that $MC = 2x$. By the Pythagorean theorem, $OB^2 = OM^2 + MB^2$, or $100 = x^2 + 4x^2$, from which

$x^2 = 20$. Therefore, $S_{ABC} = \frac{1}{2} AB \cdot OM = 2x^2 = 40$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-01.jpg?height=301&width=509&top_left_y=890&top_left_x=777)

## Answer

40.

## [ Projections of the bases, sides, or vertices of a trapezoid Problem 54302 Topics: Area of a trapezoid [ $\underline{\text{ T }}$ trigonometric ratios in a right triangle ]

Difficulty: $3+$

Grades: 8,9

Given the larger base $a$ of an isosceles trapezoid, its height $h$, and the angle $\alpha$ at the base, find the area of the trapezoid.

## Hint

The projection of the diagonal of an isosceles trapezoid on the base is equal to the midline of the trapezoid.

## Solution

The projection of the lateral side on the larger base is $h \operatorname{ctg} \alpha$, and the midline of the trapezoid is $a - h \operatorname{ctg} \alpha$.

Therefore, the area of the trapezoid is $h(a - h \operatorname{ctg} \alpha)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-01.jpg?height=349&width=592&top_left_y=2072&top_left_x=733)

## Answer

$h(a - h \operatorname{ctg} \alpha)$