Question

даю 50 баллов. около окружности описана бедренная трапеция, средняя линия которой равна 5 см, а синус острого угла при основании 4/5. Найдите площадь трапеции.

Answer 1

надеюсь ты разберёшь мой почерк

Answer 2

Ответ на фотографии).

Answer 3

Answer in detail. I give 50 points. Around the circle is inscribed a trapezoid, the midline of which is 5 cm, and the sine of the acute angle at the base is 4/5. Find the area of the trapezoid.

To find the area of the trapezoid, we need to use the following formula:

$$A = \frac{h(a + b)}{2}$$

where $A$ is the area, $h$ is the height, $a$ and $b$ are the bases.

We know the midline of the trapezoid is 5 cm, which is the average of the bases:

$$\frac{a + b}{2} = 5$$

We can solve for $a + b$:

$$a + b = 10$$

We also know the sine of the acute angle at the base is 4/5, which means the opposite side over the hypotenuse is 4/5. We can use the Pythagorean theorem to find the adjacent side:

$$c^2 = a^2 + b^2$$

where $c$ is the hypotenuse, $a$ is the opposite side, and $b$ is the adjacent side.

We plug in the values we know:

$$(5)^2 = (4)^2 + b^2$$

We solve for $b$:

$$b^2 = 9$$

$$b = 3$$

Now we have the adjacent side, which is also the height of the trapezoid. We can plug it into the area formula:

$$A = \frac{h(a + b)}{2}$$

$$A = \frac{3(10)}{2}$$

$$A = 15$$

The area of the trapezoid is 15 cm$^2$.