Question

Сторона АD прямокутника АВСD дорівнює а і утворює з діагоналлю АС кут a. Знайти радіус кола, описаного навколо прямокутника.

Answer 1

Ответ:     r =  a/( 2cosα ) .

Объяснение:

 ABCD - прямокутник ; AD = a ;  ∠CAD = α .

 Як відомо , r = 1/2 AC .  Із прямок. ΔACD   cosα = AD/AC ;

 AC = AD/cosα = a/cosα . Знайдемо радіус кола, описаного навколо прямокутника :

     r = 1/2 * a/cosα = a/( 2cosα ) ;   r =  a/( 2cosα ) .

Answer 2

Problem Analysis

We are given a rectangle ABCD, where the length of one side, AD, is equal to 'a'. The angle between the diagonal AC and side AD is denoted as 'a'. We need to find the radius of the circle circumscribed around the rectangle.

Solution

To find the radius of the circle, we can use the properties of a rectangle and the relationship between the radius of the circumscribed circle and the rectangle's diagonal.

Let's denote the radius of the circumscribed circle as 'R' and the length of the diagonal AC as 'd'.

From the properties of a rectangle, we know that the diagonals of a rectangle are equal in length. Therefore, we have:

AC = BD

Since ABCD is a rectangle, we can use the Pythagorean theorem to find the length of the diagonal AC:

AC^2 = AB^2 + BC^2

Since AB = AD = a (given), and BC = CD (opposite sides of a rectangle are equal), we can rewrite the equation as:

AC^2 = a^2 + CD^2

Now, let's consider the right triangle ACD. The angle between the diagonal AC and side AD is 'a'. Therefore, we can use trigonometric functions to relate the sides of the triangle:

cos(a) = AD / AC

Rearranging the equation, we get:

AC = AD / cos(a)

Substituting the value of AC from the previous equation, we have:

a^2 + CD^2 = (AD / cos(a))^2

Simplifying the equation, we get:

CD^2 = (AD / cos(a))^2 - a^2

Since CD = BC, we can rewrite the equation as:

BC^2 = (AD / cos(a))^2 - a^2

Now, let's consider the right triangle BCD. The hypotenuse BC is equal to the radius of the circumscribed circle R. Using the Pythagorean theorem, we have:

R^2 = BC^2 + CD^2

Substituting the value of BC^2 from the previous equation, we get:

R^2 = [(AD / cos(a))^2 - a^2] + (AD / cos(a))^2

Simplifying the equation, we have:

R^2 = 2(AD / cos(a))^2 - a^2

Finally, we can find the radius R by taking the square root of both sides:

R = sqrt(2(AD / cos(a))^2 - a^2)

Therefore, the radius of the circle circumscribed around the rectangle is sqrt(2(AD / cos(a))^2 - a^2).

Answer

The radius of the circle circumscribed around the rectangle ABCD, where the length of one side AD is equal to 'a' and the angle between the diagonal AC and side AD is 'a', is sqrt(2(AD / cos(a))^2 - a^2).