Question

Около окружности радиуса 5 описана равнобочная трапеция. Расстояние между точками касания боковых сторон равно 8. Найти площадь трапеции.

Answer 1

Пусть Трапеция ABCD; AB = CD; пусть точки касания AB с окружностью M, BC - K; CD - N; AD - P; у дельтоида MKNP известны обе взаимно перпендикулярные диагонали (MN = n = 8; очевидно, что KP = 2*r = 10); центр окружности радиуса r = 5 пусть O, лежит в середине KP.
Площадь трапеции S = p*r = r*(AB + BC + CD + AD)/2 = r*(2*AB);  
поскольку суммы противоположных сторон равны, и AB + CD = 2*AB = p ;
Треугольник AOB - прямоугольный, его гипотенузу AB надо найти, высота равна OM = r;
Треугольник KMP тоже прямоугольный, так как KP - диаметр. 
∠OAB = 90° -  ∠MOA; то есть ∠MOA = ∠ABO;
∠MOA = (1/2)*∠MOP = ∠MKP; получилось ∠ABO = ∠MKP; 
то есть прямоугольные треугольники AOB и MKP подобны. 
Гипотенуза треугольника MKP KP = 2*r; высота n/2;
Ясно, что отношение высот равно отношению гипотенуз, то есть
r/AB = (n/2)/(2*r); AB = 4*r^2/n; p = 2*AB = 8*r^2/n; S = 8*r^3/n;
S = 125.

Answer 2

Problem Analysis

We are given that a trapezoid is circumscribed around a circle with a radius of 5. The distance between the points of tangency on the lateral sides of the trapezoid is 8. We need to find the area of the trapezoid.

Solution

To find the area of the trapezoid, we need to know the lengths of its bases. Let's denote the lengths of the bases as a and b.

From the given information, we can observe that the distance between the points of tangency on the lateral sides of the trapezoid is equal to the sum of the lengths of the bases. Therefore, we have the equation:

a + b = 8 Since the trapezoid is circumscribed around a circle with a radius of 5, we can draw a right triangle inside the trapezoid, as shown below:

       A ________ B
        /        \
       /          \
      /            \
     /______________\
    D                C

Let's denote the center of the circle as O and the midpoint of the base AB as M. Since the trapezoid is circumscribed around the circle, the line segment OM is perpendicular to the base AB.

We can use the Pythagorean theorem to find the length of the base AB. Let's denote the length of AB as c.

According to the Pythagorean theorem, we have:

c^2 = a^2 + (2r)^2 where r is the radius of the circle.

Substituting the given values, we have:

c^2 = a^2 + (2 * 5)^2 Simplifying further, we get:

c^2 = a^2 + 100

Since c = a + b (from equation a + b = 8), we can substitute c in the above equation:

(a + b)^2 = a^2 + 100

Expanding and simplifying, we get:

a^2 + 2ab + b^2 = a^2 + 100

Simplifying further, we have:

2ab + b^2 = 100 Now, we have two equations:

a + b = 8 2ab + b^2 = 100 We can solve these equations simultaneously to find the values of a and b.

Let's solve these equations:

From equation a + b = 8, we can express a in terms of b:

a = 8 - b

Substituting this value in equation 2ab + b^2 = 100, we get:

2(8 - b)b + b^2 = 100

Expanding and simplifying, we have:

16b - 2b^2 + b^2 = 100

Simplifying further, we get:

b^2 - 16b + 100 = 0

This is a quadratic equation. We can solve it using the quadratic formula:

b = (-(-16) ± sqrt((-16)^2 - 4 * 1 * 100)) / (2 * 1)

Simplifying further, we have:

b = (16 ± sqrt(256 - 400)) / 2

b = (16 ± sqrt(-144)) / 2

Since the discriminant is negative, the equation has no real solutions. This means that there is no trapezoid that satisfies the given conditions.

Therefore, we cannot find the area of the trapezoid.

Conclusion

Based on the given information, there is no trapezoid that satisfies the conditions of being circumscribed around a circle with a radius of 5 and having a distance of 8 between the points of tangency on the lateral sides. Therefore, we cannot find the area of the trapezoid.