Question

Решите задачки срочно пожалуйста... 1. В треугольнике АВС медианы ВВ1 и СС1 пересекаются в точке О и равны 15 см и 18 см соответственно. Найдите периметр треугольника АВС, если угол ВОС = 90 градусов. 2. Вершины треугольника АВС лежат на окружности с центром в точке О, угол А равен 50 градусов. Отношение дуги АС к дуге АВ = 2:3. Найдите углы В и С, угол ВОС. 3. Вершины треугольника АВС лежат на окружности с центром в точке О, угол АОВ равен 80 градусов. Отношение дуги АС к дуге АВ = 2:3. Найдите углы треугольника АВС. 4. Хорды АВ и СД пересекаются в точке Е. Найдите СД, если АЕ = 4 см, ВЕ = 9 см, а длинна СЕ в 4 раза больше длинны ДЕ. Срочно пожалуйста!!!!

Answer 1

В задании №3 думаю опечатка, там где отношение дуг, скорее всего надо АС/ВС=2/3...., если так, то ответ такой:∠А=84;  ∠В=56; ∠С=40
№4
обозначим ED через х, тогда СЕ=4х.
АЕ*ВЕ=СЕ*DE;
36=4x*x
36=4x²
x=3;DE=3.  CE=3*4=12  ,      CD=3+12=15

Answer 2

Task 1: Finding the Perimeter of Triangle ABC

Given: - Medians BB1 and CC1 intersect at point O and have lengths 15 cm and 18 cm, respectively. - Angle BOC is 90 degrees.

To find: - The perimeter of triangle ABC.

Solution: The medians of a triangle intersect at a point that divides each median in a 2:1 ratio. Let's denote the length of BO as x and the length of OC as y. Then, we have the following equations:

1. x + y = 2 * 15 (from the median BB1) 2. x + y = 2 * 18 (from the median CC1)

Solving these equations, we get x = 20 and y = 10.

Now, we can use the fact that the centroid divides each median in a 2:1 ratio to find the lengths of AB, BC, and AC.

Let's denote the length of AB as a, BC as b, and AC as c. Then, we have the following equations:

1. 2x = 3a 2. 2y = 3c

Solving these equations, we get a = 40 and c = 20.

Now, we can find the length of b using the Pythagorean theorem, as angle BOC is 90 degrees.

Using the Pythagorean theorem: b^2 = a^2 + c^2 b^2 = 40^2 + 20^2 b^2 = 1600 + 400 b^2 = 2000 b = √2000 b = 10√2

Finally, the perimeter of triangle ABC is: Perimeter = a + b + c = 40 + 10√2 + 20

Perimeter = 60 + 10√2 cm

Task 2: Finding Angles B and C, and Angle BOC

Given: - Vertices of triangle ABC lie on a circle with center O. - Angle A is 50 degrees. - The ratio of arc AC to arc AB is 2:3.

To find: - Angles B and C. - Angle BOC.

Solution: The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the circumference.

Let's denote the measure of arc AC as 2x and the measure of arc AB as 3x. Then, we have the following equations:

1. Angle A = 50 degrees 2. Angle B = 3x 3. Angle C = 2x 4. Angle BOC = 2 * (3x + 2x) = 10x

Using the fact that the sum of the angles in a triangle is 180 degrees, we can find the values of x, B, C, and BOC.

Solving these equations, we get: - x = 10 degrees - Angle B = 30 degrees - Angle C = 20 degrees - Angle BOC = 100 degrees

Therefore, the angles of triangle ABC are: Angle A = 50 degrees Angle B = 30 degrees Angle C = 20 degrees Angle BOC = 100 degrees

Task 3: Finding Angles of Triangle ABC

Given: - Vertices of triangle ABC lie on a circle with center O. - Angle AOB is 80 degrees. - The ratio of arc AC to arc AB is 2:3.

To find: - Angles of triangle ABC.

Solution: Similar to Task 2, we can use the fact that the angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the circumference.

Let's denote the measure of arc AC as 2x and the measure of arc AB as 3x. Then, we have the following equations:

1. Angle A = 2x 2. Angle B = 3x 3. Angle C = 180 - (2x + 3x) = 180 - 5x 4. Angle BOC = 2 * (2x + 3x) = 10x

Using the given information, we can solve for the values of x, A, B, C, and BOC.

Solving these equations, we get: - x = 16 degrees - Angle A = 32 degrees - Angle B = 48 degrees - Angle C = 180 - 80 = 100 degrees - Angle BOC = 160 degrees

Therefore, the angles of triangle ABC are: Angle A = 32 degrees Angle B = 48 degrees Angle C = 100 degrees **