Question

1.Перимет параллелограмма равен 16см. Чему равен стороны параллеограмма, если известно, что одна его сторона в 3 раза больше другой.2. Вромбе ABCD <D=140 градусов. Определите углы треугольника AOD (О-точка пересичения диагоналей.3. На диагонале МР прямоугольника МNPQ отложены равные отрезки МА и РВ. Докажите что ANBQ-параллелограмм.

Answer 1

Пусть одна сторона- x см, тогда вторая 3x см P= 2(a+b) 16= 2(x+3x) 16= 2x+6x 16=8x x=2 Первая сторона 2 см, вторая 6 см


угол ОДА =140/2=70 градусов
угол ДОА=90 градусов, так как диагонали ромба перпендикулярны
угол ОАД=90-70=20 градусов
Ответ: 90, 70, 20

пусть O- точка пересечения диагоналей прямоугольника. Тогда MO=PO, NO=QO.
т. к. МА=РВ, то МО-МА=РО-РВ, т. е. АО=ВО
а дальше признак параллелограмма - диагонали делятся точкой О пополам

Answer 2

I have searched the web for your question and found some relevant results. Here is a summary of the answers:

1. To find the sides of the parallelogram, we need to use the formula for the perimeter: P = 2(a + b), where a and b are the lengths of the adjacent sides. Since one side is three times longer than the other, we can write: P = 2(x + 3x), where x is the shorter side. Substituting the given value of P = 16 cm, we get: 16 = 2(4x), or x = 2 cm. Therefore, the sides of the parallelogram are 2 cm and 6 cm. You can see a detailed solution with a diagram at [this link](https://uchi.ru/otvety/questions/1-perimet-parallelogramma-raven-16sm-chemu-raven-storoni-paralleogramma-esli-izvestno-cht).

2. To find the angles of the triangle AOD, we need to use the properties of the rhombus ABCD. The diagonals of a rhombus are perpendicular and bisect the opposite angles. Therefore, angle AOD is a right angle (90 degrees), angle ADO is half of angle ADC (70 degrees), and angle DAO is the complement of angle ADO (20 degrees). You can see a detailed solution with a diagram at [this link](https://bit.ly/2N4DmTD).

3. To prove that ANBQ is a parallelogram, we need to show that its opposite sides are parallel and equal. We can do this by using the congruence of triangles NPB and QMA. These triangles are congruent by the SAS criterion, since they have a common side MP, equal sides MA and BP (given), and equal angles NPB and QMA (alternate interior angles). Therefore, their corresponding sides are equal: NB = AQ and AN = BQ. Also, their corresponding angles are equal: angle NPB = angle QMA (given), angle NBP = angle QAM (vertical angles), and angle PNB = angle MAQ (corresponding angles). Thus, ANBQ is a parallelogram by the definition. You can see a detailed solution with a diagram at [this link](https://bit.ly/2IjEuSB).

I hope this helps you with your geometry homework.