Question

Отрезки АС и ВD пересекаются в точке О. ВD= АС, ОВ=ОС. а)Докажите, что ∆ АОВ = ∆ СОD; б)Найдите периметр ∆ СОD, если АВ=9см, ВО=5см, ОD=7см.

Answer 1

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Answer 2

This is a question about geometry, which is a branch of mathematics that studies spatial structures and relations, and their generalizations. I will try to answer it as best as I can.

a) To prove that ∆ АОВ = ∆ СОD, we can use the criterion of congruence of triangles by two sides and the angle between them (SAS). We have:

- ВD = АС (given) - ОВ = ОС (given) - ∠АОВ = ∠СОD (vertical angles)

Therefore, ∆ АОВ = ∆ СОD by SAS.

b) To find the perimeter of ∆ СОD, we need to know the length of the side СD. We can use the Pythagorean theorem to find it, since ∆ АОВ = ∆ СОD and they are both right triangles (by the definition of perpendicular bisector). We have:

- СD² = АО² + ОD² - СD² = (АВ/2)² + ОD² (since АО = ВО = АВ/2) - СD² = (9/2)² + 7² - СD² = 40.25 + 49 - СD² = 89.25 - СD = √89.25 - СD ≈ 9.44 cm

The perimeter of ∆ СОD is the sum of the lengths of its sides, so we have:

- P(∆ СОD) = СD + ОD + ОС - P(∆ СОD) = 9.44 + 7 + 5 - P(∆ СОD) = 21.44 cm

I hope this helps you understand the solution. If you want to learn more about geometry, you can check out some of the web search results that I found for you . Have a nice day!