Question

В треугольнике АВС высота СН делит сторону АВ на отрезки АН и ВН. Нйдите: 1. стороны треугольника АВС, если: А) АН=9м, ВН=16м, СН=12м. Б) АН=15дм, ВН=6дм, СН=8дм. 2. сторону АВ, если: А) АС=17см, ВС=10см, СН=8см. В) АС=20дм, ВС=15дм, СН=12дм.

Answer 1

Пункт Б по аналогии.
2 нужно объяснять?

Answer 2

To solve the given problems, we can use the properties of a right-angled triangle and apply the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

1. In triangle ABC with height CH, dividing side AB into segments AN and BN:

A) Given: AN = 9m, BN = 16m, CH = 12m.

We can use the Pythagorean theorem to find the lengths of sides AC and BC.

AC^2 = AN^2 + CH^2 AC^2 = 9^2 + 12^2 AC^2 = 81 + 144 AC^2 = 225 AC = √225 AC = 15m

BC^2 = BN^2 + CH^2 BC^2 = 16^2 + 12^2 BC^2 = 256 + 144 BC^2 = 400 BC = √400 BC = 20m

B) Given: AN = 15dm, BN = 6dm, CH = 8dm.

We'll convert dm (decimeters) to meters (1 dm = 0.1 m) for consistency.

AN = 15dm = 15 * 0.1m = 1.5m BN = 6dm = 6 * 0.1m = 0.6m CH = 8dm = 8 * 0.1m = 0.8m

Using the Pythagorean theorem:

AC^2 = AN^2 + CH^2 AC^2 = 1.5^2 + 0.8^2 AC^2 = 2.25 + 0.64 AC^2 = 2.89 AC = √2.89 AC ≈ 1.70m

BC^2 = BN^2 + CH^2 BC^2 = 0.6^2 + 0.8^2 BC^2 = 0.36 + 0.64 BC^2 = 1 BC = √1 BC = 1m

2. Find side AB (hypotenuse), given:

A) AC = 17cm, BC = 10cm, CH = 8cm.

Using the Pythagorean theorem:

AB^2 = AC^2 + BC^2 AB^2 = 17^2 + 10^2 AB^2 = 289 + 100 AB^2 = 389 AB = √389 AB ≈ 19.72cm

B) AC = 20dm, BC = 15dm, CH = 12dm.

Convert dm (decimeters) to meters (1 dm = 0.1 m):

AC = 20dm = 20 * 0.1m = 2m BC = 15dm = 15 * 0.1m = 1.5m CH = 12dm = 12 * 0.1m = 1.2m

Using the Pythagorean theorem:

AB^2 = AC^2 + BC^2 AB^2 = 2^2 + 1.5^2 AB^2 = 4 + 2.25 AB^2 = 6.25 AB = √6.25 AB = 2.5m

Remember to check whether the given values satisfy the properties of a right-angled triangle before applying the Pythagorean theorem.