Question

ВАРИАНТ 2 1) ΔАВС~ΔА1В1С1, АВ и А1В1 сходственные стороны треугольников, АВ:А1В1=3:5, А1В1=25 см; А 1С 1=30 см; В 1С 1=35 см. Найдите стороны ΔАВС. 2) ΔMNK~ΔM1N1K1 , M 1N 1=20 см, M 1K 1=45 см, N 1K 1=25см. Найдите периметр ΔMNK . Вычислите площадь ΔMNK, если известно, что площадь ΔM1N1K1 равна 180 см2. 3) Площади подобных треугольников равны 100дм2 и 25 дм2, сумма их периметров равна 117 дм. Найдите периметры обоих треугольников. ПОМОГИТЕ, СРОЧНО НАДО!

Answer 1

1) ΔАВС~ΔА1В1С1, АВ и А1В1 сходственные стороны треугольников, АВ:А1В1=3:5, А1В1=25 см; А 1С 1=30 см; В 1С 1=35 см. Найдите стороны ΔАВС.
коэффициент подобия
k = 3/5
AB = k*A₁B₁ = 3/5*25 = 15 см
АС = k*А₁С₁ = 3/5*30 = 18 см
BC = k*B₁C₁ = 3/5*35 = 21 см
2) ΔMNK~ΔM1N1K1 , M 1N 1=20 см, M 1K 1=45 см, N 1K 1=25см. Найдите периметр ΔMNK . Вычислите площадь ΔMNK, если известно, что площадь ΔM1N1K1 равна 180 см2.
ошибка в условии M₁N₁ + N₁K₁ = M₁K₁, это не треугольник
3) Площади подобных треугольников равны 100дм2 и 25 дм2, сумма их периметров равна 117 дм. Найдите периметры обоих треугольников.
Пусть коэффициент подобия большего треугольника к меньшему равен k
Тогда их площади относятся как k²
k² = 100/25 = 4
k = 2
Пусть периметр меньшего P
Периметр большего K*P
P+k*P = 117
P(1+2) = 117
P = 117/3 = 39 дм
и периметр большего
k*P = 2*39 = 78 дм

Answer 2

Problem 1: ΔABC ~ ΔA1B1C1, AB and A1B1 are corresponding sides of the triangles, AB:A1B1=3:5, A1B1=25 cm, A1C1=30 cm, B1C1=35 cm. Find the sides of ΔABC.

To find the sides of ΔABC, we can use the concept of corresponding sides in similar triangles. Since ΔABC is similar to ΔA1B1C1, the ratio of the corresponding sides will be the same.

Let's assume that the sides of ΔABC are AB, AC, and BC. We know that AB:A1B1=3:5, and A1B1=25 cm. Using this information, we can find the length of AB:

AB/A1B1 = 3/5 AB/25 = 3/5 AB = (3/5) * 25 AB = 15 cm

Similarly, we can find the lengths of AC and BC using the ratios:

AC/A1C1 = AB/A1B1 AC/30 = 15/25 AC = (15/25) * 30 AC = 18 cm

BC/B1C1 = AB/A1B1 BC/35 = 15/25 BC = (15/25) * 35 BC = 21 cm

Therefore, the sides of ΔABC are AB = 15 cm, AC = 18 cm, and BC = 21 cm.

Problem 2: ΔMNK ~ ΔM1N1K1, MN and M1N1 are corresponding sides of the triangles, M1N1 = 20 cm, M1K1 = 45 cm, N1K1 = 25 cm. Find the perimeter of ΔMNK. Calculate the area of ΔMNK if the area of ΔM1N1K1 is known to be 180 cm².

To find the perimeter of ΔMNK, we need to know the lengths of all three sides. Since ΔMNK is similar to ΔM1N1K1, the ratio of the corresponding sides will be the same.

Let's assume that the sides of ΔMNK are MN, MK, and NK. We know that MN/M1N1 = MK/M1K1 = NK/N1K1. Using this information, we can find the lengths of MN, MK, and NK:

MN/M1N1 = MK/M1K1 = NK/N1K1 MN/20 = MK/45 = NK/25

Let's solve for MN first:

MN/20 = MK/45 MN = (20/45) * MK

Next, let's solve for NK:

NK/25 = MK/45 NK = (25/45) * MK

Now, we can substitute these values into the equation MN/M1N1 = MK/M1K1 = NK/N1K1:

(M20/45) * MK/20 = MK/45 = (25/45) * MK/25

Simplifying this equation, we get:

MK = 20 cm

Now that we know the length of MK, we can find the lengths of MN and NK:

MN = (20/45) * 20 = 8.89 cm NK = (25/45) * 20 = 11.11 cm

To find the perimeter of ΔMNK, we add the lengths of all three sides:

Perimeter = MN + MK + NK = 8.89 cm + 20 cm + 11.11 cm = 40 cm

To calculate the area of ΔMNK, we can use the formula for the area of a triangle:

Area = (1/2) * base * height

Since we know the lengths of two sides, MN and NK, we can use them as the base and height respectively. Therefore, the area of ΔMNK is:

Area = (1/2) * MN * NK = (1/2) * 8.89 cm * 11.11 cm = 49.39 cm²

Problem 3: The areas of two similar triangles are 100 dm² and 25 dm², and the sum of their perimeters is 117 dm. Find the perimeters of both triangles.

Let's assume that the perimeters of the two triangles are P1 and P2, and their areas are A1 and A2 respectively.

We know that the areas of similar triangles are proportional to the squares of their corresponding sides. Therefore, we can write the following equation:

(A1/A2) = (P1/P2)²

Substituting the given values, we get:

(100/25) = (P1/P2)² 4 = (P1/P2)²

Taking the square root of both sides, we get:

2 = P1/P2

Since the sum of their perimeters is 117 dm, we can write another equation:

P1 + P2 = 117

Using the equation P1 = 2P2, we can substitute this into the second equation:

2P2 + P2 = 117 3P2 = 117 P2 = 39 dm

Substituting this value back into P1 = 2P2, we get:

P1 = 2 * 39 dm P1 = 78 dm

Therefore, the perimeters of the two triangles are P1 = 78 dm and P2 = 39 dm.

I hope this helps! Let me know if you have any further questions.