ABC and A1B1C1 triangles are similar and their similar parties refer to as 6: 5. The absorber of the ABC triangle is larger than the area of the triangle A1B1C1 by 7.7 cm2. Find the area of these triangles.
We introduce the variable x and we denote the area of a smaller triangle A1B1C1, then the area of the ABC triangle will be equal to x + 7.7.
The likea factor, according to the condition of the problem, is 6/5.
We use the theorem on the areas of such triangles:
SABC / SA1B1C1 = k².
(x + 7.7) / x = 36/25
36x = 25x + 192.5
11x = 192.5
x = 17.5 – the area of the triangle A1B1C1;
17.5 + 7.7 = 25.2 – the area of the ABC triangle.
25.2 / 17.5 = 1.44;
36/25 = 1.44;
1.44 = 1.44.
Answer; 25.2 and 17.5 triangles area.
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