O maior desses triângulos equiláteros tem área 14. Qual é a área total dos outros quatro?
(The largest of these equilateral triangles has area 14. What is the total area of the other four?)
PCFilho
(pescado no Twitter da professora Catriona Agg)
sábado, 17 de abril de 2021
5 comentários:
Regras para postar comentários:
I. Os comentários devem se ater ao assunto do post, preferencialmente. Pense duas vezes antes de publicar um comentário fora do contexto.
II. Os comentários devem ser relevantes, isto é, devem acrescentar informação útil ao post ou ao debate em questão.
III. Os comentários devem ser sempre respeitosos. É terminantemente proibido debochar, ofender, insultar e/ou caluniar quaisquer pessoas e instituições.
IV. Os nomes dos clubes devem ser escritos sempre da maneira correta. Não serão tolerados apelidos pejorativos para as instituições, sejam quais forem.
V. Não é permitido pedir ou publicar números de telefone/Whatsapp, e-mails, redes sociais, etc.
VI. Respeitem a nossa bela Língua Portuguesa, e evitem escrever em CAIXA ALTA.
Os comentários que não respeitem as regras acima poderão ser excluídos ou não, a critério dos moderadores do blog.
Assinar:
Postar comentários (Atom)
Given: All of the triangles in the diagram are equilateral, and the area of the largest one is 14.
ResponderExcluirQuestion: What is the total area of the other four triangles?
Let’s call the side length of the largest triangle s. This means that (√3/4)s² = 14. We draw a line segment from the leftmost vertex of the largest triangle to the leftmost vertex of the yellow triangle. This segment is orthogonal to the segment connecting the top and leftmost vertices of the yellow triangle, so these two segments and the one whose side length is s form a right triangle. So, if we call the measure of the angle between the side of the yellow triangle and the side of the largest triangle θ, this means that the lengths of the segment we drew and the side of the yellow triangle are (s sin θ) and (s cos θ), respectively. So, the area of the yellow triangle is (√3/4)(s cos θ)² = [(√3/4)s²] cos² θ = 14 cos² θ. The segment we drew is √3 times the side length of one of the small triangles, which means that that side length is (1/√3)(s sin θ) and the area of one of these triangles is (√3/4)[(1/√3)(s sin θ)]² = [(√3/4)s²](1/3) sin² θ = (14/3) sin² θ. This means that the total area of the three small triangles is 14 sin² θ. Therefore, the total area of the four colored triangles is 14(cos² θ + sin² θ) = (14)(1) = 14.
My answer: 14 square units (!!)
It still blows my mind that the four smaller triangles have a total area equal to that of the large triangle.
ExcluirThere is an amazing "visual" proof, that I will try to explain. It is possible to use the Pythagorean theorem to prove that the four smaller triangles have a total area equal to that of the large triangle.
ExcluirThe idea is to create a right triangle in the center of the figure, with each side shared with a different equilateral triangle.
The hypotenuse would be the lower side of the large triangle. One cathetus would be the left side of the yellow triangle. The other cathetus would be the side of an equilateral triangle created by a smart rearrangement of the three small green triangles.
When three regular polygons have their sides connected in a right triangle, the Pythagorean theorem states that the areas of the two cathetus-polygons sum to the area of the hypotenuse-polygon.
Maybe I should make a drawing of this. :)
Did you notice that I also posted solutions for some other recent geometry posts?
ExcluirYes, Jake. Thank you! I am going to check them out soon!
Excluir