tag:blogger.com,1999:blog-38193052.post6558927118815310356..comments2024-03-20T08:57:17.447-03:00Comments on Jornalheiros: Um triângulo no octógonoPC Filhohttp://www.blogger.com/profile/16547063456626761789noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-38193052.post-41937304405609151322015-12-13T21:57:15.282-02:002015-12-13T21:57:15.282-02:00:):)PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-6461490983296409312015-12-13T15:07:34.465-02:002015-12-13T15:07:34.465-02:00That actually occurred to me after I had posted my...That actually occurred to me <i>after</i> I had posted my solution.jrh150482https://www.blogger.com/profile/10502831081969372299noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-51476432433511274442015-12-10T19:22:12.461-02:002015-12-10T19:22:12.461-02:00Well done, Jake!
I have another solution: divide ...Well done, Jake!<br /><br />I have another solution: divide the octagon in 8 equal triangular pieces, like we do with a pizza. The area of each of these triangles is, obviously, 1/8 of the area of the octagon.<br /><br />The triangle of the problem has the same base as one of these 1/8-triangles (a side of the octagon), and twice the height. Therefore, this triangle has twice the area of a 1/8-triangle.<br /><br />Then, the area of the blue triangle is 1/4 of the area of the octagon. :)PC Filhohttps://www.blogger.com/profile/16547063456626761789noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-31876483167315126442015-12-10T12:19:46.355-02:002015-12-10T12:19:46.355-02:00A solução do Jake é interessante. Outra opção que ...A solução do Jake é interessante. Outra opção que me ocorre seria:<br /><br />(i) dividir o octógono regular em 8 triângulos isósceles.<br />(ii) calcular a área de um triângulo isósceles pela apótema do triângulo, usando ceno e cosseno.<br />(iii) calcular a área do triângulo azul tendo por base a área dos triângulos isósceles. Gustavo Salleshttps://www.blogger.com/profile/11173805136363123186noreply@blogger.comtag:blogger.com,1999:blog-38193052.post-56680566657375345082015-12-10T11:07:13.706-02:002015-12-10T11:07:13.706-02:00For the sake of simplicity, I will set 1 as the le...For the sake of simplicity, I will set 1 as the length of each of the octagon's sides.<br /><br />If we draw two vertical lines connecting the top two vertices with the bottom two, we divide the octagon into five regions: the blue triangle, two smaller triangles, and two trapezoids. Note that the three triangles make a rectangle and that the area of the blue triangle is half the area of the rectangle.<br /><br />Now we calculate the area of each trapezoid. For each trapezoid, the height is the leg of an isosceles right triangle whose hypotenuse is a side of the octagon. Since the octagon's sides have length 1, the height of the trapezoid is 1/√2. The short base of the trapezoid is one of the octagon's sides, so its length is 1. The long base of the trapezoid can be partitioned into three segments, one of which is the same length as the short base and the other two of which are the same length as the height of the trapezoid. Therefore, its length is 1+2(1/√2)=1+√2. Therefore, each trapezoid has an area of (1/2)(1/√2)(1+1+√2)=(2+√2)/(2√2)=(1+√2)/2 square units.<br /><br />Back to the rectangle. Its height is 1+√2 and its base is 1, so its area is 1+√2 square units. This is equal to twice the area of each trapezoid, so the rectangle's area is equal to half the area of the octagon. Since the area of the blue triangle is half the area of the rectangle, it must be one-fourth the area of the octagon.<br /><br />My answer: The area of the blue triangle is one-fourth the area of the octagon.jrh150482https://www.blogger.com/profile/10502831081969372299noreply@blogger.com