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Making a right-angle mark using the `calc` package in `TikZ`
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Numerical conditional within tikz keys?Labeling sides and angles of a right triangle for an argument of the Pythagorean TheoremDrawing and labeling a triangle for a demonstration for the Law of CosinesTangent to circle given point on circleA drawing of a right triangle is not displayed properlyTikZ: Drawing an arc from an intersection to an intersectionShading inside a triangle when the coordinates of one vertex are calculated by TikZA right angle not drawnLine up nested tikz enviroments or how to get rid of themRight-angle mark not drawn correctly
I have a right triangle and the altitude from the vertex of its right angle drawn. A=(-4,0), B=(0,3), and C=(0,0) are the vertices of the triangle, and F=(-36/25,48/25) is the foot of the altitude on the hypotenuse AB. I use the following commands to draw a right-angle mark.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
Why doesn't it look like three sides of a square with edge length 3mm?
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
begin{tikzpicture}
coordinate (circle_1) at (-8/5, 4/5);
draw[blue, fill=blue=25!] (circle_1) circle (4/5);
coordinate (circle_1) at (-3/5, 9/5);
draw[orange, fill=orange=25!] (circle_1) circle (3/5);
path (-4,0) coordinate (A) (0,3) coordinate (B) (0,0) coordinate (C);
coordinate (F) at (-36/25,48/25);
draw (A) -- (B) -- (C) -- cycle;
draw (F) -- (C);
%A right-angle mark is drawn at F.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
%A right-angle mark is drawn at C.
coordinate (U) at ($(C)!3mm!-45:(A)$);
draw ($(A)!(U)!(C)$) -- (U) -- ($(B)!(U)!(C)$);
end{tikzpicture}
end{document}
tikz-pgf
add a comment |
I have a right triangle and the altitude from the vertex of its right angle drawn. A=(-4,0), B=(0,3), and C=(0,0) are the vertices of the triangle, and F=(-36/25,48/25) is the foot of the altitude on the hypotenuse AB. I use the following commands to draw a right-angle mark.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
Why doesn't it look like three sides of a square with edge length 3mm?
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
begin{tikzpicture}
coordinate (circle_1) at (-8/5, 4/5);
draw[blue, fill=blue=25!] (circle_1) circle (4/5);
coordinate (circle_1) at (-3/5, 9/5);
draw[orange, fill=orange=25!] (circle_1) circle (3/5);
path (-4,0) coordinate (A) (0,3) coordinate (B) (0,0) coordinate (C);
coordinate (F) at (-36/25,48/25);
draw (A) -- (B) -- (C) -- cycle;
draw (F) -- (C);
%A right-angle mark is drawn at F.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
%A right-angle mark is drawn at C.
coordinate (U) at ($(C)!3mm!-45:(A)$);
draw ($(A)!(U)!(C)$) -- (U) -- ($(B)!(U)!(C)$);
end{tikzpicture}
end{document}
tikz-pgf
Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago
add a comment |
I have a right triangle and the altitude from the vertex of its right angle drawn. A=(-4,0), B=(0,3), and C=(0,0) are the vertices of the triangle, and F=(-36/25,48/25) is the foot of the altitude on the hypotenuse AB. I use the following commands to draw a right-angle mark.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
Why doesn't it look like three sides of a square with edge length 3mm?
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
begin{tikzpicture}
coordinate (circle_1) at (-8/5, 4/5);
draw[blue, fill=blue=25!] (circle_1) circle (4/5);
coordinate (circle_1) at (-3/5, 9/5);
draw[orange, fill=orange=25!] (circle_1) circle (3/5);
path (-4,0) coordinate (A) (0,3) coordinate (B) (0,0) coordinate (C);
coordinate (F) at (-36/25,48/25);
draw (A) -- (B) -- (C) -- cycle;
draw (F) -- (C);
%A right-angle mark is drawn at F.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
%A right-angle mark is drawn at C.
coordinate (U) at ($(C)!3mm!-45:(A)$);
draw ($(A)!(U)!(C)$) -- (U) -- ($(B)!(U)!(C)$);
end{tikzpicture}
end{document}
tikz-pgf
I have a right triangle and the altitude from the vertex of its right angle drawn. A=(-4,0), B=(0,3), and C=(0,0) are the vertices of the triangle, and F=(-36/25,48/25) is the foot of the altitude on the hypotenuse AB. I use the following commands to draw a right-angle mark.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
Why doesn't it look like three sides of a square with edge length 3mm?
documentclass{amsart}
usepackage{amsmath}
usepackage{tikz}
usetikzlibrary{calc,intersections}
begin{document}
begin{tikzpicture}
coordinate (circle_1) at (-8/5, 4/5);
draw[blue, fill=blue=25!] (circle_1) circle (4/5);
coordinate (circle_1) at (-3/5, 9/5);
draw[orange, fill=orange=25!] (circle_1) circle (3/5);
path (-4,0) coordinate (A) (0,3) coordinate (B) (0,0) coordinate (C);
coordinate (F) at (-36/25,48/25);
draw (A) -- (B) -- (C) -- cycle;
draw (F) -- (C);
%A right-angle mark is drawn at F.
coordinate (U) at ($(F)!3mm!45:(A)$);
draw ($(A)!(U)!(F)$) -- (U) -- ($(C)!(U)!(F)$);
%A right-angle mark is drawn at C.
coordinate (U) at ($(C)!3mm!-45:(A)$);
draw ($(A)!(U)!(C)$) -- (U) -- ($(B)!(U)!(C)$);
end{tikzpicture}
end{document}
tikz-pgf
tikz-pgf
asked 13 mins ago
A gal named DesireA gal named Desire
6981411
6981411
Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago
add a comment |
Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago
Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago
Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago
add a comment |
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Nice figure! I may use it to accompany a problem on a test in Geometry.
– user74973
7 mins ago