Measure the length of the ruler of 3 feet, 4 feet and 5 feet, draw a right triangle on the paper according to the following method, and fold it in half. Then put the right angle point of the right triangle on the arc, draw two diameters of the garden, and get the intersection o of the two diameters, which is the center of the garden.
Let any point on the garden O be a, connect the AO, put a right angle side of the right triangle on the AO, and draw the tangent of any point A on the garden O.
If you don't know the center o of the garden, use a ruler to cross any point A on the garden O and draw any two strings AE and AF. According to Pythagorean theorem, we can draw the center O. Method:
Measure AE, measure AG=AE/2 on AE, let AG=4Y, draw a garden, one end of the ruler is fixed at point A, with A as the center, a pen fixed at 5Y as the radius, draw another garden, with G as the center and 3Y as the radius, and get the intersection point H of the two gardens. Connect GH, and GH will hang AE. You can also draw the vertical line of AF, and the intersection of the two vertical lines is the center of the garden.
The method of drawing tangent is as follows:
Measure the radius of the garden with a ruler, AO=R, and let R=4X. Then, according to Pythagorean theorem, draw a garden with O as the center and 5X=5R/4 as the radius with a ruler, and then draw a garden with A as the center and 3X=3R/4 as the radius, so as to get the intersection point of C and connect AC, which is the tangent of any A on the garden O. 。
because
CO^2=(5X)^2
ao^2+ac^2=(4x)^2+(3x)=(5x)^2=co^2
So the triangle AOC is a right triangle, and AO is perpendicular to AC.
Or let R=3X, take the center of the garden as O, 5X=5R/3 as the radius, take the center of the garden as A, 4X=4R/3 as the radius, draw the garden and get the intersection point D.
Connect AD, then AD is the tangent of any a on the circle o.