We apply the Pythagorean theorem, which tells us that, in every rectangle triangle, the square of the hypotenuse is equal to the sum of the squares of the Hicks.
Two perpendicular lines are plotted.
on them, take the measurements of two of the sides of the squares.
The squares of these measurements can be made. But your drawing is not necessary to find the square solution.
We join the ends of these sides, which will be the Hicks of a rectangle triangle, to find their hypotenuse.
On one end of this hypotenuse, we trace another perpendicular to take the measure from the side of the third square.
Joining the ends of these segments, now Hicks, we find the hypotenuse of the sum of the three squares.
We repeat the same operation with the side of the fourth square.
We found the hypotenuse of the formed rectangle triangle.
This will be the side of the square, of the sum of the four dice.
We check that, the sum of the areas of these four squares, is equal to the area of the square solution.
Simplified representation of the exercise without plotting the squares.
Draw the circumference of the area with the sum of three, of radii: 10, 15 and 20 mm.
We use the same procedure.
We form triangles rectangles, where the Hicks are the diameters of the circumferences, and the hypotenuse will be the diameters of the circumferences sum.
Draw a regular pentagon, which has twice the area of another radius die 25 mm.
This case is as if Sumáramos two circumferences of equal radius.
We draw a rectangle triangle, where the Hicks will be the diameter of the radius circumference 25, and the hypotenuse will be the diameter of the double area circumference.
In it we'll draw the Pentagon, which will have twice the area of the Pentagon given.