#1
First construct a square OABC (S) before anything as shown below. To carry out this construction, first we consider the ratio of the one of the side of the given square (S) to that of the square to be constructed(whose area is three the area of S). Note that for the constructed square to be three times of S, the ratio of their sides should be 1 to square root of 3, where as their respective areas should be 1 to 3 which we need to construct. By using Pythagoras theorem, a^2 + b^2 = c^2, which means if we need the ratio of the sides of the two squares to be 1 to 3^1/2, then we have to choose the sides carefully. Consider taking a = 1 = OA and c = 2 = OH then b = {c^2 - a^2}^1/2 = {2^2 - 1^2}^1/2 = 3^1/2 = AH as illustrated in the figure 1 below.
Therefore it means AH is the side of the square whose area is three times the area of S. To construct this, start at point O and construct a a circle (c1) center O and radius AH as shown in figure 2. From point O through A construct a line to meet circle c1 at D and also from O through C construct a line to meet the same circle c1 at F.
At F construct a line (L1) perpendicular to OF, then at D construct a line (L2) perpendicular to OD to meet line L1 at Point E. Using the points ODEF construct the square as shown in figure 2. The square constructed ODEF is three times area of the given square S = OABC to see the file that created the above image, click here.
By the bases of considering the ratio of side of square S (OABC) to the constructed square (ODEF) which is 1 to 3^1/2, the following was noted. Note that this was done by the aid of Pythagoras theorem to construct square root of 3 unites to be used to construct square ODEF. Now given the ratio of sides of the two squares is 1 to 3^1/2. Let us take the sides of square S be "a" units then as per the ratio of the two squares, sides of square ODEF (T)will be "a*3^1/2.
Which means area of square S is a*a = a^2
Also area of square ODEF is a*3^1/2*a*3^1/2 = 3*a^2
which show that area of square ODEF (T) is three times that of S (OABC).
#2
In this construction, a circle C center O and point P are given, therefore construct them first before anything. Construct a segment from O to P and construct its midpoint D. Using D as the center and DP as the radius of the circle, construct circle c2. Using circle C and c2, construct a point of intersection A and B, then construct a line L1 and L2 through PA and PB respectively. Join O to A and O to B by segment OA and OB respectively. The angle constructed PAO and PBO is 90 degrees respectively. To play with the gsp file that created the image below, click here.
A line L1 is tangent to the circle C with center O if and only if there's a point A on C such that L1 passes through A and L1 is perpendicular to the radius OA. Now using the figure below as per its construction, D is the midpoint of OP hence OD = DP = DA = radius of a circle. Angle OPA is equal to angle PAD and equal to x degrees ( isosceles triangle). Also angle DOA = angle DAO = y degrees (isosceles triangle), since DA is equal to DO. By considering triangle PAO the sum of its three angles should add to 180 degrees. Therefore with this in mind, then angle POA + angle OAP + angle APO = 180 degrees. Which means y + y + x + x = 180 degrees. Therefore, 2y + 2x = 180. By simplifying the expression, we obtain y + x = 90 degrees. Given angle OAP is equal to y + x, then from the expression, we can conclude that angle OAP is 90 degrees which means line L1 is perpendicular to OA. PAO is congruent to PBO since OA = OB and both triangle share OP, the it means AP is equal to BP. Therefore by side, side, side we can conclude that PAO is congruent to PBO. Given triangle OPA is congruent to triangle OPB, then by the same argument line L2 is perpendicular to OB then angle PBO is also 90 degrees. By definition of a line of tangency to a circle, we can conclude that line L1 and L2 are tangent to circle C at A and B respectively.
