he quadratic equation +4x 21 = 0. 3.11 TRANSFORMATION OF AREAS (a) Draw an irregular hexagon and then construct, with straightedge and compasses, a square having the same area. (b) With straightedge and compasses, divide a quadrilateral ABCD into 3 equivalent parts by straight lines drawn through vertex A. (c) Bisect a trapezoid by a line drawn from a point P in the smaller base. (d) Transform triangle ABC so that the angle A is not altered, but the side opposite the angle A becomes parallel to a given line MN. (e) Transform a given triangle into an isosceles triangle having a given vertex angle. 312 REGULA 3.11 - ly 22xr+al+rsly-21-4y] = 0. a pair of lines on the intersections of the circle with the lines AR and AS. It follows that the second factor set equal to 0 represents the line. RS. Setting y 0, we find OLrs/le + s) h/g setting y-2, we find AT 4/(r+s)-4/g (b) First trisect the diagonal BD by points E and F. Then the broken lines AEC and AFC divide the figure into 3 equivalent parts. Transform these parts so as to fulfull the conditions by drawing parallels to AC through E and F. (d) Through B draw BD parallel to MN to cut AC in D. Then, if the required triangle is AB'C', AC' is a mean proportional between AC and AD. (e) Let ABC be the given triangle. Draw AB' making the given vertex angle arallel to AC through B in B. Now lat it cut