A Math Forum Project

Geometry Forum - Problem of the Week Solutions - The yard and the walkway, May 15-19, 1995 Annie says: John Conway sent me this problem, and while he stated that obviously the rhombus and the walkway have equal areas, not so obvious is the fact that the perimeters of the two are equal. Well, it wasn't obvious enough that any of these folks figured it out. One set of student hypothesized it in a Sketchpad drawing, but failed to show any work for either part, so they aren't listed here (and couldn't get their drawing to show that the perimeters were equal, but they thought there was a chance). Perhaps John will enlighten us with a proof of this fact! In any event, students used a couple of different methods to show the areas were equal. Some used the fact that the area of a rhombus is half the product of the diagonals, which in this case happen to be the edges of the original rectangle, so that works out neatly. Others (including Keith and Ellen, whose answer is not included here) drew in the diagonals of the rhombus, and then pointed out that the area of the rhombus was half of each of the smaller rectangles formed. Susan Quan Susan Quan Grade 8 Masterman School Philadelphia, PA Since the area of the rectangle is DC times DA and the area of the rhombus is 1/2(DC)(DA), the area of the rhombus equals the area of EFGH Eric Wahl Eric Wahl Grade 11 Masterman School Philadelphia, PA The area of the rhombus is the area of ABCD - area of 4 congruent triangular regions: AB(AD) - 4(1/2)(1/2 AB)(1/2(AD) = 1/2 (AB)(AD). So the area of the rhombus equals the area of EFGH. Brad Warner Brad Warner, grade 9, Fairfield HS If the area of the walkway is 1/2 the area of all of ABCD, then the area of EFGH is also 1/2 the area of all of ABCD. Let lw be the area of ABCD and 1/2 lw be the area of EFGH. The rhombus is inscribed in ABCD so each corner touches the midpoint of a side. By taking away the area of the 4 triangles, which are all congruent because 2 sides and an included angle are congruent in all of them, you will find the area of the rhombus. The area of each triangle is 1/2 bh, where b=1/2 l and h=1/2 w. Area of the rhombus = lw - 4(1/2)(1/2 l)(1/2 w), A = lw - 2(1/2l)(1/2 w), A = lw - l(1/2 w), A = 1/2 lw. The area of the rhombus is 1/2 lw which is also the area of EFGH. The relationship between EFGH and the rhombus is that their areas are equal. Julie Black Julie Black, grade 9, Fairfield HS EFGH is similar to the rhombus because they both have the same area. It is given that the area of EFGH is half of the area of ABCD. Then, the area of a rhombus is A=1/2 d1*d2. The diagonals of the rhombus are each congruent to a side of the rectangle. So, the area of the rhombus (A=1/2 d1*d2) is really equal to 1/2 the area of ABCD (A=1/2 lw). So both the rhombus and EFGH are 1/2 the area of ABCD. Jennifer Abrahamsen Jennifer Abrahamsen, grade 9, Fairfield HS Let AB=y and AD=x. The area of rectangle ABCD=xy. Area of rectangle EFGH=area of ABCD-area of the walkway. Area of the walkway is 1/2 area of ABCD=1/2 xy. Area of EFGH=xy-1/2 xy=1/2 xy. Area of a rhombus is 1/2 the product of the diagonals. Since the diagonals are x and y, the area of the rhombus is 1/2 xy. Therefore, the area of EFGH is equal to the area of the rhombus. Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page