SOLUTION: Gold Medal Problem ABCD is a square. POints EFGH are teh midpoints of the sides. AB =1 inch. what is the area of the inside square region

Algebra ->  Parallelograms -> SOLUTION: Gold Medal Problem ABCD is a square. POints EFGH are teh midpoints of the sides. AB =1 inch. what is the area of the inside square region      Log On


   

Question 847014: Gold Medal Problem
ABCD is a square. POints EFGH are teh midpoints of the sides. AB =1 inch. what is the area of the inside square region
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!


Use the Pythagorean theorem on ΔDCG

DG² = DC² + CG²
DG² = 1² + %281%2F2%29%5E2
DG² = 1 + 1%2F4
DG² = 5%2F4
DG = sqrt%285%29%2F2

ΔDIH ∽ ΔDCG, therefore

ID%2F%28DH%29%22%22=%22%22DC%2FDG

ID×DG = DC×DH

ID%22%22%2Aexpr%28sqrt%285%29%2F2%29%22%22=%22%221%2Aexpr%281%2F2%29

Multiply through by 2

IDsqrt%285%29%22%22=%22%221

ID = 1%2Fsqrt%285%29 

Rationalize the denominator:

ID = 1%2Fsqrt%285%29%22%22%2A%22%22sqrt%285%29%2Fsqrt%285%29

sqrt%285%29%2F5 

ΔCJD ∽ ΔDCG, therefore,

JG%2F%28CG%29%22%22=%22%22CG%2FDG

JG×DG = CG×CG

JG%22%22%2Aexpr%28sqrt%285%29%2F2%29%22%22=%22%22expr%281%2F2%29%2Aexpr%281%2F2%29

JG%22%22%2Asqrt%285%29%2F2%22%22=%22%22expr%281%2F4%29 

Multiply through by 4

2JGsqrt%285%29%22%22=%22%221

JG = 1%2F%282%2Asqrt%285%29%29 = sqrt%285%29%2F10  (rationalizing)

DI + IJ + JG = DG

sqrt%285%29%2F5 + IJ + sqrt%285%29%2F10 = sqrt%285%29%2F2

Multiply through by 10

2%2Asqrt%285%29 + 10·IJ + sqrt%285%29 = 5%2Asqrt%285%29

3%2Asqrt%285%29 + 10·IJ = 5%2Asqrt%285%29

                  10·IJ = 2%2Asqrt%285%29

                     IJ = 2%2Asqrt%285%29%2F10

                     IJ = sqrt%285%29%2F5

So the area of the inner square is IJ² = %28sqrt%285%29%2F5%29%5E2 = 5%2F25 = 1%2F5 
 
Edwin