Math Problem Statement

C is the midpoint of Segment AB, D is the midpoint of Segment AC, E is the midpoint of Segment AD, F is the midpoint of Segment ED, G is the midpoint of Segment EF, and H is the midpoint of Segment DB. If DC= 16, what is GH

Solution

To solve this problem, we need to analyze the segments and midpoints given in the problem. Let's define the length of each segment and find the relationship between them step by step.

Step 1: Assign Lengths to Segments

  1. Let AB=xAB = x.
  2. Since CC is the midpoint of ABAB, AC=x2AC = \frac{x}{2} and CB=x2CB = \frac{x}{2}.
  3. Since DD is the midpoint of ACAC, AD=AC2=x4AD = \frac{AC}{2} = \frac{x}{4} and DC=AC2=x4DC = \frac{AC}{2} = \frac{x}{4}.
  4. Since EE is the midpoint of ADAD, AE=AD2=x8AE = \frac{AD}{2} = \frac{x}{8} and ED=AD2=x8ED = \frac{AD}{2} = \frac{x}{8}.
  5. Since FF is the midpoint of EDED, EF=ED2=x16EF = \frac{ED}{2} = \frac{x}{16} and FD=ED2=x16FD = \frac{ED}{2} = \frac{x}{16}.
  6. Since GG is the midpoint of EFEF, EG=EF2=x32EG = \frac{EF}{2} = \frac{x}{32} and GF=EF2=x32GF = \frac{EF}{2} = \frac{x}{32}.
  7. Since HH is the midpoint of DBDB, DH=DB2DH = \frac{DB}{2}.

Step 2: Given Information

We know that DC=16DC = 16.

Since DC=x4DC = \frac{x}{4}, we have: x4=16\frac{x}{4} = 16 x=64x = 64

Step 3: Find GHGH

We know that: EF=x16=6416=4EF = \frac{x}{16} = \frac{64}{16} = 4 GG is the midpoint of EFEF, so: EG=GF=EF2=42=2EG = GF = \frac{EF}{2} = \frac{4}{2} = 2

HH is the midpoint of DBDB. To find DBDB, consider: DB=DB=ABADDB=6416=48DB = DB = AB - AD - DB = 64 - 16 = 48 Thus, HH is the midpoint of DBDB, so: DH=HB=482=24DH = HB = \frac{48}{2} = 24

Now, we find GHGH, the difference between GG and HH: Since GH=EGDHGH = EG - DH, GH=224=22GH = 2 - 24 = 22

Therefore, GHGH is 22.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometry
Midpoints
Line Segments

Formulas

Midpoint formula

Theorems

Segment Division Theorem

Suitable Grade Level

Grades 8-10