prove that a square ABCD is half the area of a square ACEF, when the segment AB is half of the segment AE

Question

              F

 

A______B______E

 

C______D

 

Answer 1

How do we know which square is larger? ACEF (because its diagonal AE is longer than its side AC, which happens to be the diagonal of ABCD). Comparing the large square to the small square the ratio of similarity is √2, so the ratios of the areas is √2 squared, 2. So the larger square has twice the area of the small square.

Answer 2

I'm going to do this based on the way I think your drawing should've looked (as it is now, ACEF does not make a square...). It seems like it should look like this, based on your tag about congruent triangles.

               F


A             B             E


D             C 

If you draw this on paper, you'll see it breaks up into 5 congruent triangles. The area of square ABCD is twice the area of triangle ABC, and the area of triangle ACEF is four times the area of triangle ABC. Area of a triangle is half the base times the height.

The area of triangle ABC: 0.5*AB*BC

Area of square ABCD: 2*0.5*AB*BC but we know AB=BC as it's a square. So Area of square ABCD = 2*0.5*AB*AB = AB^2 

AB = 1/2*AE, and the area of ACEF is twice the area of triangle ACE.

Area of triangle ACE = 0.5*AE*BC, but BC=AB and AE = 2*AB, so area of ACE = 0.5*2*AB*AB = AB^2

Area of square ACEF = 2*AB^2 which is twice the area of ABCD, so the area of ABCD is half the area of ACEF.

Answer 3

Statement                          Reason

ABCD is a square                  Given

AECF is a square                  Given

AB = (1/2)AE                       Given

Area ABCD = AB·AC              Def. of area of a square

Area AECF = AE·AC               Def. of area of a square

Area ABCD = [(1/2)AE]·AC     Substitution

Area of ABCD = (1/2) AE·AC   Substitution