Two triangles that have equal base lengths and equal heights, have equal areas
The ratio of the areas of ΔAEB and ΔDEC is one
The reason the ratio is one is that the area of ΔAEB is equal to the area of ΔDEC
The reasons for giving the above values are are follows;
The given parameters are;
Triangle ABC and DBC overlap
The base length of triangle ABC = The base length of triangle DBC
The height of triangle ABC = The height of triangle DBC = 8
Required:
To find the ratio of the areas of the triangle AEB and DEC
Solution:
The area of a triangle = (1/2) × Base length × Height
Given that the base, and height of triangles ΔABC and ΔDBC are equal, we have;
Area of triangle ΔABC = Area of triangle ΔDBC
The area of ΔABC = Area of ΔAEB + Area of ΔECB
Area of triangle ΔDBC = Area of ΔDEC + Area of ΔECB
Therefore;
Area of ΔAEB + Area of ΔECB = Area of ΔDEC + Area of ΔECB by substitution property of equality
Subtracting area of ΔECB from both sides gives;
Area of ΔAEB = Area of ΔDEC by subtraction property of equality
Therefore;
[tex]\dfrac{\Delta AEB}{\Delta DEC} = 1[/tex]
ΔAEB:ΔDEC = 1:1
The ratios of the areas of the triangle AEB and DEC is 1
- The reason why the ratio of the areas of triangle ΔAEB and ΔDEC is one
The reason why the ratio of the area of the two triangles is 1 is that ΔAED and ΔDEC are each obtained from ΔABC and ΔDBC which are equal to each other by subtracting ΔBEC from each, and therefore, by subtraction property of equality, the two triangles are equal
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