Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 617.
s-a=AD=2s-c=CD=3a-c=(s-c)-(s-a)=1b=2+3=52ac=(a^2+c^2)-(a-c)^2=b^2-(a-c)^2=25-1=24, ac=12Follows Area of triangle ABC=(1/2)ac=6 sq units
Let BF=BE=x; then AB=AE+BE=AD+BE=2+x and BC=CE+BF=CD+BF=3+x. By Pythageorean Theorem, we have AB^2+BC^2=AC^2, or (2+x)^2+(x+3)^2=5^2, which gives the quadratic equation x^2+5x=6. Thus, discarding negative solutions, we have x=1. Then, AB=3, BC=4 and the area of triangle ABC is (3*4)/2=6.
I have posted my solution, which is the same as above, on my blog.
We write the area in 2 ways(2+r)(3+r) = r(2+r+3+r+5)Simplifying the algebra r^2 + 5r-6 = 0 from which r= 1 after solving the quadratic and disregarding the negative rootSo area = 3 X 4 /2 = 6Sumith PeirisMoratuwaSri Lanka