SOLUTION: two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other triangle is 3 units less than its base. Find the altitudes,

Question 770089: two triangles have equal bases. The altitude of one triangle is 3 units more than its base and the altitude of the other triangle is 3 units less than its base. Find the altitudes, if the areas of triangles differ by 21 square units.
Answer by Cromlix(4381) (Show Source): You can put this solution on YOUR website!
Triangle A: Base = x. Altitude = x + 3
Triangle B: Base = x. Altitude = x - 3
Area of triangle = 1/2(base*altitude)
Triangle A: Area = 1/2(x(x + 3))= 1/2x^2 + 3/2x
Triangle B: Area = 1/2(x(x - 3))= 1/2x^2 - 3/2x
The areas differ by 21 units^2
1/2x^2 + 3/2x -(1/2x^2 - 3/2x) = 21
1/2x^2 + 3/2x - 1/2x^2 + 3/2x = 21
3/2x + 3/2x = 21
6/2x = 21
x = 2*21/6
x = 7 units
Triangle A: altitude = x + 3 = 10 units
Triangle B: altitude = x - 3 = 4 units
Area: Triangle A = 35 units
Area: Triangle B = 14 units
Hope this helps.
:-)

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