SOLUTION: The altitude of a triangle is 2 metres longer than its base what are the dimensions of the altitude and the base if the area of the triangle is 40 metres squared. can you please

Question 694541: The altitude of a triangle is 2 metres longer than its base what are the dimensions of the altitude and the base if the area of the triangle is 40 metres squared.
can you please help me I don't understand this. Thanks in advance :)
Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Area of triangle = (1/2) *base * altitude
let base be x m
altitude = (x+2) m
Area = (1/2) *x*(x+2)
40= (1/2) *x^2+2x
80=x^2+2x
x^2+2x-80=0
x^2+10x-8x-80=0
x(x+10)-8(x+10)=0
(x+10)(x-8)=0
x=-10, OR 8
length cannot be negative
so base = 8m
Altitude = 8+2 = 10 m
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
The altitude of a triangle is 2 metres longer than its base what are the dimensions of the altitude and the base if the area of the triangle is 40 metres squared.

Formula for area of a right-triangle: , where A is the area of the triangle, H is the height (altitude) and B, the base

Since H (altitude) is 2 metres longer than B (base), then it can be said that B = H - 2. As A or area = 40 becomes:

-----

---- Cross-multiplying

(H + 8)(H - 10) = 0

H = - 8 (ignore as ray/segment/line CANNOT be negative)

H, or height = metres

Base = 10 - 2, or metres

You can do the check!!

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