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

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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) About Me  (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) About Me  (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: A+=+%281%2F2%29BH, 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 A+=+%281%2F2%29BH becomes: 40+=+%281%2F2%29%28H+-+2%29H

40+=+%281%2F2%29%28H%5E2+-+2H%29 ----- 40+=+%28H%5E2+-+2H%29%2F2

H%5E2+-+2H+=+80 ---- Cross-multiplying

H%5E2+-+2H+-+80+=+0

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

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

H, or height = highlight_green%2810%29 metres

Base = 10 - 2, or highlight_green%288%29 metres

You can do the check!!

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