ABCD is a plane quadrilateral and E is any point on AD. EF is drawn parallel to DB to meet AB in F, and EG is drawn parallel to
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Answer:
5 cm
Step-by-step explanation:
We khow that the altitude of this triangle is 1cm shorter than the base
- Let H be our altitude and B our base and A the area of the triangle
- A= (B*H)/2 ⇒ 15=(B*H)/2
- H is 1cm shorter than B ⇒ B=H+1
- H*(H+1)/2=15 ⇒ H*(H+1)=30⇒ H²+H=30⇒H²+H-30+0
that's a quadratic equation . Let's calculate the dicriminant .
Let Δ be the dicriminant
- a=1
- b=1
- c= -30
- Δ=b²-4*a*c = 1²-4*1*(-30)=1+4*30=121≥0
- Δ≥0⇔ that we have two solutions x and y
- x= (-1-
)/2= (-1-11)/2= -6
- y= (-1+
We have a negative value and a positive one
The altitude is a distance so it can't be negative
H= 5cm