Respuesta :
The area of a triangle is always:
A=bh/2, we are told that A=48 and h=b-4 so:
48=(b/2)(b-4)
48=(b^2-4b)/2
b^2-4b=96
b^2-4b-96=0
b^2-12b+8b-96=0
b(b-12)+8(b-12)=0
(b+8)(b-12)=0, and since b>0
b=12m
A=bh/2, we are told that A=48 and h=b-4 so:
48=(b/2)(b-4)
48=(b^2-4b)/2
b^2-4b=96
b^2-4b-96=0
b^2-12b+8b-96=0
b(b-12)+8(b-12)=0
(b+8)(b-12)=0, and since b>0
b=12m
let the base be x metres.
1/2 × (x) × (x - 4) = 48
(x) × (x - 4) = 48 ÷ 1/2
(x) × (x - 4) = 96
x^2 - 4x = 96
x^2 - 4x - 96 = 0
(x - 12) (x + 8) = 0
x = 12 or -8 (rejected, since length of base must be positive)
therefore the length of the base (x) = 12m
-
hope this helps :)))
1/2 × (x) × (x - 4) = 48
(x) × (x - 4) = 48 ÷ 1/2
(x) × (x - 4) = 96
x^2 - 4x = 96
x^2 - 4x - 96 = 0
(x - 12) (x + 8) = 0
x = 12 or -8 (rejected, since length of base must be positive)
therefore the length of the base (x) = 12m
-
hope this helps :)))