find the area of a rectangle that has a length of 8.3 inches. and a width that is 0.6 times the length. show work please
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Hey there :)
We know the area formula of a triangle
We are given:
The area = 56 m²
The base = let p represent this
The height = 9 m less than the base = p - 9
Apply the formula
56 =
× ( p ) × ( p - 9 )
56 = × ( p² - 9p )
Divide on both sides
56 ÷ = p² - 9p
112 = p² - 9p
Bring 112 to the other side and equate equation to 0
p² - 9p - 112
Factorise { Sum = -9 , Product = - 112 , therefore suitable factors = - 16 × 7 )
( p - 16 )( p + 7 ) = 0
↓ ↓
p = 16 p = - 7 ← Reject p = - 7 since length cannot be negative
So, the length of the base is 16 m
Check:
56 m² = × 16 × ( 16 - 9 )
56 m² = 8 × 7
56 m² = 56 m²
We know the area formula of a triangle
We are given:
The area = 56 m²
The base = let p represent this
The height = 9 m less than the base = p - 9
Apply the formula
56 =
56 =
Divide
56 ÷
112 = p² - 9p
Bring 112 to the other side and equate equation to 0
p² - 9p - 112
Factorise { Sum = -9 , Product = - 112 , therefore suitable factors = - 16 × 7 )
( p - 16 )( p + 7 ) = 0
↓ ↓
p = 16 p = - 7 ← Reject p = - 7 since length cannot be negative
So, the length of the base is 16 m
Check:
56 m² =
56 m² = 8 × 7
56 m² = 56 m²