Answer: The dimensions of the rectangle are 6ft and 16ft.
Step-by-step explanation:
Let the width of the rectangle be y
Let the length of the rectangle be y+10
Area= 96ft
Since area of a rectangle is length multiplied by width
y(y+10) = 96
y^2 +10y =96
y^2 +10y -96= 0
Solving algebraically,
y^2 +16y - 6y - 96= 0
y(y+16) - 6(y+16)= 0
(y-6)(y+16)= 0
y-6= 0
y =6ft
Recall that the width was denoted as y. That means the width is 6ft.
Since length is y+10= 6+10 = 16ft
Length is 16ft, width is 6ft.
Answer:
Dimensions of the rectangle = 16 by 6
Step-by-step explanation:
Let the length and breadth represent L and B
Hence L= B+10=------eqn 1
LB = 96 -----eqn2
From eqn1 L=B+10, it's substituted into eqn2
B(B+10)=96
B²+10B-96=0
It's now solved quadratically
B²+16B-6B-96
B(B+16)-6(B+16)=0
(B-6)(B+16)=0
B=6 and -16
Hence B=6 ft since it's the positive value
L= 96/B = 96/6 = 16 ft
Dimensions of the rectangle = 16 by 6
