The length of a rectangle is nine inches less than twice the width If the perimeter of the rectangle is 84 inches find the dimensions of the rectangle
Answer:
Dimensions of the rectangle are 25 inches × 17 inches.
Step-by-step explanation:
Let the length of the rectangle = l inches and width = w inches
The length of the rectangle is 9 inches less than twice the width.
Therefore, l = (2w - 9)
Perimeter of the rectangle is = 84 inches
And perimeter of the rectangle = 2(Length + width)
= 2(l + w)
= 2[(2w - 9) + w]
= 2[3w - 9]
Now 84 = 2{3w - 9)
3w - 9 = \(\frac{84}{2}\)
3w - 9 = 42
3w = 42 + 9
3w = 51
w = \(\frac{51}{3}\)
w = 17 inches
Since l = 2w - 9
l = 2×17 - 9
= 34 - 9
= 25 inches
Therefore, dimensions of the rectangle are 25 inches × 17 inches.