The length of rectangle is 5 more than the width. The perimeter is 22 feet. How do you find the length and width?

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The length of rectangle is 5 more than the width. The perimeter is 22 feet. How do you find the length and width?

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Answer 1 · 2017-04-06T14:47:52

$W = \frac{11}{6} ft = 1 ft 10 inches = 22 inches$ $W = 11/6 " ft" = 1 " ft " 10 " inches" = 22 " inches"$
$L = 9 ft 2 inches = 110 inches$ $L = 9 " ft " 2 " inches" = 110 " inches"$

Explanation:

Given: $L = 5 W$ $L = 5W$ , Perimeter $= 22 ft$ $= 22 " ft"$

Perimeter, $P = 2 L + 2 W$ $P = 2L + 2W$

Substitute in your given values:
$22 = 2 (5 W) + 2 W$ $22 = 2(5W) + 2W$

Distribute and solve for $W$ $W$ :

$22 = 10 W + 2 W$ $22 = 10W + 2 W$
$22 = 12 W$ $22 = 12W$

$W = \frac{22}{12} = \frac{11}{6} ft = 1 ft 10 inches = 22 inches$ $W = 22/12 = 11/6 " ft" = 1 " ft " 10 " inches" = 22 " inches"$

$L = 5 W = 5 \cdot \frac{11}{6} = \frac{55}{6} = 9 ft 2 inches = 110 inches$ $L = 5W = 5*11/6 = 55/6 = 9 " ft " 2 " inches" = 110 " inches"$

Answer 2 · 2017-04-06T14:52:59

length: 8
width: 3

Explanation:

$a$ $a$ =length;
$b$ $b$ =width;

$a = b + 5$ $a=b+5$ --> length is 5 more than the width

$22 = 2 a + 2 b$ $22=2a+2b$ --> div all by 2
$11 = a + b$ $11=a+b$
$b = 11 - a$ $b=11-a$

$b = 11 - b - 5$ $b=11-b-5$
$2 b = 6$ $2b=6$

$b = 3$ $b=3$
$a = b + 5 = 3 + 5 = 8$ $a=b+5=3+5=8$

$2 \cdot 8 + 2 \cdot 3 = 16 + 6 = 22$ $2*8+2*3=16+6=22$ --> CORRECT

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The length of rectangle is 5 more than the width. The perimeter is 22 feet. How do you find the length and width?

2 Answers

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