Question

The length of a rectangle is 3 times the width. If the perimeter is to be greater than 56 meter. What are the lossible values for the width?( use w as the width)

Answer 1

Answer:

w > 7m

Step-by-step explanation:

let l is the length of rectangle and w is the width of rectangle,

Perimeter of rectangle is 2(l+w).

From the question statement, we observe that

l=3w

2(l+w)>56

Put l=3w in above inequality,

2(3w+w)>56

2(4w)>56

8w>56

Multiplying by 1/8 on both sides of equation we get

(1/8)8w>(1/8)56

w>7m

We get the conclusion that width must be greater than 7m.

Answer 2

Answer:

w > 7

Step-by-step explanation:

L = 3w

2L + 2w > 56

sub for L

2(3w) + 2w > 56

6w + 2w > 56

8w > 56

w > 7

Check: L = 3(8)

L = 24

2(24) + 2(8) > 56

48 + 16 > 56

64 > 56