Question

The length of a rectangle is 8cm more than 3 times it’s width. The perimeter of rectangle is 64 cm.
A) What are the dimensions of rectangle?
B) What is the area of the rectangle?

Answer 1

Find what you know.

Perimeter = 64cm

Length = width*3 + 8cm

Width = unknown (use w)

Perimeter of a rectangle formula is:

P = 2*L + 2*W or P = 2(L + W)

We know P and we have a formula for L

64cm = 2[(w*3 + 8cm) + w]

64cm = 2(3w + 8cm) + 2w

64cm = 6w + 16cm + 2w

 

Combine like terms:

64cm = 8w + 16cm

 

Subtract 16cm from both sides:

64cm - 16cm = 8w

48cm = 8w

 

Divide both sides by 8 to isolate the variable.

48cm/8 = w

6cm = w or swap them and w = 6cm

 

Now take the formula for perimeter again and plug in the value for width

 

64cm(P) = 2(L + 6cm(W))

64cm = 2L + 12cm

 

Subtract 12cm from both sides:

64cm - 12cm = 2L

52cm = 2L

 

Divide both sides by 2 to isolate the variable:

52cm/2 = L

26cm = L or swap them and L = 26cm

 

Rectangle is 26cm in length and 6cm in width.  There is no height variable in a 2D rectangle, but sometimes (rarely) width is referred to as height. If this is a 3D box, than the height is equal to the width, in which case the information given might be slightly misleading.

Answer 2

Let x be the width of the tectangle,
Then, lenghth of the rectangle= 8+3x

Given Perimeter =64
But, Perimeter= 2(l+b)
Then, 2(l+b)=64

l+b=32. {Dividing both sides by 2} (eqn.1)

Now, Substituting values of l & b in eqn.1

$x + (8 + 3x) = 32$
$x + 8 + 3x = 32$
$4x = 24$
$x = 24 \div 4$
$x = 6$





Now, Width =6
Length=

$8 + 3(6)$
$26$