What is the % change in the area of a rectangle when its length increases by 10% and its width decreases by 10%?

Question

Skylar Adams · Accepted Answer

I tried this:

Let us call the length #l# and width #w#; we get for the area #A#: #A=l*w# let us change the two to get: #A'=(l+0.1l)*(w-0.1w)# rearrange: #A'=lwcancel(-0.1lw)+cancel(0.1lw)-0.01lw# #A'=0.99lw# but #A=lw# so substituting: #A'=0.99A# so the new area is #99%# of #A#. For example; imagine a rectangle where: #l=10# and #w=5# Area#=10*5=50# Now we increase the length and decrease the width: #l=10+0.1*10=11# #w=5+0.1*5=4.5# Area'#=11*4.5=49.5# that represents #99%# of #50#.

Autumn Arnold · Answer

undefined When the length of a rectangle increases by 10% and its width decreases by 10%, the change in area can be calculated using the formula for the percentage change:

Percentage change = ((New value - Old value) / Old value) * 100

Let's denote:
- $L$ as the original length of the rectangle,
- $W$ as the original width of the rectangle,
- $A$ as the original area of the rectangle.

After the changes:
- The new length $L_{\text{new}} = 1.10L$ (increased by 10%),
- The new width $W_{\text{new}} = 0.90W$ (decreased by 10%).

The new area $A_{\text{new}} = L_{\text{new}} \times W_{\text{new}} = (1.10L) \times (0.90W) = 0.99LW$.

Using the percentage change formula:

$$
\text{% Change} = \left( \frac{A_{\text{new}} - A}{A} \right) \times 100
$$

$$
= \left( \frac{0.99LW - LW}{LW} \right) \times 100
$$

$$
= \left( \frac{-0.01LW}{LW} \right) \times 100
$$

$$
= -1\%
$$

Therefore, the percentage change in the area of the rectangle is a decrease of 1%.