To find the perimeter of a rectangle, you must first understand that a rectangle's opposite sides are equal. Therefore if you know the length of one side it is the identical length of the opposite side.

\(\displaystyle 12cm + 12cm = 24cm\)

\(\displaystyle 4cm +4cm=8cm\)

\(\displaystyle 24cm+8cm=32cm\)

32cm is your answer.

First you need to understand that opposite sides of a rectangle are equal. Then you are able to add the value of each side to itself and then add your sums together to get the perimeter of a rectangle.

\(\displaystyle 10cm+10cm=20cm\)

\(\displaystyle 7cm+7cm=14cm\)

\(\displaystyle 20cm+14cm=34cm\)

To find the perimeter of a rectangle, we will use the following formula:

\(\displaystyle \text{perimeter of rectangle} = a+b+c+d\)

Where a, b, c, and d are the lengths of the sides of the rectangle.

Now, let's look at the rectangle.

We can see the length is 6in.  Because it is a rectangle, the opposite side is also 6in. 

We can also see the width of the rectangle is 3in. Because it is a rectangle, the opposite side is also 3in.

Knowing this, we can substitute into the formula.  We get

\(\displaystyle \text{perimeter of rectangle} = 6\text{in} + 6\text{in} + 3\text{in} + 3\text{in}\)

\(\displaystyle \text{perimeter of rectangle} = 18\text{in}\)

To find the perimeter of a rectangle, we will use the following formula:

\(\displaystyle P = a+b+c+d\)

Where a, b, c, and d are the lengths of the sides of the rectangle.

Now, given the rectangle

we can see the length is 11in.  Because it is a rectangle, the opposite side is also 11in.

The width is 5in.  Because it is a rectangle, the opposite side is also 5in.

So, we can substitute.

\(\displaystyle P = 11\text{in} + 11\text{in} + 5\text{in} + 5\text{in}\)

\(\displaystyle P = 32\text{in}\)

The formula for perimeter of a rectangle is \(\displaystyle P=2(l + w)\)

To solve for the width we can plug our known values into the equation. 

\(\displaystyle 10=2(2+w)\)

\(\displaystyle 10=4+2w\)

Subtract \(\displaystyle 4\) from both sides

\(\displaystyle 6=2w\)

Divide \(\displaystyle 2\) by both sides

\(\displaystyle 3=w\)

\(\displaystyle P=2l+ 2w\)

We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown. 

\(\displaystyle 54=2l+2(17)\)

\(\displaystyle 54=2l+34\)

Subtract \(\displaystyle 34\) from both sides

\(\displaystyle 54-34=2l+34-34\)

\(\displaystyle 20=2l\)

Divide \(\displaystyle 2\) by both sides

\(\displaystyle \frac{20}{2}=\frac{2l}{2}\)

\(\displaystyle 10=l\)

The formula for perimeter of a rectangle is \(\displaystyle P=2(l + w)\)

To solve for the perimeter we can plug our known values into the equation. 

\(\displaystyle P=2(7+4)\)

\(\displaystyle P=2(11)\)

\(\displaystyle P=22\)

The fence is going around the garden, so this is a perimeter problem. 

\(\displaystyle P=2l+2w\)

\(\displaystyle P=2(6)+2(3)\)

\(\displaystyle P=12+6\)

\(\displaystyle P=18\)

The fence is going around the backyard, so this is a perimeter problem. 

\(\displaystyle P=2l+2w\)

\(\displaystyle P=2(5)+2(4)\)

\(\displaystyle P=10+8\)

\(\displaystyle P=18\)

The area of a triangle is given by \(\displaystyle 0.5\times altitude\times base\).

altitude \(\displaystyle = 10\)

base \(\displaystyle = 3\times a\) = \(\displaystyle 30\)

Therefore:

Area \(\displaystyle = 0.5\times10\times30 = 150\)