Calculate the perimeter of special quadrilaterals Geometry

The perimeter of a rectangle with width w and height h equals:

P=2\left(w+h\right)

P=2\left(AB+10\right)

Here, the width \overline{AB} is unknown and needs to be determined. Using the Pythagoras theorem, we have:

AB=\sqrt{13^2-5^2}=12

Therefore, the perimeter of the rectangle equals:

P=2\left(AB+5\right) = 2\left(12 + 5\right) =2 \times 17=34

The perimeter of a square with side x equals:

P=4x

Here, the side x is unknown and needs to be determined. Using the Pythagoras theorem, we have:

x^2+x^2=6^2

2x^2=36

x^2=18

x=\sqrt{18} =3\sqrt{2}

Therefore:

P=4x=4\times 3\sqrt{2} = 12\sqrt{2}

The perimeter of a trapezoid is the sum of its sides:

P=AB + BC + CD +DE

We have:

• AB = AE + 11 + 5 = AE+16
• AD=15
• DC = 11
• BC is unknown.

The length of \overline{AE} and \overline{BC} need to be determined.

Using the Pythagoras theorem, we have:

AE=\sqrt{15^2-12^2}=\sqrt{225-144} =9

We have:

BC=\sqrt{12^2+5^2}=\sqrt{144+25} =13

Therefore:

P=\left(9+16\right)+ 13+ 11 + 15=64

The perimeter of a trapezoid is the sum of its sides:

P=AB + BC + CD +DE

We have:

• AB = 12
• AD=12
• DC = 12+5=17
• BC is unknown.

The length of \overline{BC} needs to be determined.

Using the Pythagoras theorem, we have:

BC=\sqrt{12^2+5^2}=\sqrt{169} =13

Therefore:

P=12+13+17+12=54

The perimeter of a parallelogram with base b and side s equals:

P=2\left(b+s\right)

P=2\left(8+5\right)

p=26

The perimeter of a kite with short side length x and long side length y equals:

P=2\left(x+y\right)

Here, we have:

x=7

Using the Pythagoras theorem, we have:

y=\sqrt{12^2+5^2}=13

Therefore, the perimeter of the rectangle equals:

P=2\left(7+13\right) =40

The perimeter of a rhombus with side x equals:

P=4x

Using the Pythagoras theorem, we have:

x=\sqrt{6^2+8^2}= 10

Therefore, the perimeter of the rhombus equals:

P=4\times 10 = 40