Perimeter of a Trapezoid - Examples, Exercises and Solutions

To find the perimeter we will add all the sides:

$4+5+9+6=9+9+6=18+6=24$

In order to calculate the perimeter of the trapezoid we must add together the measurements of all of its sides:

7+10+7+12 =

36

And that's the solution!

First, we calculate the long base from the existing data:

Multiply the short base by 1.5:

$5\times1.5=7.5$

Now we will add up all the sides to find the perimeter:

$2+5+3+7.5=7+3+7.5=10+7.5=17.5$

We replace the data in the formula to find the perimeter:

$25=4+7+11+x$

$25=22+x$

$25-22=x$

$3=x$

Since this is an isosceles trapezoid, and the two legs are equal, we can claim that:

$AB=CD=6$

Now let's add all the sides together to find the perimeter

$6+6+10+12=$

$12+22=34$

To calculate the perimeter, we add up all the sides:

$10+x+(6+x)+(x+1)$

Now, given that x equals 3, we substitute in the appropriate places:

$10+3+(6+3)+(3+1)=$

$10+3+9+4=$

$13+13=26$

Since ABCD is a trapezoid, one can determine that:

$AD=BC=7$

Thus the formula to find the area will be

$$$S_{ABCD}=\frac{(AB+DC)\times h}{2}$

Since we are given the perimeter of the trapezoid, we can find$AB+DC$

$P_{ABCD}=7+AB+7+DC$

$34=14+AB+DC$

$34-14=AB+DC$

$20=AB+DC$

Now we will place the data we obtained into the formula in order to calculate the area of the trapezoid:

$S=\frac{20\times5}{2}=\frac{100}{2}=50$

We can find the height BE by calculating the trapezoidal area formula:

$S=\frac{(AB+CD)}{2}\times h$

We replace the known data: $9=\frac{(3+6)}{2}\times BE$

We multiply by 2 to get rid of the fraction:

$9\times2=9\times BE$

$18=9BE$

We divide the two sections by 9:

$\frac{18}{9}=\frac{9BE}{9}$

$2=BE$

If we draw the height from A to CD we get a rectangle and two congruent triangles. That is:

$AF=BE=2$

$AB=FE=3$

$ED=CF=1.5$

Now we can find one of the legs through the Pythagorean theorem.

We focus on triangle BED:

$BE^2+ED^2=BD^2$

We replace the known data:

$2^2+1.5^2=BD^2$

$4+2.25=DB^2$

$6.25=DB^2$

We extract the root:

$\sqrt{6.25}=DB$

$2.5=DB$

Now that we have found DB, it can be argued that:

$AC=BD=2.5$

We calculate the perimeter of the trapezoid:$6+3+2.5+2.5=$

$9+5=14$

To find the perimeter of the trapezoid, all its sides must be added:

We will focus on finding the bases.

To find GF we use the Pythagorean theorem: $A^2+B^2=C^2$in the triangle AFG

We replace

$3^2+GF^2=5^2$

We isolate GF and solve:

$9+GF^2=25$

$GF^2=25-9=16$

$GF=4$

We perform the same process with the side DB of the triangle ABD:

$8^2+DB^2=17^2$

$64+DB^2=289$

$DB^2=289-64=225$

$DB=15$

We start by finding FB:

$FB=AB-AF=17-5=12$

Now we reveal EF and CB:

$GF=GE=4$

$DB=DC=15$

This is because in an isosceles triangle, the height divides the base into two equal parts so:

$EF=GF\times2=4\times2=8$

$CB=DB\times2=15\times2=30$

All that's left is to calculate:

$30+8+12\times2=30+8+24=62$

We calculate the perimeter of the left trapezoid:

$P=10+12+x+y$

$P=22+x+y$

We calculate the perimeterof the right trapezoid:

$P=x+5+17+\frac{y}{3}+\frac{2y}{3}$

$P=x+22+\frac{2y+y}{3}$

$P=x+22+\frac{3y}{3}$

$P=x+22+y$

It can be seen that the two perimeters are identical to each other.

We calculate the perimeter of the left trapezoid:

$P=6+10+7+\frac{5}{2}x+5$

$P=28+\frac{5}{2}x$

We calculate the perimeter of the right trapezoid:

$P=7+x+x+16+\frac{x}{2}+5$

$P=2\frac{1}{2}x+28$

$P=\frac{5}{2}x+28$

The perimeters of the two trapezoids are equal to each other.