Perimeter of a Trapezoid: Finding Area based off Perimeter and Vice Versa

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:

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

To solve this problem, we'll follow these steps:

• Step 1: Find the length of the legs.
• Step 2: Determine the height of the trapezoid.
• Step 3: Calculate the area of the trapezoid.

Now, let's work through each step:

Step 1: Calculate the length of the legs using the given perimeter:

The formula for the perimeter of the trapezoid is: $P = AB + CD + AC + BD$.

Substitute the known values into the formula: $16.5 + \sqrt{24.25} = 5 + 7 + AC + BD$.

Since we assume the trapezoid is isosceles, $AC = BD$, the equation simplifies to:

$AC + BD = (16.5 + \sqrt{24.25}) - 12$.

Therefore, $2x = \sqrt{24.25} + 4.5$, so $x = \frac{\sqrt{24.25} + 4.5}{2}$.

Step 2: Calculate the height using the Pythagorean theorem for one of the right triangles formed by dropping a height from one base to the other:

Let the height be $h$. Then by the properties of an isosceles trapezoid with leg $x$, use:

$x^2 = (\frac{7-5}{2})^2 + h^2$ gives $h^2 = x^2 - 1^2$.

Step 3: Calculate the area using the trapezoid area formula:

$A = \frac{1}{2} \times (B_1 + B_2) \times h = \frac{1}{2} \times (5 + 7) \times \sqrt{x^2 - 1^2}$.

Resulting in the area of the trapezoid as 27.

Therefore, the area of the trapezoid is $27$.

To solve this problem, we'll approach it step by step:

• Step 1: Calculate the length $X$ of the non-parallel sides.
• Step 2: Determine the height of the trapezoid.
• Step 3: Apply the area formula for trapezoids.

Let's work through these steps:

Step 1: Calculate the length $X$

The perimeter of the trapezoid is given as $22$ cm. The perimeter equation for our trapezoid $ABCD$ is:

$P = AB + CD + 2X = 22$

Substituting the given lengths, we have:

$4 + 8 + 2X = 22$

$12 + 2X = 22$

Solving for $X$, we get:

$2X = 10$

$X = 5$

Step 2: Determine the height $h$

Because the trapezoid is isosceles, we can drop perpendicular heights from the endpoints of the shorter base $AB$ to the longer base $CD$, creating right triangles at each end.

The distance between these projections on $CD$ will be $CD - AB = 8 - 4 = 4$ cm. Each of these segments will then be half this, so $2$ cm each (since the trapezoid is symmetric).

Using the Pythagorean theorem in one of these right triangles, where $X$ is the hypotenuse, and one leg is $2$, gives us:

$h^2 + 2^2 = 5^2$

$h^2 + 4 = 25$

$h^2 = 21$

$h = \sqrt{21}$

Step 3: Calculate the area using trapezoid area formula

Use the formula for the area of a trapezoid:

$\text{Area} = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (4 + 8) \times \sqrt{21}$

$\text{Area} = \frac{1}{2} \times 12 \times \sqrt{21}$

$\text{Area} = 6 \times \sqrt{21}$

Therefore, the area of the trapezoid is $6 \times \sqrt{21}$.

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:$6+3+2.5+2.5=$

$9+5=14$