Area of a Trapezoid = ½(a+b)*h

= ½ (2+6) * 4

= ½ (8) * 4

= 4 * 4 = 16

Write the formula to find the area of a trapezoid.

\(\displaystyle A= \frac{1}{2} (b1+b2)h\)

Substitute the givens and evaluate the area.

\(\displaystyle A= \frac{1}{2} (b1+b2)h=\frac{1}{2} (2x+4x)(2x) = 2x^2+4x^2=6x^2\)

The formula for the area of a trapezoid is:

\(\displaystyle A = \frac{b_1 + b_2}{2}* h\) 

We have here the height and one of the bases, plus the area, and we are being asked to find the length of base \(\displaystyle AD\). Plug in known values and solve.

\(\displaystyle 525 = \frac{37 + b_2}{2}* 21\)

\(\displaystyle 25 = \frac{37 + b_2}{2}\)

\(\displaystyle 50 = 37 + b_2\)

\(\displaystyle b_2 = 13\)

Thus, \(\displaystyle AD = 13cm\)

To solve, simply use the formula for the area of a trapezoid.

Substitute 

\(\displaystyle B1=6, B2=7, h=2\)

into the area formula.

Thus,

\(\displaystyle A=\frac{1}{2}(B1+B2)h=\frac{1}{2}(6+7)*2=6+7=13\)

The altitude of the trapezoid splits the trapezoid into two right triangles and a rectangle. Choose one of those right triangles. The base length of that right triangle is necessary to solve for the diagonal.

Using the base lengths of the trapezoid, the length of the base of the right triangle can be solved.  The length of the rectangle is 1 unit.  The longer length of the trapezoid base is 5 units.  

Since there are 2 right triangles bases that lie on the longer base of the trapezoid, we will assume that their base lengths are \(\displaystyle x\) since their lengths are unknown. Combining the lengths of the right triangles and the rectangle, write the equation to solve for the length of the right triangle bases.

\(\displaystyle x+1+x=5\)

\(\displaystyle 2x=4\)

\(\displaystyle x=2\)

The length of each triangular base is 2.  

The diagonal of the trapezoid connects from either bottom angle of the trapezoid to the far upper corner of the rectangle.  This diagonal connects to form another right triangle, where the sum of the solved triangular base and the rectangle length is a leg, and the altitude of the trapezoid is another leg.

\(\displaystyle \textup{Sum of a triangular base and the rectangle length= }2+1=3\)

\(\displaystyle \textup{Altitude of trapezoid= }4\)

Use the Pythagorean Theorem to solve for the diagonal.

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 3^2+4^2=c^2\)

\(\displaystyle 25=c^2\)

\(\displaystyle c=5\)

To find the diagonal, we must subtract the top base from the bottom base:

\(\displaystyle 14-10=4\)

This leaves us with 4, which is the sum of the distance to the left and right of the top base. Taking half of that\(\displaystyle \left(\frac{4}{2}=2\right)\)gives us the length of the distance to only the left side. 

This means that the base of a triangle that includes that diagonal is equal to

\(\displaystyle 2+10=12\).

Since the height is \(\displaystyle 5\), we can solve this problem either using the Pythagorean Theorum or by remembering that this is a special right triangle (\(\displaystyle 5-12-13\) triangle). Therefore, the hypotenuse is \(\displaystyle 13\). 

See the figure below for clarification:

The upper limit of a trapezoid's diagonal length is determined by the lengths of the larger base and larger side because the larger base, larger side and longest diagonal form a triangle, meaning you can use a triangle's side length rule.

Specifically, the non-inclusive upper limit will be the sum of the larger base and larger side.

In this case, \(\displaystyle 10 + 8 = 18\), meaning that the diagonal length can go up to but not including \(\displaystyle 18\).

Write the formula used to find the area of a trapezoid.

\(\displaystyle A=\frac{1}{2}h(b1+b2)\)

Substitute the given information to the formula and solve for the unknown base.

\(\displaystyle 10=\frac{1}{2}(1)(1+b2)\)

\(\displaystyle 20=1+b2\)

\(\displaystyle b2=19\)

We know that all three horizontal lines are parallel to one another. By definition, we can set up a ratio between the lengths of the sides provided to us in the question and the lengths of the two parallel lines:

\(\displaystyle \frac{AE}{ED} = \frac{EF}{DC}\)

Once we substitute the given information, we get 

\(\displaystyle \frac{4}{6}=\frac{EF}{18}\)

We cross multiply to solve for EF

\(\displaystyle EF = \frac{(4)(18)}{6}\)

We can start this problem by drawing the height and labeling the lengths with the given values.

When we do this, we can see that we have drawn a triangle inside the paralellogram including \(\displaystyle \measuredangle A\). Because we know the lengths of two sides of this triangle, we can use trigonometry to find \(\displaystyle \measuredangle A\).

With respect to \(\displaystyle \measuredangle A\), we know the values of the opposite and hypotenuse sides of the triangle. Thus, we can use the sine function to solve for \(\displaystyle \measuredangle A\).

\(\displaystyle \sin \left (\measuredangle A\right ) = \frac{\textup{opposite}}{\textup{hypotenuse}}\)

\(\displaystyle \sin \left (\measuredangle A\right ) = \frac{12}{15}= 0.8\)

\(\displaystyle \measuredangle A = \arcsin\left ( 0.8\right ) = 53.1^{\circ}\)