Diagonals of an isosceles trapezoid - Examples, Exercises and Solutions

Let's recall that in an isosceles trapezoid, the sum of the two angles on each of the trapezoid's legs equals 180 degrees.

In other words:

$A+C=180$

$B+D=180$

Since angle D is known to us, we can calculate:

$180-50=B$

$130=B$

Given that the trapezoid is isosceles and the angles on both sides are equal, it can be argued that:

$∢C=∢D$

$∢A=∢B$

We know that the sum of the angles of a quadrilateral is 360 degrees.

Therefore we can create the formula:

$∢A+∢B+∢C+∢D=360$

We replace according to the existing data:

$120+120+2x+2x=360$

 $240+4x=360$

$4x=360-240$

$4x=120$

We divide the two sections by 4:

$\frac{4x}{4}=\frac{120}{4}$

$x=30$

True: in every isosceles trapezoid the base angles are equal to each other.

To answer the question, we must know an important rule about isosceles trapezoids:

The sum of the angles that define each of the trapezoidal sides (not the bases) is equal to 180

Therefore:

∢B+∢D=180

3X+X=180

4X=180

X=45

It's important to remember that this is still not the solution, because we were asked for angle B,

Therefore:

3*45 = 135

And this is the solution!

The diagonals of an isosceles trapezoid are always equal to each other,

but they do not necessarily bisect each other.

(Reminder, "bisect" means that they meet exactly in the middle, meaning they are cut into two equal parts, two halves)

For example, the following trapezoid ABCD, which is isosceles, is drawn.

Using a computer program we calculate the center of the two diagonals,

And we see that the center points are not G, but the points E and F.

This means that the diagonals do not bisect.

Since we are given the perimeter of the trapezoid and not the length of CD, we can calculate:

$22=3+3+7+CD$

$22=CD+13$

$22-13=CD$

$9=CD$

To answer the exercise, certain information is needed:

1. In a quadrilateral the sum of the interior angles is 180.

2. The isosceles trapezoid has equal angles.

3. From here it is we know that the sum of the angles adjacent to a side of the trapezoid is 180°.

We turn this conclusion into an exercise:

2y+20+60=180

We add up the relevant angles

2y+80=180

We move the sections:

2y=180-80

2y=100

Divided by 2

y=50

When we substitute Y we get:

2(50)+20=120

And this is the solution!

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:

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