Diagonal of a Trapezoid – Formulas and Examples

We can use any definition for the sides of the trapezoid. The important thing when using the law of cosines is that the given angle is the angle that lies between the two given sides.

In this case, it is, so we can use the following definitions:

• Side 1, $latex a=10$ m
• Side 2, $latex d=12$ m
• Angle, $latex \beta= 60$°

And we replace these values in the formula:

$latex d_{1}=\sqrt{{{a}^2}+{{d}^2}-2ad~\cos(\beta)}$

$$d_{1}=\sqrt{{{10}^2}+{{12}^2}-2(10)(12)~\cos(60°)}$$

$latex d_{1}=\sqrt{100+144-240(0.5)}$

$latex d_{1}=\sqrt{244-120}$

$latex d_{1}=\sqrt{124}$

$latex d_{1}=11.14$

The length of the diagonal of the trapezoid is 11.14 m.

We can recognize the following information:

Side 1, $latex a=6$ m

Side 2, $latex d=8$ m

Angle, $latex \beta=60$°

Now, we use the diagonal formula with these values:

$latex d_{1}=\sqrt{{{a}^2}+{{d}^2}-2ad~\cos(\beta)}$

$$d_{1}=\sqrt{{{6}^2}+{{8}^2}-2(6)(8)~\cos(60°)}$$

$latex d_{1}=\sqrt{36+64-96(0.5)}$

$latex d_{1}=\sqrt{100-48}$

$latex d_{1}=\sqrt{52}$

$latex d_{1}=7.2$

The length of the diagonal of the trapezoid is 7.2 m.

We can identify the following information:

Side 1, $latex a=9$ m

Side 2, $latex d=15$ m

Angle, $latex \beta=45$°

Now, we use the diagonal formula with these values:

$latex d_{1}=\sqrt{{{a}^2}+{{d}^2}-2ad~\cos(\beta)}$

$$d_{1}=\sqrt{{{9}^2}+{{15}^2}-2(9)(15)~\cos(45°)}$$

$latex d_{1}=\sqrt{81+225-270(0.707)}$

$latex d_{1}=\sqrt{306-190.89}$

$latex d_{1}=\sqrt{115.11}$

$latex d_{1}=10.73$

The length of the diagonal of the trapezoid is 10.73 m.

We have the following data:

• Side 1, $latex a=20$ m
• Side 2, $latex d=22$ m
• Angle, $latex \beta =30$°

We use these values in the diagonal formula:

$latex d_{1}=\sqrt{{{a}^2}+{{d}^2}-2ad~\cos(\beta)}$

$$d_{1}=\sqrt{{{20}^2}+{{22}^2}-2(20)(22)~\cos(30°)}$$

$latex d_{1}=\sqrt{400+484-880(0.866)}$

$latex d_{1}=\sqrt{884-762.1}$

$latex d_{1}=\sqrt{121.9}$

$latex d_{1}=11.04$

The length of the diagonal of the trapezoid is 11.04 m.