**Facts about Quadrilaterals - Interior Angles**

Every quadrilateral has certain properties, depending on the “species.” Here are some useful facts concerning these four-sided plane figures.

**Sum Of Measures Of Interior Angles**

No matter what the shape of a quadrilateral, as long as all four sides are straight line segments of positive and finite length, and as long as all four vertices lie in the same plane, the sum of the measures of the interior angles is 360° (2 *π* rad). Figure 3-7 shows an example of an irregular quadrilateral. The interior angles are denoted *w, x, y* , and *z* . In this example, angle *w* measures more than 180° ( *π* rad). If you use your imagination, you might call this type of quadrilateral a “boomerang,” although this is not an official geometric term.

**Fig. 3-7** . In any plane quadrilateral, the sum of the measures of the interior angles *w, x, y* , and *z* is 360° (2 *π* rad).

**Parallelogram Diagonals**

Suppose we have a parallelogram defined by four points *P, Q, R* , and *S* . Let *D* be a line segment connecting *P* and *R* as shown in Fig. 3-8A. Then *D* is a *minor diagonal* of the parallelogram, and the following triangles defined by *D* are congruent:

**Fig. 3-8** . Triangles defined by the minor diagonal ( **A** ) or the major diagonal ( **B** ) of a parallelogram are congruent.

Let *E* be a line segment connecting *Q* and *S* (Fig. 3-8B). Then *E* is a *major diagonal* of the parallelogram, and the following triangles defined by *E* are congruent:

**Bisection Of Parallelogram Diagonals**

Suppose we have a parallelogram defined by four points *P, Q, R* , and *S* . Let *D* be the diagonal connecting *P* and *R* ; let *E* be the diagonal connecting *Q* and *S* (Fig. 3-9). Then *D* and *E* bisect each other at their intersection point *T* . In addition, the following pairs of triangles are congruent:

The converse of the foregoing is also true: If we have a plane quadrilateral whose diagonals bisect each other, then that quadrilateral is a parallelogram.

**Fig. 3-9** . The diagonals of a parallelogram bisect each other.

**Rectangle**

Suppose we have a parallelogram defined by four points *P, Q, R* , and *S* . Suppose any of the following statements is true for angles in degrees:

*∠QRS* = 90° = *π* /2 rad

*∠RSP* = 90° = *π* /2 rad

*∠SPQ* = 90° = *π* /2 rad

*∠PQR* = 90° = *π* /2 rad

Then all four interior angles are right angles, and the parallelogram is a *rectangle* : a four-sided plane polygon whose interior angles are all congruent. The converse of this is also true: If a quadrilateral is a rectangle, then any given interior angle is a right angle. Figure 3-10 shows an example of a parallelogram *PQRS* in which *∠QRS* = 90° = *π* /2 rad. Because one angle is a right angle and opposite pairs of sides are parallel, all four of the angles must be right angles.

**Fig. 3-10** . If a parallelogram has one right interior angle, then the parallelogram is a rectangle.

**Rectangle Diagonals**

Suppose we have a parallelogram defined by four points *P, Q, R* , and *S* . Let *D* be the diagonal connecting *P* and *R* ; let *E* be the diagonal connecting *Q* and *S* . Let the length of *D* be denoted by *d* ; let the length of *E* be denoted by *e* (Fig. 3-11). If *d* = *e* , then the parallelogram is a rectangle. The converse is also true: if a parallelogram is a rectangle, then *d* = *e* . Thus, a parallelogram is a rectangle if and only if its diagonals have equal lengths.

**Fig. 3-11** . The diagonals of a rectangle have equal length.

**Rhombus Diagonals**

Suppose we have a parallelogram defined by four points *P, Q, R* , and *S* . Let *D* be the diagonal connecting *P* and *R* ; let *E* be the diagonal connecting *Q* and *S* . If *D* is perpendicular to *E* , then the parallelogram is a rhombus (Fig. 3-12). The converse is also true: If a parallelogram is a rhombus, then *D* is perpendicular to *E* . A parallelogram is a rhombus if and only if its diagonals are perpendicular.

**Fig. 3-12** . The diagonals of a rhombus are perpendicular.

**Trapezoid Within Triangle**

Suppose we have a triangle defined by three points *P, Q* , and *R* . Let *S* be the midpoint of side *PR* , and let *T* be the midpoint of side *PQ* . Then line segments *ST* and *RQ* are parallel, and the figure defined by *STQR* is a trapezoid (Fig. 3-13). In addition, the length of line segment *ST* is half the length of line segment *RQ* .

**Fig. 3-13** . A trapezoid is formed by “chopping off” the top of a triangle.

**Median Of A Trapezoid**

Suppose we have a trapezoid defined by four points *P, Q, R* , and *S* . Let *T* be the midpoint of side *PS* , and let *U* be the midpoint of side *QR*. Line segment *TU* is called the *median* of trapezoid *PQRS*. The median of a trapezoid is always parallel to both the base and the top, and always splits the trapezoid into two other trapezoids. That is, polygons *PQUT* and *TURS* are both trapezoids (Fig. 3-14). In addition, the length of line segment *TU* is half the sum of the lengths of line segments *PQ* and *SR*. That is, the length of *TU* is equal to the average, or *arithmetic mean* , of the lengths of *PQ* and *SR*.

**Fig. 3-14** . The median of a trapezoid.

**Median With Transversal**

Look again at Fig. 3-14. Suppose *L* is a transversal line that crosses both the top of the large trapezoid (line segment *PQ* ) and the bottom (line segment *SR* ). Then *L* also crosses the median, line segment *TU* . Let *A* be the point at which *L* crosses *PQ* , let *B* be the point at which *L* crosses *TU* , and let *C* be the point at which *L* crosses *SR* . Then the lengths of line segments *AB* and *BC* are equal.

There is a second fact that should also be mentioned. Again, refer to Fig. 3-14. Suppose *PQRS* is a trapezoid, with sides *PQ* and *RS* parallel. Suppose *TU* is a line segment parallel to both *PQ* and *RS* , and that intersects both of the non-parallel sides of the trapezoid, that is, sides *PS* and *QR* . Let *L* be a transversal line that crosses all three parallel line segments *PQ, TU* , and *RS* , at the points *A, B* , and *C* respectively, as shown. In this scenario, if line segments *AB* and *BC* are equally long, then line segment *TU* is the median of the large trapezoid *PQRS*.

**Fig. 3-14** . The median of a trapezoid, also showing a transversal line.

**Facts about Quadrilaterals Practice Problems**

**PROBLEM 1**

Suppose a particular plane figure has diagonals that are the same length, and in addition, they intersect at right angles. What can be said about this polygon?

**SOLUTION 1**

From the above rules, this polygon must be a rectangle, because its diagonals are the same length. But it must also be a rhombus, because its diagonals are perpendicular to each other. There’s only one type of polygon that can be both a rectangle and a rhombus, and that is a square. A square is a rhombus in which both pairs of opposite interior angles happen to have the same measure. A square is also a rectangle in which both pairs of opposite sides happen to be equally long.

**PROBLEM 2**

Suppose a sign manufacturing company gets tired of making rectangular billboards, and decides to put up a trapezoidal billboard instead. The top and the bottom of the billboard are horizontal, but neither of the other sides is vertical. The big sign measures 20 meters across the top edge, and 30 meters across the bottom edge. Two different companies want to advertise on the billboard, and both of them insist on having portions of equal height. What is the length of the line that divides the spaces allotted to the two advertisements? Does this represent a fair division of the sign area?

**SOLUTION 2**

The line segment that divides the two portions is the median of the sign. Its length, therefore, is the average of 20 meters and 30 meters, which, as you should be able to guess right away, is 25 meters. Whether or not this represents a fair split of the sign area can be debated. The advertiser on the bottom gets more area than the advertiser on the top, but the ad on top is likely to be the one that drivers in passing cars and trucks look at first. By the time drivers are finished with the ad on the top, they might be passing the sign.

Practice problems for these concepts can be found at: Quadrilaterals Practice Test.