11.03 Properties of rectangles, rhombuses, and squares

Solving for missing sides or angles in rectangles, rhombuses, and squares

Once we are familiar with the properties of shapes, we can use them as extra conditions when finding unknown lengths or angles.

Here are some properties that we will find useful.

• A rectangle is a parallelogram with four right angles. The diagonals of a rectangle are congruent.
• A rhombus is a parallelogram with four congruent sides. The diagonals of a rhombus are perpendicular, and each diagonal bisects a pair of opposite angles.
• A square is a parallelogram with four right angles and four congruent sides. Squares have all the properties of a rectangle and all the properties of a rhombus.

 

Worked example

Question 1

Consider the square $ABCD$ABCD. If $AC=14$AC=14 m, find $BD$BD.

Think: Squares have all the properties of a rectangle. We can use the fact that the diagonals of a rectangle are congruent.

Do: From the properties of rectangles we know that $\overline{AC}\cong\overline{BD}$AC≅BD. If two segments are congruent, then they have the same length, so $AC=BD=14$AC=BD=14 m.

 

Using algebraic expressions

It can be useful to express the side lengths or angle measures of a shape in terms of an unknown variable like $x$x. Similarly to the approach above, we can use the properties of the shape to find the value of $x$x.

 

Worked example

Question 2

Consider the rhombus $PQRS$PQRS. If $PS=5x-4$PS=5x−4 and $PQ=2x+8$PQ=2x+8, find the value of $x$x.

Think: A rhombus has four congruent sides. This means that $\overline{PS}\cong\overline{PQ}$PS≅PQ, and so $PS=PQ$PS=PQ.

Do: We set up an equation in terms of $x$x using the equality $PS=PQ$PS=PQ.

$PS$PS	$=$=	$PQ$PQ
$5x-4$5x−4	$=$=	$2x+8$2x+8	(Substitute the expression for each side length)
$3x$3x	$=$=	$12$12	(Collect the like terms on either side of the equation)
$x$x	$=$=	$4$4	(Divide both sides by $3$3 to isolate $x$x)

So the value of $x$x is $4$4.

Reflect: Now that we know $x$x, we can determine the side length of the rhombus. When $x=4$x=4, then $PS=5\times4-4=16$PS=5×4−4=16 and $PQ=2\times4+8=16$PQ=2×4+8=16. So $PS=PQ=16$PS=PQ=16.

 

Practice questions

Question 3

Question 4

Question 5

 

 

Prove the properties of rectangles, rhombuses, and squares

A parallelogram is defined as a quadrilateral whose opposite sides are parallel. If we add more conditions onto this definition we will be able to describe several other shapes.

A rectangle is a parallelogram with four right angles. Rectangles have all the properties of parallelograms, but not all parallelograms are rectangles.

Rectangle

We summarize the properties of rectangles as theorems below.

A rhombus is a parallelogram with all four sides congruent. Rhombi have all the properties of parallelograms, but not all parallelograms are rhombi.

Rhombus

A square is a parallelogram with four right angles and four congruent sides. We can think of a square as being like a rhombus with four right angles, or a rectangle with four sides congruent. That is, squares share all the properties of both rhombi and rectangles.

Square

We summarize the properties of rhombi and squares as theorems below.

 

Exploration

Consider the following theorem.

If a parallelogram is a rhombus, then it is composed of four congruent triangles.

To begin proving the above theorem, we can first label the rhombus by its set of vertices, $ABCD$ABCD, and include another vertex $E$E that is the intersection of the two diagonals.

Rhombus $ABCD$ABCD with $E$E the intersection of the diagonals.

 

From the definition of a rhombus, we know that all four sides are congruent. We can write this as a geometric statement, $\overline{AB}\cong\overline{BC}\cong\overline{CD}\cong\overline{DA}$AB≅BC≅CD≅DA.

Next, we can use the fact that the diagonals of a parallelogram bisect each other to state that $\overline{AE}\cong\overline{EC}$AE≅EC and $\overline{DE}\cong\overline{EB}$DE≅EB.

Finally, we notice that the segment $\overline{AE}$AE is common to $\triangle AEB$△AEB and $\triangle AED$△AED. Clearly a segment is congruent to itself, so we can use the reflexive property of congruence we can state that $\overline{AE}\cong\overline{AE}$AE≅AE, and by the same argument we also have $\overline{BE}\cong\overline{BE}$BE≅BE, $\overline{CE}\cong\overline{CE}$CE≅CE, and $\overline{DE}\cong\overline{DE}$DE≅DE.

Hence, $\triangle AEB$△AEB, $\triangle BEC$△BEC, $\triangle CED$△CED, and $\triangle DEA$△DEA have side-side-side congruence. This means that rhombus $ABCD$ABCD is composed of four congruent triangles.

We can formalize the above steps into a two-column proof where each line contains a geometric statement in the left column and a corresponding reason in the right column.

Two-column proof

Given the rhombus $ABCD$ABCD, show that it is composed of four congruent triangles.
Statements	Reasons
$ABCD$ABCD is a rhombus	Given
$ABCD$ABCD is a parallelogram $\overline{AB}\cong\overline{BC}\cong\overline{CD}\cong\overline{DA}$AB≅BC≅CD≅DA	Definition of a rhombus
$\overline{AE}\cong\overline{EC}$AE≅EC and $\overline{DE}\cong\overline{EB}$DE≅EB	If a quadrilateral is a parallelogram, then its diagonals bisect each other.
$\overline{AE}\cong\overline{AE}$AE≅AE, $\overline{BE}\cong\overline{BE}$BE≅BE, $\overline{CE}\cong\overline{CE}$CE≅CE, and $\overline{DE}\cong\overline{DE}$DE≅DE	Reflexive property of congruence
$\triangle AEB\cong\triangle CEB\cong\triangle CED\cong\triangle AED$△AEB≅△CEB≅△CED≅△AED	Side-side-side congruence theorem

The final line contains the statement that the rhombus is composed of four congruent triangles, which is what we wanted to show.

We may be able to change the order of some lines without changing the validity of the proof, while there are other lines that require certain preceding statements. For example, this line

$\overline{AE}\cong\overline{AE}$AE≅AE, $\overline{BE}\cong\overline{BE}$BE≅BE, $\overline{CE}\cong\overline{CE}$CE≅CE, and $\overline{DE}\cong\overline{DE}$DE≅DE

Reflexive property of congruence

does not depend on any other line in the proof. We can introduce this line at any point, so long as it appears before the final line. This is because the statement $\triangle AEB\cong\triangle CEB\cong\triangle CED\cong\triangle AED$△AEB≅△CEB≅△CED≅△AED depends on the preceding statements that the corresponding sides of each triangle are congruent.

Practice questions

Question 6

Question 7