Geometry · G.CO.C.11 · 5.G.B.3

Rectangles, rhombuses and squares

A parallelogram already has equal opposite sides and diagonals that cut each other in half. What more do you get when every corner is a right angle, or every side is the same length?

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Properties of a parallelogramGeometry

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Rectangles, rhombuses and squares

Rhombuses, rectangles and squares Shapes in more than one category, and quadrilaterals in none ▾ Show all 7 (2 more)

Why this exists

Every parallelogram comes with a set of guarantees: its opposite sides are parallel and congruent, its opposite angles are congruent, its neighboring angles add to 180^\circ, and its two diagonals bisect each other. Those facts hold for every parallelogram, however slanted it is.

Ask the parallelogram for more and it gives more back. Make every corner a right angle, and the two diagonals come out the same length. Make every side the same length, and the diagonals cross at right angles and cut every corner exactly in half. Ask for both at once and you have a square, which has all of these together.

These extra facts are what let you find a length or an angle inside a rectangle or a rhombus from very little information: one side and one diagonal, or one angle between a side and a diagonal. They are also the facts that the tests in proving a parallelogram is a rectangle or a rhombus turn round the other way.

Definition

All three shapes are parallelograms with something extra.

A rectangle is a parallelogram with four right angles.

A rhombus is a parallelogram whose four sides are all congruent.

A square is a parallelogram that is both a rectangle and a rhombus: four right angles and four congruent sides.

For a quadrilateral ABCD, named in order around it, “rectangle” means that each of \angle A, \angle B, \angle C and \angle D measures 90^\circ, and “rhombus” means AB = BC = CD = DA.

These definitions are inclusive. A square has four right angles, so it is a rectangle. It has four congruent sides, so it is a rhombus. Every one of the three is a parallelogram. So anything proved for every rectangle is true of every square, and anything proved for every rhombus is true of every square as well. How the whole family nests is drawn out in the quadrilateral family tree and in shapes in more than one category.

The word “parallelogram” could be left out of both definitions without changing which shapes they describe. A quadrilateral with four right angles has \overline{AD} and \overline{BC} both perpendicular to \overline{AB}, and two lines perpendicular to the same line are parallel; the same goes for the other pair of sides, so it is a parallelogram. A quadrilateral with four congruent sides has both pairs of opposite sides congruent, which makes it a parallelogram by the tests for a parallelogram.

Intuition

Fold the shapes. A rectangle folds onto itself along the line through the midpoints of two opposite sides, and it has two such lines. Take the fold that runs up the middle: it swaps A with B and D with C, so it lays the diagonal \overline{AC} exactly on top of the diagonal \overline{BD}. Two segments that fold onto each other are the same length. That is the reason a rectangle’s diagonals are equal.

Figure 1Where each shape folds onto itself. The square has the rectangle’s two folds and the rhombus’s two.

A rhombus folds along each of its diagonals instead. Fold along \overline{AC}: the point B lands on D, because AB = AD and CB = CD. The fold swaps the two halves of the angle at A, so the diagonal cuts that angle in half, and the same at C. And the segment \overline{BD} folds onto itself end to end, so the fold line meets it at its midpoint and at right angles. That is the reason a rhombus’s diagonals are perpendicular.

A square folds both ways, four lines in all, and a quarter turn about its center also carries it onto itself. A parallelogram that is neither a rectangle nor a rhombus has no line of symmetry at all, only the half turn about its center, which is where its own fact about the diagonals bisecting each other comes from. Line symmetry and rotational symmetry say more about folds and turns.

A fold is a picture, not a proof. The proofs below use congruent triangles, and they say exactly the same thing.

Results

Inherited properties. A rectangle, a rhombus and a square are all parallelograms, so each has every property of a parallelogram: opposite sides parallel and congruent, opposite angles congruent, neighboring angles supplementary, diagonals that bisect each other.

Rectangle Diagonals Theorem. The diagonals of a rectangle are congruent, AC = BD. If they meet at E, the four halves are all equal:

AE = BE = CE = DE

The second statement follows from the first because the diagonals also bisect each other, so each half is half of the same length. The point E is therefore the same distance from all four corners, and the diagonals cut the rectangle into four isosceles triangles.

Figure 2The diagonals of a rectangle are congruent, so the four halves from E are all equal.

The diagonal is the hypotenuse of a right triangle whose legs are two sides, so by the Pythagorean theorem, for a rectangle of length l and width w:

AC = BD = \sqrt{l^2 + w^2}

Rhombus Diagonals Theorem. The diagonals of a rhombus are perpendicular, \overline{AC} \perp \overline{BD}, and each diagonal bisects the two angles at its ends:

\begin{aligned} \angle BAC &\cong \angle DAC \\ \angle BCA &\cong \angle DCA \\ \angle ABD &\cong \angle CBD \\ \angle ADB &\cong \angle CDB \end{aligned}

Figure 3The diagonals of a rhombus cross at right angles and cut each corner angle in half.

Since the diagonals also bisect each other, each one is the perpendicular bisector of the other, and together they cut the rhombus into four congruent right triangles. Each triangle has legs half of one diagonal and half of the other, and a side of the rhombus as its hypotenuse. So for diagonals d_1 and d_2 and side s:

s^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2

The four right triangles each have area \tfrac12 \cdot \tfrac{d_1}{2} \cdot \tfrac{d_2}{2}, so the rhombus has area

\text{Area} = \frac{d_1 d_2}{2}

Square. A square is a rectangle and a rhombus, so it has every result above. Its diagonals are congruent, perpendicular and bisect each other, and each diagonal splits a right angle into two 45^\circ angles. For a square of side s, the diagonal is

d = s\sqrt{2}

Figure 4A square has every property of both. Each diagonal splits a right angle into two 45^\circ angles.

Put together, this is what each shape always has:

Every parallelogram: opposite sides parallel and congruent, opposite angles congruent, diagonals that bisect each other.
Every rectangle: all of that, plus four right angles and congruent diagonals, and at least two lines of symmetry.
Every rhombus: all of the parallelogram’s facts, plus four congruent sides, perpendicular diagonals, and diagonals that bisect the corner angles, and at least two lines of symmetry.
Every square: everything in the three lines above, and four lines of symmetry.

A shape can have more than its kind promises. A particular parallelogram may have congruent diagonals, for instance, but then it is a rectangle.

Why it is true

The rectangle’s diagonals. Look at the two triangles \triangle DAB and \triangle CBA in the rectangle figure. They share the side \overline{AB}. The sides \overline{AD} and \overline{BC} are opposite sides of a parallelogram, so they are congruent. The angles \angle DAB and \angle CBA are both right angles, so they are congruent. That is two sides and the angle between them, so the triangles are congruent by SAS (proving triangles congruent):

\triangle DAB \cong \triangle CBA

Corresponding sides of congruent triangles are congruent, and \overline{DB} corresponds to \overline{CA}. So BD = AC.

The rhombus’s diagonals are perpendicular. Let the diagonals meet at E and look at \triangle ABE and \triangle ADE. AB = AD, because all four sides of a rhombus are congruent. BE = DE, because the diagonals of a parallelogram bisect each other. \overline{AE} is shared. So the triangles are congruent by SSS, and the corresponding angles \angle AEB and \angle AED are congruent. Those two angles sit side by side along the straight line BD, so they form a linear pair and add to 180^\circ. Two equal angles that add to 180^\circ are each 90^\circ, so \overline{AC} \perp \overline{BD}.

The rhombus’s diagonals bisect its angles. Look at \triangle ABC and \triangle ADC. AB = AD and CB = CD, all being sides of the rhombus, and \overline{AC} is shared. By SSS the triangles are congruent, so \angle BAC \cong \angle DAC and \angle BCA \cong \angle DCA: the diagonal \overline{AC} bisects the angles at A and at C. The same argument with \triangle BAD and \triangle BCD, which share \overline{BD}, shows that \overline{BD} bisects the angles at B and at D.

The square. A square is a rectangle, so its diagonals are congruent, and a rhombus, so they are perpendicular and bisect its angles. Half of a right angle is 45^\circ. For the diagonal, the triangle \triangle ABC has a right angle at B and two legs of length s, so

\begin{aligned} AC^2 &= s^2 + s^2 = 2s^2 \\ AC &= s\sqrt{2} \end{aligned}

Every step leaned on a parallelogram property (opposite sides congruent, diagonals bisecting each other) or on the extra condition that makes the shape special. Nothing depended on the particular shape drawn, so the results hold for every rectangle and every rhombus.

Worked examples

Example 1

A half diagonal and a whole one

Rectangle PQRS has diagonals meeting at T, with PT = 4x - 3 and QS = 6x + 2. Find x, PT and the length of each diagonal.

Solution

The two lengths given are not the same kind of length: PT is half of the diagonal \overline{PR}, and QS is a whole diagonal. Since the diagonals of a parallelogram bisect each other, PR = 2 \cdot PT = 2(4x - 3). Since the diagonals of a rectangle are congruent, QS = PR:

\begin{aligned} 6x + 2 &= 2(4x - 3) \\ 6x + 2 &= 8x - 6 \\ 8 &= 2x \\ x &= 4 \end{aligned}

Then PT = 4(4) - 3 = 13, so PR = 26, and QS = 6(4) + 2 = 26.

Check. Both diagonals are 26, as a rectangle requires, and PT is half of one of them.

Example 2

Angles cut off by the diagonals of a rectangle

Rectangle ABCD has diagonals meeting at E, and m\angle BAC = 34^\circ. Find m\angle EBA, m\angle AEB, m\angle BEC and m\angle ECB.

Figure 5One angle between a side and a diagonal, and the rest of the figure follows.

Solution

In a rectangle the four half-diagonals are equal, so EA = EB and \triangle AEB is isosceles. The angles opposite its equal sides are equal, by the base angles theorem. \angle EAB is the same angle as \angle BAC, since E lies on \overline{AC}, so

m\angle EBA = m\angle EAB = 34^\circ

The angles of a triangle add to 180^\circ (the triangle angle sum), so

m\angle AEB = 180^\circ - 34^\circ - 34^\circ = 112^\circ

A, E and C lie on one line, so \angle AEB and \angle BEC form a linear pair:

m\angle BEC = 180^\circ - 112^\circ = 68^\circ

\triangle ABC has its right angle at B, so its two other angles add to 90^\circ, and \angle ECB is the same angle as \angle ACB:

m\angle ECB = 90^\circ - 34^\circ = 56^\circ

Check. \triangle BEC is isosceles too, with EB = EC, so m\angle EBC = 56^\circ as well, and 56 + 56 + 68 = 180.

Example 3

A brace across a gate

A rectangular gate frame is 3.5 m wide and 1.2 m tall. Two braces run along its diagonals and are bolted together where they cross. How long is each brace, and how far is the bolt from each corner?

Solution

The braces are the diagonals of a rectangle, which are congruent, so one length answers for both. Each diagonal is the hypotenuse of a right triangle whose legs are the width and the height:

\begin{aligned} d^2 &= 3.5^2 + 1.2^2 \\ &= 12.25 + 1.44 = 13.69 \\ d &= 3.7 \end{aligned}

Each brace is 3.7 m long. The bolt is where the diagonals cross, which is the midpoint of each, and all four half-diagonals of a rectangle are equal. So the bolt is 3.7 \div 2 = 1.85 m from every corner.

Example 4

Angles in a rhombus

Rhombus PQRS has diagonals meeting at T, and m\angle QPR = 32^\circ. Find m\angle QPS, m\angle PQR, m\angle PQS, m\angle PTQ and m\angle PRQ.

Figure 6A rhombus with one angle given between a side and a diagonal.

Solution

The diagonal \overline{PR} bisects the angle at P, so the angle at P is twice 32^\circ:

m\angle QPS = 2 \cdot 32^\circ = 64^\circ

\angle QPS and \angle PQR are neighboring angles of a parallelogram, so they are supplementary:

m\angle PQR = 180^\circ - 64^\circ = 116^\circ

The diagonal \overline{QS} bisects the angle at Q, so m\angle PQS = 116^\circ \div 2 = 58^\circ.

The diagonals of a rhombus are perpendicular, so m\angle PTQ = 90^\circ.

Check. In \triangle PTQ, 32 + 58 + 90 = 180.

For the last angle, \triangle PQR has PQ = QR, two sides of the rhombus, so it is isosceles and its base angles are equal: m\angle PRQ = m\angle QPR = 32^\circ. The other route agrees: the angle at R is opposite the angle at P, so it is 64^\circ, and the diagonal \overline{PR} cuts it in half.

Example 5

A rhombus from its diagonals

A rhombus has diagonals of length 18 and 24. Find its side, its perimeter and its area.

Solution

The diagonals bisect each other at right angles, so they cut the rhombus into four right triangles, each with legs 18 \div 2 = 9 and 24 \div 2 = 12 and a side of the rhombus as its hypotenuse:

\begin{aligned} s &= \sqrt{9^2 + 12^2} \\ &= \sqrt{81 + 144} \\ &= \sqrt{225} = 15 \end{aligned}

All four sides are equal, so the perimeter is 4 \cdot 15 = 60.

Each of the four triangles has area \tfrac12 \cdot 9 \cdot 12 = 54, so the rhombus has area 4 \cdot 54 = 216. The formula gives the same: \tfrac12 \cdot 18 \cdot 24 = 216.

Example 6

A rhombus with unknowns

Rhombus ABCD has AB = 3y - 4 and BC = y + 10, and its diagonal \overline{AC} has length 16. Find y, the side length, the length of \overline{BD} and the area.

Solution

The sides of a rhombus are all congruent, so AB = BC:

\begin{aligned} 3y - 4 &= y + 10 \\ 2y &= 14 \\ y &= 7 \end{aligned}

The side is 3(7) - 4 = 17, and 7 + 10 = 17 agrees.

Let the diagonals meet at E. They bisect each other, so AE = 16 \div 2 = 8, and they are perpendicular, so \triangle AEB has a right angle at E with hypotenuse AB = 17:

\begin{aligned} BE^2 &= 17^2 - 8^2 \\ &= 289 - 64 = 225 \\ BE &= 15 \end{aligned}

So BD = 2 \cdot 15 = 30, and the area is \tfrac12 \cdot 16 \cdot 30 = 240.

Example 7

A square’s diagonal, both ways round

(a) A square has side 6. Find the length of a diagonal, the distance from the center to each corner, and the angle a diagonal makes with a side.

(b) A square has diagonal 10. Find its side and its area.

Solution

(a) A diagonal cuts the square into two right triangles with both legs 6:

\begin{aligned} d^2 &= 6^2 + 6^2 = 72 \\ d &= \sqrt{72} = 6\sqrt{2} \end{aligned}

(simplifying the square root: \displaystyle \sqrt{72} = \sqrt{36 \cdot 2}.) That is about 8.49. The diagonals bisect each other and are congruent, so the center is \displaystyle 3\sqrt{2}, about 4.24, from every corner. A square is a rhombus, so each diagonal bisects a right angle, and the angle it makes with a side is 45^\circ.

(b) Now the diagonal is known and the side is not. From \displaystyle s\sqrt{2} = 10,

s = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2}

which is about 7.07. The area is \displaystyle s^2 = (5\sqrt{2})^2 = 25 \cdot 2 = 50. Because a square is a rhombus, the rhombus formula gives the area directly as well: \tfrac12 \cdot 10 \cdot 10 = 50.

Example 8

The midpoints of a rectangle’s sides

Rectangle ABCD has midpoints P on \overline{AB}, Q on \overline{BC}, R on \overline{CD} and S on \overline{DA}. Show that PQRS is a rhombus, and find its area as a fraction of the rectangle’s.

Figure 7Join the midpoints of a rectangle’s sides and the four corner triangles are congruent.

Solution

Opposite sides of a rectangle are congruent, so write AB = CD = 2a and BC = DA = 2b. The midpoints halve each side:

\begin{aligned} AP = PB = CR = RD &= a \\ BQ = QC = DS = SA &= b \end{aligned}

Each corner triangle, \triangle PBQ, \triangle QCR, \triangle RDS and \triangle SAP, has a right angle at a corner of the rectangle, between legs of length a and b. Two sides and the angle between them match, so all four triangles are congruent by SAS. Their hypotenuses are corresponding sides, so

PQ = QR = RS = SP

A quadrilateral with four congruent sides is a parallelogram (its opposite sides are congruent), and a parallelogram with four congruent sides is a rhombus. So PQRS is a rhombus. Each of its sides is \displaystyle \sqrt{a^2 + b^2}, which is half the rectangle’s diagonal.

For the area, the diagonals of the rhombus are \overline{PR} and \overline{QS}. \overline{PR} runs straight across between the midpoints of two opposite sides, so it is as long as BC, which is 2b, and \overline{QS} is as long as AB, which is 2a. The rhombus formula gives

\text{Area} = \frac{(2a)(2b)}{2} = 2ab

The rectangle’s area is (2a)(2b) = 4ab, so the rhombus fills exactly half of it. The four corner triangles, each of area \tfrac12 ab, fill the other half: 4ab - 4 \cdot \tfrac12 ab = 2ab.

The rhombus is a square only when a = b, that is, when the rectangle is itself a square.

Boundaries

The results run one way. “A rectangle has congruent diagonals” does not say that a quadrilateral with congruent diagonals is a rectangle. An isosceles trapezoid has congruent diagonals too (properties of an isosceles trapezoid), and a kite has perpendicular diagonals, one of which bisects two of its angles, without being a rhombus. Inside the parallelogram family the reverse statements do hold, and they are the tests in proving a parallelogram is a rectangle or a rhombus.

Each shape keeps only its own extras. The diagonals of a rectangle are perpendicular only when the rectangle is a square, and the diagonals of a rhombus are congruent only when the rhombus is a square. A rectangle’s diagonal bisects a corner only when it is a square too: if \overline{AC} bisected the right angle at A, then m\angle BAC = 45^\circ, the right triangle \triangle ABC would have two 45^\circ angles, and AB = BC.

Isosceles, but not all congruent. The diagonals cut a rectangle into four isosceles triangles, but only opposite ones are congruent: \triangle AEB \cong \triangle CED and \triangle BEC \cong \triangle DEA. Neighboring triangles have apex angles that add to 180^\circ, so they match only when both are 90^\circ, which happens only in a square. In a rhombus all four triangles are congruent, but they are right triangles, not isosceles ones, unless the rhombus is a square.

Inclusive words. Every statement about rectangles here includes squares. A book that uses “rectangle” to mean “a rectangle that is not a square” is using a different definition, and its statements need reading with that in mind.

The flat plane. Every proof above leaned on the parallelogram properties, which rest on the parallel postulate. On the surface of a sphere, with great circles as the straight lines, the angles of a four-sided figure add to more than 360^\circ, so no four-sided figure there has four right angles, and there are no rectangles at all.

Where this goes next

The theorems here start from a rectangle or a rhombus and say what follows. In a proof the question usually runs the other way: you have a parallelogram and want to show it is a rectangle or a rhombus. Those reverse statements, that congruent diagonals or perpendicular diagonals are enough inside a parallelogram, are the tests in proving a parallelogram is a rectangle or a rhombus.

The equal half-diagonals of a rectangle say that a circle centered where the diagonals cross passes through all four corners. Every rectangle fits exactly inside a circle, with its diagonals as diameters, and that is the first step toward the right angle in a semicircle.

The area \tfrac12 d_1 d_2 of a rhombus sits beside the formulas in area of parallelograms and trapezoids, and it works for any quadrilateral whose diagonals are perpendicular, kites included, by the same four right triangles. And the whole family, with trapezoids and kites in their places, is laid out in the quadrilateral family tree.

Exercises

In every exercise, ABCD is named in order around the shape, and its diagonals meet at E.

What each shape guarantees

For each of 1 to 3, say which of parallelogram, rectangle, rhombus and square always have the property.

The diagonals are congruent.
The diagonals are perpendicular.
Each diagonal bisects the angles at both of its ends.

True or false? Explain each answer.

Every square is a rhombus.
Every rhombus is a rectangle.
The diagonals of a rectangle bisect its angles.

Rectangles

Rectangle ABCD has AC = 18. Find BD, BE and DE.
Rectangle ABCD has AE = 2x + 7 and BD = 6x - 2. Find x, AE and BD.
A rectangle is 12 by 35. Find the length of a diagonal.
A rectangle has a diagonal of 25 and one side of 7. Find the other side, the perimeter and the area.
Rectangle ABCD has m\angle BDC = 28^\circ. Find m\angle ECD, m\angle CED and m\angle ADB.
Rectangle ABCD has m\angle AEB = (5x + 10)^\circ and m\angle BEC = (3x + 10)^\circ. Find x, both angles and m\angle EAB.

Rhombuses

A rhombus has diagonals of 14 and 48. Find its side, its perimeter and its area.
A rhombus has side 10 and one diagonal of 12. Find the other diagonal and the area.
Rhombus ABCD has m\angle BCA = 25^\circ. Find m\angle BCD, m\angle ABC, m\angle BAC, m\angle CBD and m\angle BEC.
Rhombus ABCD has AB = 5x - 8 and CD = 2x + 13, and m\angle DAC = (4y + 2)^\circ and m\angle BAC = (6y - 12)^\circ. Find x, the side, y, m\angle DAB and BD.
A rhombus has perimeter 52 and one diagonal of 24. Find the other diagonal and the area.

Squares

A square has side 9. Find a diagonal, and the distance from the center to a corner, exactly and to two decimal places.
A square has diagonal 14. Find its side exactly and its area.
Square ABCD has its diagonals meeting at E. Find m\angle EAB and m\angle AEB, and say what kind of triangle \triangle AEB is.
A square and a rhombus both have side 10, and one angle of the rhombus is 60^\circ. Which has the larger area, and by how much?

Reasoning

Show that a rhombus whose diagonals are congruent is a square.
Explain why there is a circle that passes through all four corners of any rectangle, and say where its center is and what its radius is.
A kite has perpendicular diagonals but is not a rhombus. Explain why this does not contradict the Rhombus Diagonals Theorem.
Rhombus ABCD has midpoints P on \overline{AB}, Q on \overline{BC}, R on \overline{CD} and S on \overline{DA}. Show that PQRS is a rectangle.
Explain why a square has four lines of symmetry while a rectangle that is not a square has only two.

Answers

Rectangle and square. A rhombus with angles of 60^\circ and 120^\circ has one diagonal longer than the other, and so does a parallelogram that is neither.
Rhombus and square. The diagonals of a rectangle that is not a square are not perpendicular.
Rhombus and square. Each is a parallelogram with four congruent sides, and the SSS argument for the rhombus applies.
True. A square has four congruent sides and is a parallelogram, which is the whole definition of a rhombus.
False. A rhombus with angles of 60^\circ and 120^\circ has no right angle, so it is not a rectangle.
False, unless the rectangle is a square. If \overline{AC} bisected the right angle at A, then m\angle BAC = 45^\circ, so the right triangle \triangle ABC would have two 45^\circ angles and AB = BC.
BD = 18, BE = 9 and DE = 9. The diagonals of a rectangle are congruent, and they bisect each other, so each half is 18 \div 2 = 9.
x = 8, AE = 23 and BD = 46. BD equals AC, which is 2 \cdot AE, so 6x - 2 = 2(2x + 7) = 4x + 14 and 2x = 16. Then AE = 2(8) + 7 = 23 and BD = 6(8) - 2 = 46.
37. By the Pythagorean theorem, d^2 = 12^2 + 35^2 = 144 + 1225 = 1369, and \displaystyle \sqrt{1369} = 37.
The other side is 24, the perimeter 62 and the area 168. \displaystyle \sqrt{25^2 - 7^2} = \sqrt{576} = 24; the perimeter is 2(7 + 24) and the area 7 \cdot 24.
m\angle ECD = 28^\circ, m\angle CED = 124^\circ and m\angle ADB = 62^\circ. EC = ED, so \triangle ECD is isosceles with base angles 28^\circ; its third angle is 180^\circ - 2 \cdot 28^\circ. \angle ADB and \angle BDC fill the right angle at D, so m\angle ADB = 90^\circ - 28^\circ.
x = 20, m\angle AEB = 110^\circ, m\angle BEC = 70^\circ and m\angle EAB = 35^\circ. The two angles form a linear pair: (5x + 10) + (3x + 10) = 180, so 8x = 160. \triangle AEB is isosceles with EA = EB, so each base angle is (180^\circ - 110^\circ) \div 2.
Side 25, perimeter 100, area 336. The half-diagonals are 7 and 24, so \displaystyle s = \sqrt{49 + 576} = 25; the area is \tfrac12 \cdot 14 \cdot 48.
The other diagonal is 16 and the area is 96. Half the known diagonal is 6, so half the other is \displaystyle \sqrt{10^2 - 6^2} = 8; the area is \tfrac12 \cdot 12 \cdot 16.
m\angle BCD = 50^\circ, m\angle ABC = 130^\circ, m\angle BAC = 25^\circ, m\angle CBD = 65^\circ and m\angle BEC = 90^\circ. \overline{AC} bisects the angle at C; the neighboring angle at B is 180^\circ - 50^\circ; \triangle ABC is isosceles with BA = BC, so m\angle BAC = m\angle BCA; \overline{BD} bisects the 130^\circ angle; the diagonals are perpendicular. Check in \triangle BEC: 25 + 65 + 90 = 180.
x = 7, the side is 27, y = 7, m\angle DAB = 60^\circ and BD = 27. The sides are equal: 5x - 8 = 2x + 13 gives x = 7 and side 27. \overline{AC} bisects the angle at A: 4y + 2 = 6y - 12 gives y = 7, each half is 30^\circ, and m\angle DAB = 60^\circ. \triangle ABD has AB = AD and a 60^\circ angle between them, so its base angles are each 60^\circ too, it is equilateral, and BD = 27.
The other diagonal is 10 and the area is 120. The side is 52 \div 4 = 13, half the known diagonal is 12, so half the other is \displaystyle \sqrt{13^2 - 12^2} = 5; the area is \tfrac12 \cdot 24 \cdot 10.
The diagonal is \displaystyle 9\sqrt{2} \approx 12.73, and the center is \displaystyle \tfrac{9\sqrt{2}}{2} \approx 6.36 from each corner. d^2 = 9^2 + 9^2 = 162, and the diagonals bisect each other.
The side is \displaystyle 7\sqrt{2} and the area is 98. \displaystyle s = \tfrac{14}{\sqrt{2}} = 7\sqrt{2}, and s^2 = 49 \cdot 2 = 98, which is also \tfrac12 \cdot 14 \cdot 14.
m\angle EAB = 45^\circ and m\angle AEB = 90^\circ, and \triangle AEB is an isosceles right triangle. The diagonal bisects the right angle at A; the diagonals are perpendicular; and EA = EB because the diagonals are congruent and bisect each other.
The square, by \displaystyle 100 - 50\sqrt{3} \approx 13.4. The square’s area is 10^2 = 100. In the rhombus the 60^\circ angle makes \triangle ABD equilateral, so one diagonal is 10; half of the other is \displaystyle \sqrt{10^2 - 5^2} = 5\sqrt{3}, so the other diagonal is \displaystyle 10\sqrt{3} and the area is \displaystyle \tfrac12 \cdot 10 \cdot 10\sqrt{3} = 50\sqrt{3} \approx 86.6.
The diagonals of a rhombus bisect each other and are perpendicular. If they are also congruent, all four halves are equal, so \triangle AEB is an isosceles right triangle and m\angle EAB = m\angle EBA = 45^\circ. The same holds in \triangle AED, so m\angle DAB = 45^\circ + 45^\circ = 90^\circ. A parallelogram with one right angle has four, since neighboring angles are supplementary, so the rhombus is also a rectangle, and a rhombus that is a rectangle is a square.
The diagonals of a rectangle are congruent and bisect each other, so EA = EB = EC = ED. A circle centered at E whose radius is half a diagonal passes through all four corners.
The theorem says that if a shape is a rhombus, its diagonals are perpendicular. It says nothing about shapes that are not rhombuses, and it does not claim that perpendicular diagonals make a rhombus. The kite has a property that rhombuses share with other shapes, which is all the kite shows.
AP = AS, being halves of equal sides, so \triangle APS is isosceles and each base angle is \tfrac12(180^\circ - m\angle A). Likewise BP = BQ, so m\angle BPQ = \tfrac12(180^\circ - m\angle B). A, P and B lie on a line, so m\angle SPQ = 180^\circ - m\angle APS - m\angle BPQ, which simplifies to \tfrac12(m\angle A + m\angle B). Neighboring angles of a parallelogram add to 180^\circ, so m\angle SPQ = 90^\circ. The same holds at Q, R and S. A quadrilateral with four right angles is a parallelogram, as the Definition section shows, so PQRS is a rectangle.
A rectangle folds onto itself along the two lines joining midpoints of opposite sides, and a rhombus along its two diagonals. A square is both, so it has all four. A rectangle that is not a square cannot fold along a diagonal: that fold would send B onto D, which needs AB = AD.