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Trapezoids and kites

A four-sided shape with only one pair of parallel sides, or with two pairs of equal sides sitting side by side, looks lopsided. What do those few conditions force to be true about its angles and its diagonals?

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Why this exists

A parallelogram is generous: two pairs of parallel sides, and in return its opposite sides and opposite angles all match. Plenty of four-sided shapes are not that tidy. A roof seen from the side, a slice of a triangle, a table top that narrows toward one end: each has one pair of parallel sides and no more. A kite flown on a string has no parallel sides at all, only two pairs of equal edges meeting at the top and at the bottom.

It is fair to ask whether shapes with so little given have any properties worth knowing. They do. One pair of parallel sides is enough to tie the angles together in pairs. Add equal legs and the shape gains a mirror line, equal angles at each base and equal diagonals. Two neighboring pairs of equal sides give a kite, and its diagonals always cross at a right angle. This lesson names these shapes exactly and shows what each name guarantees.

Definition

A trapezoid is a quadrilateral with at least one pair of parallel sides. The parallel sides are its bases, and the other two sides are its legs. The two angles at the ends of one base are a pair of base angles, so a trapezoid has two pairs of base angles, one pair for each base. The height is the distance between the bases, measured along a segment perpendicular to both.

base base leg leg h A B C D
Figure 1The parallel sides are the bases and the other two are the legs. The height is the distance between the bases, measured at right angles.

In Figure 1, \overline{AB} \parallel \overline{DC}. The bases are \overline{AB} and \overline{DC}, the legs are \overline{AD} and \overline{BC}, and the base angles are \angle A and \angle B at one base, \angle D and \angle C at the other. The matching arrow marks on the two bases say they are parallel; a figure must mark it, since a drawing that only looks parallel proves nothing.

“At least one” is a deliberate choice. Under it, every parallelogram is a trapezoid, and so is every rectangle, rhombus and square, with either pair of opposite sides taken as the bases. Some books say “exactly one pair” instead and keep parallelograms out. Both choices are in use; this lesson uses “at least one”, so that every fact proved for trapezoids also holds for parallelograms.

An isosceles trapezoid is a trapezoid whose legs are congruent and are not parallel to each other. The second condition rules out a parallelogram, whose opposite sides are also congruent but which behaves differently (see Boundaries).

A B C D
Figure 2An isosceles trapezoid. Its legs are congruent, and the dashed line through the middles of the bases is a mirror line.

A kite is a quadrilateral with two pairs of congruent sides in which the two sides of each pair are neighbors: in kite ABCD,

AB = AD \qquad\text{and}\qquad CB = CD

So the sides of one pair meet at A, and the sides of the other pair meet at C. The diagonal \overline{AC} joins those two vertices. A rhombus has all four sides equal, so it is a kite, and so is a square.

A B C D E
Figure 3A kite. The equal sides come in neighboring pairs, and the diagonal \overline{AC} is a mirror line.

Intuition

Slice a triangle with a cut parallel to one of its sides, and the piece holding that side is a trapezoid: the old side and the cut are its two bases. Slice an isosceles triangle across its base like this and the trapezoid keeps the triangle’s mirror line, the dashed line in Figure 2. A mirror line carries each half onto the other, so whatever is true on the left is true on the right: the legs match, the angles at each base match, and a diagonal from one side matches the diagonal from the other.

A kite is two congruent triangles, \triangle ABC and \triangle ADC, glued along the shared side \overline{AC} as mirror images. Fold it along \overline{AC} and B lands on D. That fold is where the kite’s properties come from: the fold line meets \overline{BD} at a right angle and cuts it in half, and the angles at B and D land on each other.

Results

Angles along a leg. In trapezoid ABCD with \overline{AB} \parallel \overline{DC}, the two angles at the ends of each leg are supplementary:

\begin{aligned} m\angle A + m\angle D &= 180^\circ \\ m\angle B + m\angle C &= 180^\circ \end{aligned}

Isosceles trapezoids. In an isosceles trapezoid ABCD with \overline{AB} \parallel \overline{DC} and \overline{AD} \cong \overline{BC}:

  • the base angles at each base are congruent: \angle D \cong \angle C and \angle A \cong \angle B;
  • the diagonals are congruent: \overline{AC} \cong \overline{BD};
  • the line through the midpoints of the two bases is perpendicular to both, and it is a line of symmetry.

So an isosceles trapezoid has only two angle sizes: one pair at each base, and the two sizes add to 180^\circ.

Kites. In kite ABCD with AB = AD and CB = CD:

  • the line AC is the perpendicular bisector of \overline{BD}, so the diagonals are perpendicular and \overline{AC} cuts \overline{BD} in half;
  • \overline{AC} bisects \angle A and \angle C;
  • the angles where unequal sides meet are congruent: \angle B \cong \angle D;
  • the line AC is a line of symmetry;
  • the area is half the product of the diagonals:
\text{Area} = \tfrac{1}{2} \cdot AC \cdot BD
Why it is true

Angles along a leg. The leg \overline{AD} is a transversal crossing the parallel lines AB and DC. The angles \angle A and \angle D lie between the parallels on the same side of it, so they are same-side interior angles, which are supplementary (parallel lines and transversals). The leg \overline{BC} is another transversal and gives m\angle B + m\angle C = 180^\circ the same way.

Isosceles trapezoids. Let \overline{DC} be the longer base, and drop perpendiculars from A and B to it, with feet P and Q. Both are perpendicular to \overline{DC}, so they are perpendicular to \overline{AB} too, and ABQP has four right angles. It is a rectangle, so AP = BQ and PQ = AB.

Now \triangle APD and \triangle BQC are right triangles with congruent hypotenuses (AD = BC) and congruent legs (AP = BQ), so they are congruent by hypotenuse-leg. Matching parts give \angle D \cong \angle C and DP = QC. The angles \angle A and \angle B are the supplements of \angle D and \angle C along the legs, and supplements of equal angles are equal, so \angle A \cong \angle B.

For the diagonals, compare \triangle ADC and \triangle BCD: AD = BC, \angle D \cong \angle C, and they share \overline{DC}. By SAS (the congruence criteria) they are congruent, so AC = BD.

For the mirror line: since DP = QC, the midpoint N of \overline{DC} is also the midpoint of \overline{PQ}, and in the rectangle ABQP the midpoint M of \overline{AB} sits straight above it. So line MN is perpendicular to both bases. Reflecting in MN swaps P and Q, swaps D and C (each is the same distance from N along a line perpendicular to the mirror), and swaps A and B the same way. Each half lands on the other.

That argument takes for granted that P and Q land on the base itself; why they do, the converses (equal base angles, or equal diagonals, force equal legs) and the full two-column proofs are in proving properties of an isosceles trapezoid.

Kites. AB = AD says A is the same distance from B as from D, and CB = CD says the same of C. The points equally far from the two ends of a segment are exactly the points of its perpendicular bisector, so A and C both lie on the perpendicular bisector of \overline{BD}. Two different points fix a line, so the line AC is that perpendicular bisector. That is the first result.

Next, \triangle ABC and \triangle ADC have AB = AD, CB = CD, and the shared side \overline{AC}, so they are congruent by SSS. Matching angles give \angle ABC \cong \angle ADC, \angle BAC \cong \angle DAC and \angle BCA \cong \angle DCA: the angles at B and D are equal, and \overline{AC} bisects the angles at A and C. Reflecting in the line AC fixes A and C and, since that line is the perpendicular bisector of \overline{BD}, swaps B and D; so it carries the kite onto itself.

For the area, split the kite along \overline{AC} into \triangle ABC and \triangle ADC. Call the crossing point of the diagonals E. Because \overline{BD} \perp \overline{AC}, the height of \triangle ABC from B is BE, and the height of \triangle ADC from D is DE, and each is \tfrac{1}{2}BD. So

\begin{aligned} \text{Area} &= \tfrac{1}{2} \cdot AC \cdot \tfrac{1}{2}BD + \tfrac{1}{2} \cdot AC \cdot \tfrac{1}{2}BD \\ &= \tfrac{1}{2} \cdot AC \cdot BD \end{aligned}

Worked examples

Example 1

Telling a trapezoid from a parallelogram

Quadrilateral ABCD has m\angle A = 70^\circ, m\angle B = 85^\circ and m\angle D = 110^\circ. Find m\angle C, decide which sides are parallel, and decide whether ABCD is an isosceles trapezoid.

Solution

The angles of a quadrilateral add to 360^\circ (angle sums of polygons), so

\begin{aligned} m\angle C &= 360^\circ - 70^\circ - 85^\circ - 110^\circ \\ &= 95^\circ \end{aligned}

Now test each pair of opposite sides. The side \overline{AD} is a transversal for the lines AB and DC, and \angle A and \angle D are same-side interior angles along it. They add to 70^\circ + 110^\circ = 180^\circ, so by the converse of the same-side interior angles result (proving lines parallel), \overline{AB} \parallel \overline{DC}.

For the other pair, \angle A and \angle B are the same-side interior angles along \overline{AB}. If \overline{AD} were parallel to \overline{BC} they would add to 180^\circ, but 70^\circ + 85^\circ = 155^\circ. So \overline{AD} is not parallel to \overline{BC}.

ABCD is a trapezoid with bases \overline{AB} and \overline{DC}, and not a parallelogram. It is not isosceles: its base angles at \overline{DC} are 110^\circ and 95^\circ, and an isosceles trapezoid’s base angles match.

Check. \angle B and \angle C lie along the other leg, so they should be supplementary too: 85 + 95 = 180. They are.

Example 2

Angles of an isosceles trapezoid from algebra

PQRS is an isosceles trapezoid with \overline{PQ} \parallel \overline{SR} and legs \overline{PS} and \overline{QR}. Its base angles at \overline{SR} measure m\angle S = (3x + 10)^\circ and m\angle R = (5x - 18)^\circ. Find x and all four angles.

Solution

The base angles at one base of an isosceles trapezoid are congruent, so set the two expressions equal:

\begin{aligned} 3x + 10 &= 5x - 18 \\ 28 &= 2x \\ x &= 14 \end{aligned}

Then m\angle S = 3(14) + 10 = 52^\circ, and m\angle R = 5(14) - 18 = 52^\circ as well, which is the check that both expressions agree.

\angle P and \angle S lie along the leg \overline{PS}, so they are supplementary: m\angle P = 180^\circ - 52^\circ = 128^\circ. In the same way m\angle Q = 128^\circ.

The four angles are 52^\circ, 52^\circ, 128^\circ and 128^\circ, and they add to 360^\circ.

Example 3

The angles of a kite

Kite ABCD in Figure 4 has AB = AD and CB = CD, with m\angle A = 70^\circ and m\angle C = 50^\circ. Find m\angle B, m\angle D and m\angle BAC.

70^\circ
50^\circ
 A B C D
Figure 4Two angles of a kite are given. The other two are equal to each other.
Solution

The angles at B and D are where unequal sides meet, so they are congruent. The four angles add to 360^\circ, so the two equal ones share what is left:

\begin{aligned} m\angle B + m\angle D &= 360^\circ - 70^\circ - 50^\circ \\ &= 240^\circ \end{aligned}

so m\angle B = m\angle D = 120^\circ.

The diagonal \overline{AC} bisects \angle A, so m\angle BAC = 70^\circ \div 2 = 35^\circ.

Check. Split \angle B along the dashed diagonal \overline{BD}. \triangle ABD is isosceles with AB = AD, so its base angles are equal (isosceles triangles), and each is (180^\circ - 70^\circ) \div 2 = 55^\circ. \triangle CBD is isosceles with CB = CD, so each of its base angles is (180^\circ - 50^\circ) \div 2 = 65^\circ. Together they make m\angle ABC = 55^\circ + 65^\circ = 120^\circ, which agrees.

Example 4

The height of an isosceles trapezoid

Isosceles trapezoid ABCD in Figure 5 has \overline{AB} \parallel \overline{DC}, AB = 10, DC = 16 and legs AD = BC = 5. Find its height and its area.

P 10 16 5 5 h A B C D
Figure 5The legs are equal, so the extra length of the long base is shared equally at the two ends.
Solution

Drop perpendiculars from A and B to \overline{DC}, with feet P and Q. As in the proof above, ABQP is a rectangle, so PQ = AB = 10. The right triangles APD and BQC are congruent, so DP = QC. The three pieces make up the long base:

\begin{aligned} DP + 10 + QC &= 16 \\ DP = QC &= 3 \end{aligned}

In the right triangle APD the hypotenuse is the leg AD = 5, so by the Pythagorean theorem

\begin{aligned} AP^2 + 3^2 &= 5^2 \\ AP^2 &= 16 \end{aligned}

and since a length is positive, AP = 4. The height is 4.

The area of a trapezoid is half the sum of the bases times the height (area of a trapezoid):

\text{Area} = \tfrac{1}{2}(10 + 16)(4) = 52

Check. The figure is the rectangle ABQP, of area 10 \times 4 = 40, with two right triangles of area \tfrac{1}{2} \times 3 \times 4 = 6 at its ends, and 40 + 6 + 6 = 52.

Example 5

The diagonals of a kite

Kite ABCD in Figure 6 has AB = AD = 10, CB = CD = 17 and diagonal BD = 16. Find the other diagonal \overline{AC} and the area of the kite.

10 10 17 17 16 E A B C D
Figure 6The diagonals of a kite cut it into four right triangles.
Solution

The line AC is the perpendicular bisector of \overline{BD}, so the diagonals cross at a right angle at E, and E is the midpoint of \overline{BD}: BE = 8.

\triangle ABE has a right angle at E and hypotenuse AB = 10, so

AE = \sqrt{10^2 - 8^2} = \sqrt{36} = 6

\triangle CBE has a right angle at E and hypotenuse CB = 17, so

CE = \sqrt{17^2 - 8^2} = \sqrt{225} = 15

The diagonals cross inside the kite, so E lies between A and C, and AC = 6 + 15 = 21.

The area is half the product of the diagonals:

\text{Area} = \tfrac{1}{2} \cdot 21 \cdot 16 = 168

Check. \overline{BD} splits the kite into \triangle ABD, with base 16 and height 6, and \triangle CBD, with base 16 and height 15. Their areas are 48 and 120, and 48 + 120 = 168.

Example 6

A kite that is also a trapezoid

Show that a kite with one pair of parallel sides is a rhombus.

Solution

Let the kite be ABCD with AB = AD and CB = CD. A pair of parallel sides must be a pair of opposite sides, since neighboring sides meet. Suppose \overline{AB} \parallel \overline{DC}; the other case, \overline{AD} \parallel \overline{BC}, is the same argument with the letters B and D swapped.

ABCD is then a trapezoid with legs \overline{AD} and \overline{BC}, so the angles along the leg \overline{AD} are supplementary: m\angle A + m\angle D = 180^\circ. In a kite \angle B \cong \angle D, so substituting,

m\angle A + m\angle B = 180^\circ

But \angle A and \angle B are the same-side interior angles along \overline{AB} for the lines AD and BC. Since they are supplementary, the converse gives \overline{AD} \parallel \overline{BC}.

Both pairs of opposite sides are parallel, so ABCD is a parallelogram, and opposite sides of a parallelogram are congruent: AB = CD. Together with AB = AD and CB = CD, all four sides are equal, so ABCD is a rhombus.

So the only kites that are also trapezoids are the rhombuses, squares among them.

Boundaries

Which definition is in force matters. Under “at least one pair of parallel sides”, a parallelogram is a trapezoid; under “exactly one pair” it is not. The angle facts above hold either way, but questions such as “is every rectangle a trapezoid?” have different answers, so a problem must say which definition it uses.

Congruent legs alone do not make the base angles equal. A parallelogram that is not a rectangle has congruent opposite sides, so if one pair is called the bases, the “legs” are congruent. Yet its base angles are supplementary, not equal. That is why an isosceles trapezoid’s legs must also be non-parallel. Some books instead define an isosceles trapezoid by its equal base angles, which lets rectangles in; the two definitions agree once the legs are not parallel.

The height is not a leg. The height is measured at right angles to the bases. It equals a leg only when that leg is perpendicular to the bases, and otherwise it is shorter.

A kite needs the equal sides to be neighbors. Sides of 7, 7, 12, 12 in that order around the shape make a kite. In the order 7, 12, 7, 12 the equal sides are opposite, and the shape is a parallelogram, with none of the kite’s properties.

A kite may be dented. The definition also admits a shape with A and C on the same side of \overline{BD}, sometimes called a dart, with one angle greater than 180^\circ. The line AC is still the perpendicular bisector of \overline{BD} and still a mirror line, and the area is still \tfrac{1}{2} \cdot AC \cdot BD. But the diagonals no longer cross: the point E lies outside the segment \overline{AC}, so a length like the one in Example 5 becomes CE - AE, not AE + CE. The figures and problems in this lesson are all kites without a dent.

One diagonal of a kite is special, the other is not. \overline{AC} bisects \overline{BD}, but \overline{BD} does not bisect \overline{AC} unless all four sides are equal.

Where this goes next

The height and the two bases are exactly what the area of a trapezoid needs, and the mirror line of an isosceles trapezoid is what makes its height easy to find from its sides. The segment joining the midpoints of a trapezoid’s legs is parallel to the bases and is their average in length, a fact that grows out of the midsegment of a triangle.

The properties stated here for isosceles trapezoids, and their converses, are proved step by step in proving properties of an isosceles trapezoid. An isosceles trapezoid also has a circle through all four of its vertices, which makes it a cyclic quadrilateral.

Trapezoids and kites are two branches of the quadrilateral family tree. Parallelograms sit inside the trapezoids, rhombuses sit inside the kites, and as Example 6 shows, the rhombuses are exactly where those two branches meet.

Exercises

Naming the parts

  1. Trapezoid WXYZ has \overline{WX} \parallel \overline{ZY}. Name its bases, its legs and its two pairs of base angles.
  2. Using this lesson’s definitions, say whether each statement is true or false, and why. (a) Every parallelogram is a trapezoid. (b) Every trapezoid is a parallelogram. (c) Every rhombus is a kite. (d) Every kite is a rhombus. (e) A rectangle is an isosceles trapezoid.
  3. The sides of a kite have lengths 7, 7, 12 and 12. In what order can they go around the kite? Explain why the order 7, 12, 7, 12 does not give a kite.

Finding angles

  1. Trapezoid ABCD has \overline{AB} \parallel \overline{DC}, m\angle A = 58^\circ and m\angle B = 121^\circ. Find m\angle D and m\angle C.
  2. Isosceles trapezoid ABCD has \overline{AB} \parallel \overline{DC} and m\angle C = 73^\circ. Find the other three angles.
  3. One angle of an isosceles trapezoid measures 134^\circ. Find the other three.
  4. Kite ABCD has AB = AD and CB = CD, with m\angle A = 110^\circ and m\angle B = 95^\circ. Find m\angle D and m\angle C.
  5. Kite ABCD has AB = AD and CB = CD. Its angles at B and D each measure 105^\circ, and m\angle A is twice m\angle C. Find m\angle A and m\angle C.
  6. Kite ABCD has AB = AD and CB = CD, with m\angle BAD = 64^\circ and m\angle BCD = 42^\circ. Find m\angle ABD and m\angle ABC.

Using algebra

  1. The base angles at one base of an isosceles trapezoid measure (2x + 25)^\circ and (4x - 11)^\circ. Find x and all four angles.
  2. Isosceles trapezoid ABCD has \overline{AB} \parallel \overline{DC} and diagonals AC = 3x + 2 and BD = 5x - 12. Find x and the length of each diagonal.
  3. Trapezoid ABCD has \overline{AB} \parallel \overline{DC}, m\angle A = (5x - 4)^\circ and m\angle D = (3x + 24)^\circ. Find x and both angles.
  4. Kite ABCD has AB = AD and CB = CD, with m\angle A = 70^\circ, m\angle B = (7y - 15)^\circ and m\angle D = (4y + 30)^\circ. Find y and m\angle C.

Lengths and areas

  1. An isosceles trapezoid has bases 7 and 19 and legs 10. Find its height and its area.
  2. An isosceles trapezoid has bases 12 and 20 and height 3. Find the length of each leg and the perimeter.
  3. The diagonals of a kite have lengths 9 and 14. Find its area.
  4. Kite ABCD has AB = AD = 13, CB = CD = 20 and diagonal BD = 24. Find the diagonal \overline{AC} and the area.
  5. Kite ABCD has AB = AD = 5. Its diagonals cross at E, with BD = 8 and AC = 10. Find CB and the perimeter.
  6. The points A(0, 4), B(3, 1), C(0, -5) and D(-3, 1) are joined in order. Show that ABCD is a kite, and find its area.
  7. The points P(1, 1), Q(4, 1), R(6, 5) and S(-1, 5) are joined in order. Show that PQRS is an isosceles trapezoid, and find its area.

Reasoning

  1. Can a trapezoid have three acute angles? Explain.
  2. A trapezoid with a right angle is called a right trapezoid. Show that a right trapezoid has at least two right angles.
  3. A quadrilateral’s diagonals are perpendicular. Must it be a kite? Explain, with an example if the answer is no.
  4. In kite ABCD with AB = AD and CB = CD, explain why the line BD is not a line of symmetry unless the kite is a rhombus.
  5. The diagonals of an isosceles trapezoid are congruent. Is every quadrilateral with congruent diagonals an isosceles trapezoid? Explain.
Answers
  1. The bases are \overline{WX} and \overline{ZY}, the parallel pair, and the legs are the other two sides, \overline{XY} and \overline{ZW}. The base angles are \angle W and \angle X at \overline{WX}, and \angle Z and \angle Y at \overline{ZY}.
  2. (a) True: a parallelogram has two pairs of parallel sides, so it has at least one. (b) False: a trapezoid may have only one pair. (c) True: all four sides are equal, so both neighboring pairs match. (d) False: a kite with sides 7, 7, 12, 12 is not a rhombus. (e) False: a rectangle’s legs are parallel, and this lesson’s isosceles trapezoid has legs that are not.
  3. The two 7s must be neighbors and so must the two 12s: 7, 7, 12, 12 around the shape. In the order 7, 12, 7, 12 no two neighboring sides are equal, so it is not a kite; its opposite sides are equal, which makes it a parallelogram (proving a quadrilateral is a parallelogram).
  4. m\angle D = 122^\circ and m\angle C = 59^\circ. Along the leg \overline{AD}, 180^\circ - 58^\circ = 122^\circ; along the leg \overline{BC}, 180^\circ - 121^\circ = 59^\circ. Check: 58 + 121 + 59 + 122 = 360.
  5. m\angle D = 73^\circ and m\angle A = m\angle B = 107^\circ. The base angles at \overline{DC} are equal, and each angle at \overline{AB} is the supplement of the one along its leg: 180^\circ - 73^\circ = 107^\circ.
  6. Another 134^\circ and two of 46^\circ. The angle at the other end of its base is congruent to it, and the angles along each leg are supplementary: 180^\circ - 134^\circ = 46^\circ.
  7. m\angle D = 95^\circ and m\angle C = 60^\circ. In a kite \angle D \cong \angle B, and 360^\circ - 110^\circ - 95^\circ - 95^\circ = 60^\circ.
  8. m\angle C = 50^\circ and m\angle A = 100^\circ. The angles at A and C share 360^\circ - 105^\circ - 105^\circ = 150^\circ; with m\angle C = c, 2c + c = 150, so c = 50.
  9. m\angle ABD = 58^\circ and m\angle ABC = 127^\circ. \triangle ABD is isosceles with AB = AD, so m\angle ABD = (180^\circ - 64^\circ) \div 2 = 58^\circ. The angles at B and D are equal and share 360^\circ - 64^\circ - 42^\circ = 254^\circ, so each is 127^\circ. Check: m\angle DBC = 127^\circ - 58^\circ = 69^\circ, the base angle of isosceles \triangle CBD, (180^\circ - 42^\circ) \div 2 = 69^\circ.
  10. x = 18; the angles are 61^\circ, 61^\circ, 119^\circ and 119^\circ. Base angles at one base are equal: 2x + 25 = 4x - 11, so 36 = 2x. Then 2(18) + 25 = 61, and the angles at the other base are 180^\circ - 61^\circ = 119^\circ.
  11. x = 7, and each diagonal is 23. The diagonals are congruent: 3x + 2 = 5x - 12, so 14 = 2x. Check: 3(7) + 2 = 23 and 5(7) - 12 = 23.
  12. x = 20, m\angle A = 96^\circ and m\angle D = 84^\circ. The angles along the leg \overline{AD} are supplementary: (5x - 4) + (3x + 24) = 180, so 8x + 20 = 180.
  13. y = 15 and m\angle C = 110^\circ. \angle B \cong \angle D, so 7y - 15 = 4y + 30 and 3y = 45. Each of those angles is 90^\circ, so m\angle C = 360^\circ - 70^\circ - 90^\circ - 90^\circ = 110^\circ.
  14. Height 8, area 104. Each end of the long base sticks out by (19 - 7) \div 2 = 6, so the height is \displaystyle \sqrt{10^2 - 6^2} = 8, and the area is \tfrac{1}{2}(7 + 19)(8) = 104.
  15. Each leg is 5 and the perimeter is 42. Each end sticks out by (20 - 12) \div 2 = 4, so a leg is \displaystyle \sqrt{4^2 + 3^2} = 5, and the perimeter is 12 + 20 + 5 + 5 = 42.
  16. 63. The area is \tfrac{1}{2} \cdot 9 \cdot 14 = 63.
  17. AC = 21 and the area is 252. \overline{AC} bisects \overline{BD} at right angles, so BE = 12, \displaystyle AE = \sqrt{13^2 - 12^2} = 5 and \displaystyle CE = \sqrt{20^2 - 12^2} = 16. The area is \tfrac{1}{2} \cdot 21 \cdot 24 = 252.
  18. \displaystyle CB = \sqrt{65} and the perimeter is \displaystyle 10 + 2\sqrt{65}. E is the midpoint of \overline{BD}, so BE = 4, and \displaystyle AE = \sqrt{5^2 - 4^2} = 3. The diagonals cross, so EC = 10 - 3 = 7, and \displaystyle CB = \sqrt{7^2 + 4^2} = \sqrt{65}. The perimeter is \displaystyle 5 + 5 + \sqrt{65} + \sqrt{65}.
  19. \displaystyle AB = AD = \sqrt{3^2 + 3^2} = 3\sqrt{2} and \displaystyle CB = CD = \sqrt{3^2 + 6^2} = 3\sqrt{5}, two neighboring pairs of equal sides, so ABCD is a kite. The diagonal \overline{AC} is vertical with length 9 and \overline{BD} is horizontal with length 6, so the area is \tfrac{1}{2} \cdot 9 \cdot 6 = 27.
  20. \overline{PQ} and \overline{SR} are both horizontal, so they are parallel, with lengths 3 and 7, and the height is 5 - 1 = 4. The legs are \displaystyle PS = \sqrt{2^2 + 4^2} = \sqrt{20} and \displaystyle QR = \sqrt{2^2 + 4^2} = \sqrt{20}, so they are congruent, and their slopes are -2 and 2, so they are not parallel. The area is \tfrac{1}{2}(3 + 7)(4) = 20.
  21. No. Each angle sits at an end of exactly one leg, and the two angles along a leg add to 180^\circ, so at most one of them is acute. Two legs allow at most two acute angles.
  22. The right angle sits at an end of one leg, and the angle at the other end of that leg is its supplement, 180^\circ - 90^\circ = 90^\circ. So there are at least two right angles.
  23. No. Take A(0, 3), B(-2, 0), C(0, -1) and D(4, 0). The diagonals lie along the two axes, so they are perpendicular, but the sides are \displaystyle AB = \sqrt{13}, \displaystyle BC = \sqrt{5}, \displaystyle CD = \sqrt{17} and DA = 5, and no two neighboring sides are equal. A kite needs more than perpendicular diagonals: one of them must also bisect the other. Here the midpoint of \overline{AC} is (0, 1), which is not on \overline{BD}, and the midpoint of \overline{BD} is (1, 0), which is not on \overline{AC}.
  24. A reflection in the line BD fixes B and D, so if it carried the kite onto itself it would have to send A to C. Reflections keep lengths, so BA = BC. With AB = AD and CB = CD, all four sides would then be equal, which makes the kite a rhombus.
  25. No. Every rectangle has congruent diagonals, a 3 by 4 rectangle has both diagonals 5, but a rectangle’s legs are parallel, so under this lesson’s definition it is not an isosceles trapezoid.
kg1128 · updated 2026-10-03 · CC-BY-SA-4.0