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Areas of Quadrilaterals: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The parallel sides

A convex trapezoid has parallel bases $11$ cm and $23$ cm, perpendicular height $5$ cm, and a slanted leg of $13$ cm. Find its area.

The square mark identifies the perpendicular height.

Text description of this figure

Schematic trapezoid, not to scale. Parallel top and bottom bases are labeled 11 cm and 23 cm. A dashed perpendicular between them is labeled 5 cm, with a right-angle square. The right slanted side is labeled 13 cm.
1. Hint 1
  Identify the two parallel lengths and the distance measured perpendicular to them.
2. Hint 2
  Average the bases before multiplying by the height.
Answer

$85$ square cm.

Full solution

The height is $5$ cm; the $13$ cm leg is not perpendicular to the bases.

The area is
$A=\frac{11+23}{2}\cdot5.$

The average base length is $17$ cm, giving area $85$ square cm.

Answer

$85$ square cm.

Key idea

The trapezoid formula pairs parallel bases with their perpendicular separation.
Problem 2 The diagonal lengths

A rhombus has full diagonals $15$ cm and $22$ cm. Find its area.

Both labels give full diagonal lengths.

Text description of this figure

Schematic rhombus ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 15 cm. AC is labeled 22 cm. A single tick on every side marks all four sides congruent.
1. Hint 1
  Recall the angle between a rhombus's diagonals.
2. Hint 2
  Use both full diagonal lengths in the area formula.
Answer

$165$ square cm.

Full solution

Rhombus diagonals are perpendicular, so half their product gives the area.

Compute
$A=\frac12\cdot15\cdot22.$

The area is $165$ square cm.

Answer

$165$ square cm.

Key idea

A rhombus supplies the perpendicularity needed for half the diagonal product.
Problem 3 The divided diagonal

Convex kite $ABCD$ has equal-side pairs meeting at $A$ and $C$ . Diagonals intersect at $O$ . Given $AO$ is $7$ cm, $OC$ is $10$ cm, and $BO$ is $4$ cm, find the full lengths AC and BD, then the area.

The equal-side pairs identify AC as the bisecting diagonal.

Text description of this figure

Schematic convex kite ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. AO is 7 cm. OC is 10 cm. Single side ticks pair AB with AD; double ticks pair BC with CD.
1. Hint 1
  Which diagonal is bisected by the kite's symmetry diagonal?
2. Hint 2
  Add AO and OC for one full diagonal, and double BO for the other.
Answer

$\begin{aligned}AC&=17\text{ cm},\\BD&=8\text{ cm},\\A&=68\text{ cm}^2.\end{aligned}$

Full solution

The equal-side pairs meet at A and C, so AC perpendicularly bisects BD.
Hence
$\begin{aligned}BD&=2\cdot4, \\BD&=8\text{ cm}.\end{aligned}$

The other full diagonal has length
$\begin{aligned}AC&=7+10, \\AC&=17\text{ cm}.\end{aligned}$

Use the perpendicular full diagonals:
$\begin{aligned}A&=\frac12\cdot17\cdot8, \\A&=68\text{ cm}^2.\end{aligned}$

Answer

$\begin{aligned}AC&=17\text{ cm},\\BD&=8\text{ cm},\\A&=68\text{ cm}^2.\end{aligned}$

Key idea

A general kite need not bisect both diagonals.
Problem 4 The two altitude feet

In convex quadrilateral ABCD, diagonal $AC$ is $18$ cm. The perpendicular distances from B and D to line AC are $4$ cm and $9$ cm. Find the area. State whether you needed BD to be perpendicular to AC.

The two right-angle marks belong to separate altitude feet.

Text description of this figure

Schematic convex quadrilateral ABCD, not to scale. AC is a horizontal diagonal labeled 18 cm. Vertex B lies above AC and vertex D below it. Dashed perpendicular heights from B and D to AC are labeled 4 cm and 9 cm. Right-angle squares mark their feet. BD is not drawn or asserted perpendicular to AC.
1. Hint 1
  Consider the two triangles on opposite sides of AC.
2. Hint 2
  Each given distance is a height to AC; the feet need not coincide.
Answer

Area $117$ square cm; perpendicularity of BD and AC is not needed.

Full solution

Let $A_1$ and $A_2$ denote the areas of triangles ABC and ACD, respectively.
Their interiors do not overlap.

For triangle ABC,
$\begin{aligned}A_1&=\frac12\cdot18\cdot4, \\A_1&=36\text{ cm}^2.\end{aligned}$

For triangle ACD,
$\begin{aligned}A_2&=\frac12\cdot18\cdot9, \\A_2&=81\text{ cm}^2.\end{aligned}$

Add to obtain
$\begin{aligned}A&=36+81, \\A&=117\text{ cm}^2.\end{aligned}$

The separate perpendicular distances supply the heights.
There is no need to assume BD is perpendicular to AC.

Answer

Area $117$ square cm; perpendicularity of BD and AC is not needed.

Key idea

A chosen diagonal can be a base for two triangles with different altitude feet.
Problem 5 The joined copies

Two congruent copies of a convex trapezoid are joined along one leg by a half turn to form a parallelogram. The trapezoid's parallel bases are $b_1$ and $b_2$ , and their perpendicular separation is h. Explain why the joined parallelogram has base $b_1+b_2$ and height h, and derive the trapezoid area formula.
1. Hint 1
  Track where each parallel base goes under the half turn.
2. Hint 2
  The joined area contains two congruent trapezoids with no overlap of interiors.
Answer

The joined base is $b_1+b_2$ and height is $h$ ; the trapezoid area is $A=\tfrac12(b_1+b_2)h$ .

Full solution

The half turn sends each base to a parallel segment on the other base line.
Along either outside base of the joined figure, one original base and one copied base meet end to end, so their lengths add.

The two outer legs are half-turn images, hence parallel and congruent.
The other opposite sides lie on the two original parallel base lines, making a parallelogram with perpendicular height h.

Convexity puts the two copies on opposite sides of the shared leg, so their interiors do not overlap.
Congruence gives them equal areas.

The parallelogram area is
$2A=(b_1+b_2)h.$

Dividing by two gives
$A=\frac12(b_1+b_2)h.$

The numerical illustration shows the same joining arrangement for bases 11 and 5 units.

Answer

The joined base is $b_1+b_2$ and height is $h$ ; the trapezoid area is $A=\tfrac12(b_1+b_2)h$ .

Key idea

A rigidly moved congruent copy has equal area, so the joined total is twice one area.
Problem 6 The fabric insert

A convex kite-shaped fabric insert has full diagonals $24$ cm and $35$ cm. A rectangular piece of fabric measures $36$ cm by $25$ cm. The insert is cut from that piece without stretching. How much fabric area remains?

The insert has the stated kite side pairs.

Text description of this figure

Schematic convex kite ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 24 cm. AC is labeled 35 cm. Single side ticks pair AB with AD; double ticks pair BC with CD.
1. Hint 1
  Find the original rectangular area and the area removed.
2. Hint 2
  A kite has perpendicular diagonals; subtract its area from the original rectangular area.
Answer

$480$ square cm.

Full solution

The rectangular fabric has area
$36\cdot25=900\text{ cm}^2.$

The kite has perpendicular diagonals, so its area is
$\begin{aligned}A&=\frac12\cdot24\cdot35,\\A&=420\text{ cm}^2.\end{aligned}$

The remaining area is
$900-420=480\text{ cm}^2.$

Answer

$480$ square cm.

Key idea

Subtract an area only after checking the formula for the removed region.
Problem 7 The right trapezoid

A convex trapezoid has one leg perpendicular to its bases. Its parallel bases have lengths $13$ cm and $28$ cm. Its other leg is $17$ cm. Find the height and area.

The left leg is perpendicular to the bases.

Text description of this figure

Schematic trapezoid, not to scale. Parallel top and bottom bases are labeled 13 cm and 28 cm. A dashed perpendicular between them is labeled h, with a right-angle square. The left leg is perpendicular to the bases. The right slanted side is labeled 17 cm.
1. Hint 1
  Drop a perpendicular from the shorter base's other endpoint.
2. Hint 2
  The right triangle beside the rectangle has horizontal leg equal to the difference of the bases.
3. Hint 3
  Use the Pythagorean theorem to find the perpendicular separation.
Answer

$\begin{aligned}h&=8\text{ cm},\\A&=164\text{ cm}^2.\end{aligned}$

Full solution

The difference of the parallel bases is $15$ cm.
Together with the $17$ cm slanted leg, this forms a right triangle with the trapezoid's height h.

The Pythagorean theorem gives
$\begin{aligned}h^2&=17^2-15^2, \\h^2&=64.\end{aligned}$

The positive height is $8$ cm.
Use it with the parallel bases:
$\begin{aligned}A&=\frac{13+28}{2}\cdot8, \\A&=164\text{ cm}^2.\end{aligned}$

Answer

$\begin{aligned}h&=8\text{ cm},\\A&=164\text{ cm}^2.\end{aligned}$

Key idea

A slanted leg can help recover the height without becoming the height.
Problem 8 The diagonal condition

Test the statement: every convex quadrilateral with equal diagonals has area equal to half their product. Give a numerical counterexample using a nonsquare rectangle, including its diagonal length, actual area, and half-product value.
1. Hint 1
  A rectangle has equal diagonals, but they need not be perpendicular.
2. Hint 2
  Choose unequal side lengths from a Pythagorean triple to keep the diagonal exact.
Answer

One counterexample is a $5$ cm by $12$ cm rectangle: each diagonal is $13$ cm, actual area $60$ square cm, and the half-product is $84.5$ square cm. Other valid counterexamples are accepted.

Full solution

Choose a rectangle with sides $5$ cm and $12$ cm.
Its diagonals are equal by the rectangle theorem.

The Pythagorean theorem gives each diagonal length $13$ cm because
$5^2+12^2=13^2.$

The actual area is
$\begin{aligned}A&=5\cdot12, \\A&=60\text{ cm}^2.\end{aligned}$

Half the diagonal product is
$\frac12\cdot13\cdot13=84.5\text{ cm}^2.$

These values differ.
Equal diagonals do not supply the perpendicularity needed for the diagonal area formula.

A nonsquare rectangle has equal diagonals without a right-angle intersection.

Answer

One counterexample is a $5$ cm by $12$ cm rectangle: each diagonal is $13$ cm, actual area $60$ square cm, and the half-product is $84.5$ square cm. Other valid counterexamples are accepted.

Key idea

A formula's numerical inputs do not replace its geometric hypotheses.
Problem 9 The rhombus measurements

A rhombus has side length $17$ cm and one diagonal $16$ cm. Find its other full diagonal, its area, and its perpendicular height to a side. Give exact answers.

The side length is given in the statement, and the diagonal intersection is a right angle.

Text description of this figure

Schematic rhombus ABCD, not to scale. A is above O, B right, C below, D left. Diagonals AC and BD intersect at O with a right-angle square. BD is labeled 16 cm. AC is labeled d. A single tick on every side marks all four sides congruent.
1. Hint 1
  A diagonal and its perpendicular bisector form right triangles whose hypotenuse is a rhombus side.
2. Hint 2
  Use half of the known diagonal to find half of the other diagonal.
3. Hint 3
  After finding area from diagonals, use the side as the base in the parallelogram formula.
Answer

$\begin{aligned}d&=30\text{ cm},\\A&=240\text{ cm}^2,\\h&=\frac{240}{17}\text{ cm}.\end{aligned}$

Full solution

The known diagonal is bisected into $8$ cm parts.
The other half-diagonal is a leg of a right triangle with hypotenuse $17$ cm.

Its square is
$17^2-8^2=225.$

The half-diagonal is $15$ cm, so the full other diagonal is $30$ cm.

The perpendicular diagonal formula gives
$\begin{aligned}A&=\frac12\cdot16\cdot30, \\A&=240\text{ cm}^2.\end{aligned}$

A rhombus is a parallelogram.
Using its side as base gives
$17h=240.$

The perpendicular height is
$h=\frac{240}{17}\text{ cm}.$

Answer

$\begin{aligned}d&=30\text{ cm},\\A&=240\text{ cm}^2,\\h&=\frac{240}{17}\text{ cm}.\end{aligned}$

Key idea

Diagonal halves recover lengths, while the full diagonals give area.
Problem 10 Two area specifications

Case A. A convex trapezoid has area $175$ square cm. Its parallel bases are $(2x+3)$ cm and $(x+8)$ cm, and its perpendicular height is $7$ cm. Find x and both bases.

Case B. A convex kite ABCD has equal-side pairs meeting at A and C. Its diagonals intersect at O, its area is $175$ square cm, $BD$ is $14$ cm, and $AO$ is $9$ cm. Find the full diagonal AC and the part OC. Explain why the diagonal area formula applies.

Case A has marked parallel bases and a perpendicular height.

Text description of this figure

Schematic trapezoid, not to scale. Parallel top and bottom bases are labeled (2x+3) cm and (x+8) cm. A dashed perpendicular between them is labeled 7 cm, with a right-angle square.
1. Hint 1
  Case A: recover the base sum from the area and perpendicular height.
2. Hint 2
  Case B: recall the diagonal angle guaranteed by the kite properties.
3. Hint 3
  Case B: find the full unknown diagonal before subtracting its known part.
Answer

Case A.
$\begin{aligned}x&=13,\\b_1&=29\text{ cm},\\b_2&=21\text{ cm}.\end{aligned}$
Case B.
$\begin{aligned}AC&=25\text{ cm},\\OC&=16\text{ cm}.\end{aligned}$
The kite has perpendicular diagonals, which justify the formula.

Full solution

Case A.
Twice the area divided by the height is the base sum:
$2\cdot175\div7=50.$

Case A.
The equation for x is
$3x+11=50.$

Case A.
Solving gives $x=13$ .
Substituting into the base expressions gives $29$ cm and $21$ cm, both positive.

Case A.
The check is
$\frac{29+21}{2}\cdot7=175.$

Case B.
A convex kite has perpendicular diagonals, so its area is half the product of the full diagonal lengths.

Case B.
The full unknown diagonal is
$\begin{aligned}AC&=\frac{2\cdot175}{14}, \\AC&=25\text{ cm}.\end{aligned}$

Case B.
Subtract the known part along AC:
$\begin{aligned}OC&=25-9, \\OC&=16\text{ cm}.\end{aligned}$

Case B uses the full diagonal AC before finding its lower part.

Answer

Case A.
$\begin{aligned}x&=13,\\b_1&=29\text{ cm},\\b_2&=21\text{ cm}.\end{aligned}$
Case B.
$\begin{aligned}AC&=25\text{ cm},\\OC&=16\text{ cm}.\end{aligned}$
The kite has perpendicular diagonals, which justify the formula.

Key idea

Reverse an area formula only after verifying its geometric conditions.

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