Why is a rhombus's area 12d_1d_2 when a parallelogram's is bh?

Both are correct for a rhombus, since a rhombus is a parallelogram. They just use different measurements. The diagonal formula comes from the fact that a rhombus fits inside a d_1 by d_2 rectangle and fills exactly half of it. Use bh when you know a side and the perpendicular height; use 12d_1d_2 when you know the diagonals.

How do I find the apothem if the problem only gives me the side length?

Split the polygon into n isosceles triangles from the center, then cut one of them in half with the apothem. That right triangle has a central angle of 180^/n, an opposite leg of s/2, and an adjacent leg of a, so a=s/2(180^/n). For a regular hexagon you can skip the trig: a=ssqrt(3)2.

What is the difference between the apothem and the radius of a regular polygon?

The radius goes from the center to a vertex; the apothem goes from the center to the midpoint of a side and meets that side at a right angle. The radius is always the longer of the two. Only the apothem belongs in A=12aP. If you are given the radius, find the apothem first using a right triangle.

When should I subtract areas instead of adding them?

Subtract when the region is a large simple shape with a piece removed, such as a rectangular yard with a triangular flower bed cut out, or when a coordinate polygon has slanted sides — enclose it in an axis-aligned rectangle and remove the corner right triangles. Adding is better when the figure naturally splits into a few clean shapes. Either route gives the same answer, so pick the one with fewer pieces.

GEOM-10.1

Area: Triangles, Quadrilaterals & Regular Polygons

Master area formulas for triangles, parallelograms, trapezoids, rhombuses, kites, and regular polygons — plus how to break composite and coordinate figures into pieces.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Area: Triangles, Quadrilaterals & Regular Polygons, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Area: Triangles, Quadrilaterals & Regular Polygons free

What this lesson covers

Almost every area formula in this lesson comes from one idea: a rectangle's area is base times height. Slice a parallelogram and slide the triangle to the other end and you get a rectangle. Copy a triangle, rotate it, and you get a parallelogram. Cut a regular polygon into congruent triangles from the center and you get the apothem formula. If you understand the cutting and rearranging, you only have to memorize a handful of results — and you can rebuild any of them when your memory blanks.

In this lesson you will compute areas for triangles, parallelograms, trapezoids, rhombuses, kites, and regular polygons, then use those pieces to handle composite shapes and figures drawn on the coordinate plane. The single skill that separates correct work from near-misses is identifying the true perpendicular height, not just any convenient side length.

Base and Perpendicular Height: The Core Family

Three formulas cover most polygons you will meet, and all three depend on a height measured perpendicular to the chosen base.

Figure	Formula	What the height means
Triangle	$A=\tfrac{1}{2}bh$	Perpendicular distance from the base line to the opposite vertex
Parallelogram	$A=bh$	Perpendicular distance between the two parallel sides
Trapezoid	$A=\tfrac{1}{2}(b_1+b_2)h$	Perpendicular distance between the two parallel bases

A triangle is exactly half of a parallelogram with the same base and height, which is where the

\tfrac{1}{2}

comes from. A trapezoid is the average of its two bases times the height — if

b_1=b_2

, the formula collapses to

A=bh

, the parallelogram case.

The most common error in this whole unit is using a slanted side as the height. In a parallelogram with side lengths 10 and 6, the number 6 is a side, not necessarily a height; the height is the perpendicular segment drawn between the parallel sides, and it is always less than or equal to the slanted side. The same trap appears with trapezoid legs and with obtuse triangles, where the altitude often falls outside the triangle. When a diagram gives you a small right-angle mark on a dashed segment, that dashed segment is your height.

Also watch which base pairs with which height. A triangle has three base-height pairs, and

\tfrac{1}{2}bh

gives the same answer for all three — but only if you pair a base with the altitude drawn to that same base.

Diagonal Formulas: Rhombuses and Kites

A rhombus and a kite both have perpendicular diagonals, and that is exactly why they share a formula:

A=\tfrac{1}{2}d_1d_2

Here is why it works. Draw the rectangle that circumscribes a kite so that each side of the rectangle passes through a vertex. That rectangle has dimensions

d_1

d_2

, and the kite fills exactly half of it — each of the four right triangles inside the kite is matched by a congruent triangle outside it. The same argument works for a rhombus, since a rhombus is a special kite.

A rhombus is also a parallelogram, so

A=bh

still works if you happen to know the side and the height. Use whichever pair of measurements the problem actually hands you.

Problems often give you a side length and one diagonal instead of both diagonals. Use the fact that the diagonals of a rhombus bisect each other at right angles: the four little triangles are congruent right triangles with legs

\tfrac{d_1}{2}

and

\tfrac{d_2}{2}

and hypotenuse equal to the side. So

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

.

In a kite, the diagonals are still perpendicular, but only one diagonal is bisected — the axis of symmetry bisects the other one. Students sometimes assume both halves are equal in a kite and get the wrong second diagonal. Label the four pieces separately and add them.

Finally, remember to halve the product. Forgetting the

\tfrac{1}{2}

gives exactly double the correct area, a very common slip.

Regular Polygons and the Apothem

A regular polygon has all sides congruent and all angles congruent, so you can connect the center to every vertex and split it into

n

congruent isosceles triangles. The apothem

a

is the perpendicular distance from the center to the midpoint of a side — it is the height of each of those triangles, and the side length

s

is each triangle's base.

One triangle has area

\tfrac{1}{2}as

. Multiply by

n

triangles:

A=\tfrac{1}{2}a(ns)

, and since

ns

is the perimeter

P

A=\tfrac{1}{2}aP

When a problem gives only the side length, build the apothem with trigonometry. The central angle of each isosceles triangle is

\frac{360^\circ}{n}

; the apothem cuts it into two right triangles with angle

\frac{180^\circ}{n}

at the center, opposite leg

\tfrac{s}{2}

, and adjacent leg

a

. So

a=\frac{s}{2\tan\left(\frac{180^\circ}{n}\right)}

For a regular hexagon this is easy without trig: the six triangles are equilateral, so the radius equals the side and the apothem is

\frac{s\sqrt{3}}{2}

.

Two distinctions matter. The radius reaches a vertex; the apothem reaches a side's midpoint. The radius is always longer. Mixing them up inflates the area. Also,

\tfrac{1}{2}aP

uses the whole perimeter, not one side — if you plug in

s

where

P

belongs, your answer will be

n

times too small.

Composite Figures and Coordinate Polygons

Real problems rarely arrive as a single named shape. The strategy is always the same: decompose the region into triangles, rectangles, trapezoids, and sectors whose areas you can compute, then add. Sometimes subtracting is faster — find the area of a large rectangle and remove a triangular notch.

On the coordinate plane, choose bases that are horizontal or vertical so lengths come straight from coordinate differences and the perpendicular height is obvious. A horizontal segment from

(x_1,y)

(x_2,y)

has length

|x_2-x_1|

, and the height to a vertex is the difference in

y

-values. Avoid using the distance formula on a slanted side and then hunting for its altitude; that is where arithmetic goes wrong.

When no side is horizontal or vertical, use the box method: draw the smallest axis-aligned rectangle containing the figure, then subtract the right triangles in the corners. This turns any awkward coordinate polygon into rectangles and right triangles.

A triangle with one vertex at the origin and other vertices

(x_1,y_1)

and

(x_2,y_2)

has area

\tfrac{1}{2}|x_1y_2-x_2y_1|

, which is a fast shortcut once you trust it.

Two habits prevent the most common mistakes. First, keep units square: lengths in centimeters give an area in square centimeters. Second, sketch and label every piece before computing, so you can check that the pieces cover the region exactly once with no overlap and no gaps.

Key terms

Altitude (height).: A segment perpendicular to the chosen base, reaching the opposite vertex or the opposite parallel side; its length is the $h$ in area formulas.
Base.: The side of a polygon chosen as a reference for measuring height. Any side can serve as the base as long as the height is measured perpendicular to it.
Apothem.: In a regular polygon, the perpendicular segment from the center to the midpoint of a side; it is the height of each of the $n$ congruent triangles formed from the center.
Radius of a regular polygon.: The segment from the center to a vertex. It is always longer than the apothem and must not be substituted for it.
Central angle of a regular polygon.: The angle at the center subtended by one side, equal to $\frac{360^\circ}{n}$ .
Kite.: A quadrilateral with two distinct pairs of congruent adjacent sides; its diagonals are perpendicular, and its area is $\tfrac{1}{2}d_1d_2$ .
Composite figure.: A region formed by combining or removing simpler shapes; its area is found by adding or subtracting the areas of those pieces.
Decomposition.: Cutting a figure into non-overlapping pieces whose areas can be computed individually and summed.

Worked example

Quadrilateral

ABCD

has vertices

A(0,0)

B(6,0)

C(8,5)

, and

D(2,7)

. Find its area.

Plot the points and notice that no pair of sides is obviously parallel, so no single quadrilateral formula applies directly. Decompose instead: draw diagonal

\overline{AC}

, splitting the figure into triangle

ABC

and triangle

ACD

.

Triangle

ABC

: side

\overline{AB}

runs from

(0,0)

(6,0)

, so it is horizontal with length

6

. Because this base lies on the

x

-axis, the perpendicular height to vertex

C(8,5)

is simply its

y

-coordinate,

5

. So

A_{ABC}=\tfrac{1}{2}(6)(5)=15

square units.

Triangle

ACD

: this triangle has vertex

A

at the origin, with the other vertices

C(8,5)

and

D(2,7)

. Use the origin shortcut

\tfrac{1}{2}|x_1y_2-x_2y_1|

with

(x_1,y_1)=(8,5)

and

(x_2,y_2)=(2,7)

A_{ACD}=\tfrac{1}{2}|8\cdot 7-2\cdot 5|=\tfrac{1}{2}|56-10|=\tfrac{1}{2}(46)=23

So triangle

ACD

has area

23

square units.

Add the pieces:

15+23=38

square units.

Check it for reasonableness with the box method. The smallest axis-aligned rectangle containing all four points runs from

x=0

x=8

and

y=0

y=7

, an area of

56

. Our answer of

38

is a bit over two-thirds of that box, which fits a quadrilateral that leaves noticeable corner triangles empty. The answer is plausible.

Practice questions

A trapezoid has parallel bases of 12 cm and 18 cm, a perpendicular height of 8 cm, one leg perpendicular to the bases, and the other leg slanting and measuring 10 cm. What is its area?

96 square centimeters
120 square centimeters
150 square centimeters
240 square centimeters

Answer: 120 square centimeters

Use

A=\tfrac{1}{2}(b_1+b_2)h=\tfrac{1}{2}(12+18)(8)=\tfrac{1}{2}(30)(8)=120

square centimeters. The value 150 comes from using the 10 cm leg as the height, which is the single most common mistake here — the leg is slanted, so it is longer than the true perpendicular distance between the bases. The value 240 comes from forgetting the factor of

\tfrac{1}{2}

, and 96 comes from using only one base.

A rhombus has a perimeter of 52 inches and one diagonal measuring 24 inches. Find the area of the rhombus, showing your reasoning.

Answer: 120 square inches

All four sides of a rhombus are congruent, so each side is

52\div 4=13

inches. The diagonals bisect each other at right angles, forming four congruent right triangles. One leg is half the known diagonal,

24\div 2=12

inches, and the hypotenuse is the side, 13 inches. By the Pythagorean Theorem, the other leg satisfies

12^2+x^2=13^2

, so

x^2=169-144=25

and

x=5

. That leg is half the second diagonal, so

d_2=10

inches. Then

A=\tfrac{1}{2}d_1d_2=\tfrac{1}{2}(24)(10)=120

square inches. The 5-12-13 triple is worth recognizing on sight.

Find the area of a regular hexagon with side length 8 meters. Give both an exact answer and a decimal approximation.

Answer: $96\sqrt{3}$ square meters, about 166.3 square meters

Connect the center to all six vertices. For a hexagon the six triangles are equilateral, so each has side 8, and the apothem is the altitude of an equilateral triangle:

a=\frac{8\sqrt{3}}{2}=4\sqrt{3}

meters. The perimeter is

P=6(8)=48

meters. Then

A=\tfrac{1}{2}aP=\tfrac{1}{2}(4\sqrt{3})(48)=96\sqrt{3}\approx 166.3

square meters. If you had used the general formula

a=\frac{s}{2\tan(180^\circ/6)}=\frac{8}{2\tan 30^\circ}

, you would get the same

4\sqrt{3}

. A frequent error is using 8 as the apothem; the apothem is shorter than the side here, about 6.93 meters.

FAQ

Why is a rhombus's area $\tfrac{1}{2}d_1d_2$ when a parallelogram's is $bh$ ?: Both are correct for a rhombus, since a rhombus is a parallelogram. They just use different measurements. The diagonal formula comes from the fact that a rhombus fits inside a $d_1$ by $d_2$ rectangle and fills exactly half of it. Use $bh$ when you know a side and the perpendicular height; use $\tfrac{1}{2}d_1d_2$ when you know the diagonals.
How do I find the apothem if the problem only gives me the side length?: Split the polygon into $n$ isosceles triangles from the center, then cut one of them in half with the apothem. That right triangle has a central angle of $\frac{180^\circ}{n}$ , an opposite leg of $\frac{s}{2}$ , and an adjacent leg of $a$ , so $a=\frac{s}{2\tan(180^\circ/n)}$ . For a regular hexagon you can skip the trig: $a=\frac{s\sqrt{3}}{2}$ .
What is the difference between the apothem and the radius of a regular polygon?: The radius goes from the center to a vertex; the apothem goes from the center to the midpoint of a side and meets that side at a right angle. The radius is always the longer of the two. Only the apothem belongs in $A=\tfrac{1}{2}aP$ . If you are given the radius, find the apothem first using a right triangle.
When should I subtract areas instead of adding them?: Subtract when the region is a large simple shape with a piece removed, such as a rectangular yard with a triangular flower bed cut out, or when a coordinate polygon has slanted sides — enclose it in an axis-aligned rectangle and remove the corner right triangles. Adding is better when the figure naturally splits into a few clean shapes. Either route gives the same answer, so pick the one with fewer pieces.

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