Geometry · Chapter 30

Triangle Area and Length

One triangle, five area formulas (base-height, Heron, rs, abc/4R, shoelace) plus Pick's theorem, medians, centroids and Stewart's theorem for every cevian.

What's this chapter about?

30.1 Area of a Triangle
30.2 Triangle Properties
30.3 Practice Problems

You'll need

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30.10%

Area of a Triangle

In this section you will be able to find a triangle's area five ways (base-height, Heron, rs, abc/4R, shoelace) and pick the fastest one.

One triangle, five area formulas: ½bh, Heron, rs, abc/4R and the shoelace. Each one is the fastest for some problem. Your job is to learn which.

Look at what a problem gives you. A base and a height? Use $\frac12 bh$ . Three sides? Heron. The inradius or the circumradius? $K = rs$ or $K = \frac{abc}{4R}$ . Coordinates? The shoelace formula.

The formulas also run backwards. Know the area and the sides, and you get $r = \frac{K}{s}$ and $R = \frac{abc}{4K}$ for free.

Triangles that share a height have areas in the ratio of their bases. That one fact cracks Example 30.1.

Exploreplay first, formulas later

Try this: Drag A along the dashed line. The base and the height stay the same, so what happens to the area?
Try this: Press Heron, then Incircle, then Circumcircle. Does every formula give the same area?
Try this: Press Show me on Heron's Formula (the 13-14-15 triangle). Read r and R and check K = rs and K = abc/4R.
Try this: In the shoelace lab under Pick's Theorem, drag a vertex. Count i and b: does i + b/2 − 1 still match?

Name it

Theorem 30.1.1 · Using base and height

A triangle with base $b$ and height $h$ has an area of

\frac{1}{2} \cdot b \cdot h.

Theorem 30.1.2 · Heron's Formula

A triangle with sides $a$ , $b$ , $c$ and semiperimeter $s = \frac{a+b+c}{2}$ has an area of

\sqrt{s(s-a)(s-b)(s-c)}.

Definition 30.1.3 · Incenter

The incenter of a triangle is the intersection of all the angle bisectors. This point is also the center of the incircle, and it is equidistant from all three sides.

Definition 30.1.4 · In-radius

The inradius of a triangle is the radius of the circle inscribed in the triangle.

Theorem 30.1.5 · Inradius of a right triangle

The inradius $r$ of a right triangle is

r = \frac{1}{2}(a + b - c),

where $a$ and $b$ are the legs of the triangle and $c$ is the hypotenuse.

Theorem 30.1.6 · Using inradius

A triangle with inradius $r$ (the radius of the circle that can be inscribed in the triangle) and semiperimeter $s$ has an area of

\text{Area} = \text{inradius} \cdot \text{semiperimeter}, \qquad A = rs.

Definition 30.1.8 · Circumcenter

The circumcenter of a triangle is the intersection of all 3 perpendicular bisectors (a perpendicular bisector is the line that bisects a segment and is perpendicular to it). This point is also the center of the circumcircle, and it is equidistant from all three vertices.

Definition 30.1.9 · Circumradius

The circumradius of a triangle is the radius of the circle that the triangle is inscribed in.

Theorem 30.1.10 · Using circumradius

A triangle with circumradius $R$ (the radius of the circle that the triangle can be inscribed in) and sides $a$ , $b$ , $c$ has an area of

A = \frac{abc}{4R}.

Theorem 30.1.12 · Shoelace Theorem using coordinates

Suppose the polygon $P$ has vertices $(a_1, b_1), (a_2, b_2), \ldots, (a_n, b_n)$ , listed in clockwise order. Then the area $A$ of $P$ is

A = \frac{1}{2}\left|(a_1 b_2 + a_2 b_3 + \cdots + a_n b_1) - (b_1 a_2 + b_2 a_3 + \cdots + b_n a_1)\right|.

You can also go in counterclockwise order, as long as you take the absolute value of the answer. \nThe Shoelace Theorem gets its name from the picture you get when you list the coordinates in a column, repeat the first pair at the bottom, and mark the pairs of coordinates to be multiplied: the criss-cross lines look like the laces of a shoe.

#	x	y	↘ x·y′	↙ x′·y
A	1	1	2	7
B	7	2	42	6
C	3	6	3	6
A	1	1	Σ 47	Σ 19

Area = ½ × |47 − 19| = 14

signed: 14 (counter-clockwise)

Remark 30.1.13 · Intuitive Way of Thinking about Shoelace Theorem

Steps to the Shoelace Theorem:

Line up all of your polygon's coordinates in a vertical column.
Repeat your first coordinate at the bottom of the column.
Let $A$ be the sum of the products of all the rightward (down-right) diagonal pairs.
Let $B$ be the sum of the products of all the leftward (down-left) diagonal pairs.
Find

\frac{1}{2}|A - B|.

#	x	y	↘ x·y′	↙ x′·y
A	1	1	0	6
B	6	0	24	0
C	7	4	35	8
D	2	5	2	5
A	1	1	Σ 61	Σ 19

Area = ½ × |61 − 19| = 21

signed: 21 (counter-clockwise)

A = \tfrac12\left|\sum x_k y_{k+1} - \sum x_{k+1} y_k\right|

Theorem 30.1.14 · Pick's Theorem

If a polygon has vertices with integer coordinates (lattice points), then the area of the polygon is

i + \frac{b}{2} - 1,

where $i$ is the number of lattice points inside the polygon and $b$ is the number of lattice points on the boundary of the polygon.

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Example 30.1

AIME

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$ .

1
Call the unknown areas $x$ (next to $35$ , by $C$ ) and $y$ (next to $40$ , by $A$ ). Triangles that share a height have areas in the ratio of their bases. Which two small triangles have bases $BD$ and $DC$ and the same height?

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Problem 30.3.8(additional)

AMC 10

Given a triangle with side lengths $15$ , $20$ , and $25$ , find the triangle's shortest altitude.

Problem 30.3.12(additional)

AMC 10

A triangle has side lengths $10$ , $10$ , and $12$ . A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

Contest cornerreal contest problems

Problem 30.3.1

AMC 10

Find the area of the shaded region.

Problem 30.3.2

AMC 10

Triangle $ABC$ has a right angle at $B$ . Point $D$ is the foot of the altitude from $B$ , $AD = 3$ , and $DC = 4$ . What is the area of $\triangle ABC$ ?

Problem 30.3.9(additional)

AMC 10/12

A triangle with side lengths in the ratio $3 : 4 : 5$ is inscribed in a circle with radius $3$ . What is the area of the triangle?

Problem 30.3.10(additional)

AMC 10

The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?

Problem 30.3.11(additional)

AMC 12

Triangle $ABC$ has $AB = 13$ and $AC = 15$ , and the altitude to $\overline{BC}$ has length $12$ . What is the sum of the two possible values of $BC$ ?

Problem 30.3.20(additional)

AMC 10

A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are $3$ , $7$ , and $7$ , as shown. What is the area of the shaded quadrilateral?

Checkpoint for section 30.1

Get 2 of 3 right to master this section. One try each, no hints.

30.20%

Triangle Properties

In this section you will be able to use medians, the centroid and Stewart's theorem to find lengths inside a triangle.

A segment from a vertex to the opposite side is a cevian. Know its length and you know a lot: Stewart’s theorem turns the whole triangle into one equation.

Medians are the most famous cevians. All three meet at the centroid, which sits two thirds of the way down each one.

In a right triangle, the median to the hypotenuse is exactly half the hypotenuse. The midpoint is the center of the circumcircle.

For any cevian, Stewart’s theorem links the three sides, the two pieces $m$ and $n$ and the cevian $d$ : $man + dad = bmb + cnc$ .

Exploreplay first, formulas later

Try this: Drag D along BC. Watch man + dad and bmb + cnc. Do they ever disagree?
Try this: Press Median and change the sides. Turn on the formula: does d² = ½(b² + c²) − ¼a² hold?
Try this: Set the sides to 10, 8 and 6 with Median on. Compare AD with BD and DC. Why?
Try this: Turn on All three medians. What is AG : GM?

Name it

Definition 30.2.1 · Median

A median is a segment connecting a vertex of a triangle to the midpoint of the opposite side.

Theorem 30.2.2 · Median in a Right Triangle

In a right triangle $ABC$ with the right angle at $C$ , let the median from $C$ meet $AB$ at the point $M$ . Then

AM = BM = CM.

In a right triangle, $AB$ is a diameter of the circumcircle, and $MA$ , $MB$ and $MC$ are radii of the circumcircle.

Definition 30.2.3 · Centroid

In a triangle, the intersection of all 3 medians is the centroid.

Theorem 30.2.4 · Centroid splits each median 2 : 1

The centroid of a triangle lies on each median, $\frac{2}{3}$ of the way from the vertex to the midpoint of the opposite side.

Definition 30.2.5 · Cevian

A cevian is any segment from a vertex of a triangle to the opposite side. Medians and angle bisectors are special cases of cevians.

Theorem 30.2.6 · Stewart's Theorem

Given a triangle $ABC$ with sides of length $a$ , $b$ , $c$ opposite the vertices $A$ , $B$ , $C$ respectively. If the cevian $AD$ is drawn so that $BD = m$ , $DC = n$ and $AD = d$ , then

man + dad = bmb + cnc.

Corollary 30.2.8 · Stewart's for Angle Bisector

If $AD$ is an angle bisector, then

d^2 + mn = bc.

This follows from Stewart's Theorem and the Angle Bisector Theorem.

Corollary 30.2.9 · Stewart's Theorem for Medians

If $AD$ is a median, then

d^2 = \frac{1}{2}(b^2 + c^2) - \frac{1}{4}a^2.

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Example 30.2

AMC 12

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$ . The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles?

1
What is the area of an equilateral triangle with side $1$ ?

Predict first:
Your prediction

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Example 30.3

AMC 10

Triangle $ABC$ has side-lengths $AB = 12$ , $BC = 24$ , and $AC = 18$ . The line through the incenter of $\triangle ABC$ parallel to $\overline{BC}$ intersects $\overline{AB}$ at $M$ and $\overline{AC}$ at $N$ . What is the perimeter of $\triangle AMN$ ?

1
$I$ is the incenter, so $BI$ bisects $\angle B$ . Since $MN \parallel BC$ , $\angle MIB = \angle IBC$ (alternate angles). So what kind of triangle is $MBI$ ?

Predict first:
Click a choice or type its letter (A–E).

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Contest cornerreal contest problems

Problem 30.3.5

AMC 10

Triangle $AMC$ is isosceles with $AM = AC$ . Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV = CU = 12$ . What is the area of $\triangle AMC$ ?

Problem 30.3.7

AMC 10

Rectangle $ABCD$ has $AB = 5$ and $BC = 4$ . Point $E$ lies on $\overline{AB}$ so that $EB = 1$ , point $G$ lies on $\overline{BC}$ so that $CG = 1$ , and point $F$ lies on $\overline{CD}$ so that $DF = 2$ . Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$ , respectively. What is the value of $\dfrac{PQ}{EF}$ ?

Problem 30.3.16(additional)

AMC 12

Triangle $ABC$ is equilateral with $AB = 1$ . Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$ . Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE + FG$ ?

Problem 30.3.21(additional)

AMC 10

In $\triangle ABC$ we have $AB = 25$ , $BC = 39$ , and $AC = 42$ . Points $D$ and $E$ are on $\overline{AB}$ and $\overline{AC}$ respectively, with $AD = 19$ and $AE = 14$ . What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$ ?

Checkpoint for section 30.2

Get 2 of 3 right to master this section. One try each, no hints.

30.3

Practice Problems

In this section you will be able to mix every triangle tool from this chapter on contest problems.

Time to mix it up. These problems need area formulas, Pythagorean triples, Heron and a little algebra, sometimes all at once. Every problem has three hints if you get stuck.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 30.3.3

AMC 10

The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length $2$ and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?

Problem 30.3.4

AMC 10

Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$ . Points $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ , respectively, so that all sides of $\triangle ABP$ , $\triangle PCQ$ , and $\triangle QDA$ have integer lengths. What is the perimeter of $\triangle APQ$ ?

Problem 30.3.6

AMC 10

Points $P$ and $Q$ lie in a plane with $PQ = 8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ , $Q$ , and $R$ is a right triangle with area $12$ square units?

Problem 30.3.13(additional)

AMC 12

How many noncongruent integer-sided triangles with positive area and perimeter less than $15$ are neither equilateral, isosceles, nor right triangles?

Problem 30.3.14(additional)

AMC 10

Right triangles $T_1$ and $T_2$ have areas of $1$ and $2$ , respectively. A side of $T_1$ is congruent to a side of $T_2$ , and a different side of $T_1$ is congruent to a different side of $T_2$ . What is the square of the product of the lengths of the other (third) sides of $T_1$ and $T_2$ ?

Problem 30.3.15(additional)

AMC 12

Isosceles triangles $T$ and $T'$ are not congruent but have the same area and the same perimeter. The sides of $T$ have lengths $5$ , $5$ , and $8$ , while those of $T'$ have lengths $a$ , $a$ , and $b$ . Which of the following numbers is closest to $b$ ?

Problem 30.3.17(additional)

AMC 10

This problem also appears in the book as 30.3.6.

Problem 30.3.18(additional)

AIME

Triangle $ABC$ has $AB = 9$ and $BC : AC = 40 : 41$ . What's the largest area that this triangle can have?

Problem 30.3.19(additional)

AMC 10

Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right triangle to the area of the square is

Problem 30.3.22(additional)

AMC 12Challenge

Let $ABCD$ be a unit square. Let $Q_1$ be the midpoint of $\overline{CD}$ . For $i = 1, 2, \ldots$ , let $P_i$ be the intersection of $\overline{AQ_i}$ and $\overline{BD}$ , and let $Q_{i+1}$ be the foot of the perpendicular from $P_i$ to $\overline{CD}$ . What is

\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_iP_i\,?