24. Triangles · Mastering AMC 8

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Area of a Triangle

In this section you will be able to find the area of a triangle from its base and height, from its three sides, or from its incircle or circumcircle.

Grab the top corner of a triangle and slide it sideways, keeping it at the same height. The triangle leans and stretches. Does it get bigger? Try it and see!

The area of a triangle is half of base times height. The height is the straight-down distance from the top corner to the line of the base. Slide the top corner along a line parallel to the base, and the height stays the same. So the area stays the same too.

Know all three sides but no height? Use Heron’s formula. Know the radius of the circle inside (the incircle) or around it (the circumcircle)? There’s a formula for those too.

Exploreplay first, formulas later

Try this: Slide A left and right with the arrow keys. The shape changes. Does the area change? Why not?
Try this: Slide A past C so the triangle leans out. Where does the height land now? Is the area still the same?
Try this: Press Heron, then Incircle, then Circumcircle. Do all of them give the same area as base × height ÷ 2?

Name it

Theorem 24.1.1 · Using base and height

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

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

The height is the perpendicular from the top vertex down to the line of the base.

Theorem 24.1.2 · Heron's Formula

A triangle with sides $a, b, c$ and semiperimeter $s$ has an area of

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

The semiperimeter is half the perimeter: $s = \frac{a+b+c}{2}$ .

Definition 24.1.3 · Inradius

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

Definition 24.1.4 · Incenter

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

Theorem 24.1.5 · Area of a triangle using inradius

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

\text{inradius}\cdot\text{semiperimeter} = rs

Theorem 24.1.7

Inradius $r$ of a right triangle:

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

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

Definition 24.1.8 · Circumradius

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

Definition 24.1.9 · Circumcenter

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

Theorem 24.1.10 · Area of a triangle 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

\frac{abc}{4R}

Watch itpredict, then reveal

Example 24.1

EMCC

Given that the heights of $\triangle ABC$ have lengths $\frac{15}{7}$ , $5$ , and $3$ , what is the square of the area of $\triangle ABC$ ?

1
Call the area $K$ . Each height goes with one side: $K = \frac{1}{2}\cdot\text{side}\cdot\text{height}$ . So which side is longest: the one with the short height or the tall height?

Predict first:

open on YouTube

Try it

Problem 24.6.1

AMC 8

The twelve-sided figure shown has been drawn on $1\text{ cm}\times 1\text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$ ?

Problem 24.6.3

AMC 8

Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?

Problem 24.6.2

AMC 8

A triangle with vertices as $A = (1, 3)$ , $B = (5, 1)$ , and $C = (4, 4)$ is plotted on a $6\times 5$ grid. What fraction of the grid is covered by the triangle?

Problem 24.6.10

AMC 8

What is the area of the triangle formed by the lines $y = 5$ , $y = 1 + x$ , and $y = 1 - x$ ?

Contest cornerreal contest problems

Problem 24.6.6

AMC 8

The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?

Problem 24.6.7

AMC 8

In square $ABCE$ , $AF = 2FE$ and $CD = 2DE$ . What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$ ?

Problem 24.6.9

AMC 8

What is the area of the shaded pinwheel shown in the $5\times 5$ grid?

Problem 24.6.14(additional)

What is the radius of a circle inscribed in a triangle with side lengths 6, 8, and 10?

Checkpoint for section 24.1

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

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Special Triangles

In this section you will be able to use the side ratios of the equilateral, 45-45-90, 30-60-90 and 13-14-15 triangles.

Some triangles show up in contest after contest. If you know their side ratios by heart, you can skip pages of work. Meet the four VIP triangles!

Look: a 30-60-90 triangle is exactly half of an equilateral triangle. That is why its hypotenuse is twice its short leg. A 45-45-90 triangle is half of a square, so its hypotenuse is the square’s diagonal.

The 13-14-15 triangle is two right triangles glued together along a height of 12.

Exploreplay first, formulas later

Try this: The dashed height splits the equilateral triangle into two halves. What angles does each half have?
Try this: Pick 45-45-90 and change the leg. What do you multiply the leg by to get the hypotenuse?
Try this: Pick 30-60-90. How does the hypotenuse compare with the short leg? Is that true for every size?
Try this: Pick 13-14-15 and draw the height. Which two Pythagorean triples do you see?

Name it

Theorem 24.2.1 · Equilateral Triangle

If the side length of an equilateral triangle is $a$ :

\text{Height of the triangle} = \frac{\sqrt{3}}{2}a

This follows directly from the $30-60-90$ triangle.

\text{Area of the triangle} = \frac{\sqrt{3}}{4}a^2

Theorem 24.2.2 · 45-45-90 Triangle

If the side length of a 45-45-90 triangle is $a$ :

\text{hypotenuse of the triangle} = \sqrt{2}\times\text{side length} = \sqrt{2}a

\text{Area of the triangle} = \frac{1}{2}\times\text{side length}^2 = \frac{1}{2}a^2

Theorem 24.2.3 · 30-60-90 Triangle

If the short leg length of a 30-60-90 triangle is $a$ :

\text{Long Leg of the triangle} = \sqrt{3}\times\text{short leg} = \sqrt{3}a

\text{hypotenuse of the triangle} = 2\times\text{short leg} = 2a

\text{Area} = \frac{\sqrt{3}}{2}\times\text{short leg}^2 = \frac{\sqrt{3}}{2}a^2

Theorem 24.2.4 · 13-14-15 Triangle

If the three sides of a triangle are 13, 14, and 15, it can be divided into two right triangles with side lengths: $5, 12, 13$ and $9, 12, 15$ .

Area of this triangle $= \frac{1}{2}\times 14\times 12 = 84$

Contest cornerreal contest problems

Problem 24.6.15(additional)

MATHCOUNTS

A unit square contains four congruent non-overlapping equilateral triangles as shown in the figure. What is the largest possible side-length of one of the triangles?

Checkpoint for section 24.2

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

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Pythagorean Theorem

In this section you will be able to use a² + b² = c² to find a missing side of a right triangle, and spot Pythagorean triples fast.

Four identical right triangles sit in a big square. Shuffle them around and the empty space changes shape, but not size. That one picture proves the most famous theorem in geometry.

In a right triangle, the square on the hypotenuse equals the two squares on the legs added together. Memorize the common Pythagorean triples. They save you from square roots, and contest problems love them.

Exploreplay first, formulas later

Try this: Press Rearrange the triangles. The four triangles move, but the big square is the same. What happens to the empty space?
Try this: The empty space was c² before and a² + b² after. Why must those be equal?
Try this: Change the legs to 6 and 8, then 5 and 12. Which ones give a whole-number hypotenuse?

Name it

Theorem 24.3.1 · Pythagorean Theorem

A right triangle with legs $a$ and $b$ and hypotenuse $c$ satisfies the following relation:

c^2 = a^2 + b^2

Fact 24.3.2 · Important Pythagorean Triples

Important Pythagorean Triples

$3, 4, 5$ · $5, 12, 13$ · $7, 24, 25$ · $8, 15, 17$ · $9, 40, 41$ · $20, 21, 29$

If all numbers in a Pythagorean triple are multiplied by a constant, the resulting numbers still form a Pythagorean triple. For example, these are all Pythagorean triples: $3, 4, 5$ · $6, 8, 10$ · $9, 12, 15$ · $12, 16, 20$ · $15, 20, 25$ .

Theorem 24.3.3 · Special Properties of right triangles

In a right triangle $ABC$ where $B$ is the right angle, let $D$ be the foot of the perpendicular from $B$ to $AC$ . The following triangles are similar:

\triangle ABC \sim \triangle ADB \sim \triangle BDC

Length of the perpendicular to the hypotenuse: $BD = \frac{AB\cdot BC}{AC}$ .

Also note that:

AD\cdot CD = BD^2 \qquad AD\cdot AC = AB^2 \qquad CD\cdot CA = CB^2

Try it

Problem 24.6.8

AMC 8

Rectangle $ABCD$ and right triangle $DCE$ have the same area. They are joined to form a trapezoid, as shown. What is $DE$ ?

Problem 24.6.4

AMC 8

In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ABD$ ?

Contest cornerreal contest problems

Problem 24.6.5

AMC 8

In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB = 12$ , $BC = 4$ , $CD = 3$ , and $AD = 13$ . What is the area of quadrilateral $ABCD$ ?

Checkpoint for section 24.3

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

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Triangle Properties

In this section you will be able to work with cevians, medians and the centroid, and use the median to the hypotenuse of a right triangle.

Can you balance a cardboard triangle on the tip of a pencil? There is exactly one spot that works. It is where the three medians cross!

A median joins a corner to the middle of the opposite side. All three medians meet at the centroid. It sits two thirds of the way down each median.

In a right triangle, the midpoint of the hypotenuse is the same distance from all three corners.

Exploreplay first, formulas later

Try this: Drag A anywhere. Do the three medians always meet at one point G?
Try this: Watch the readout. What fraction of each median is the piece from the vertex to G?
Try this: Press Right-triangle median, then put A straight above B. What do you notice about PA, PB and PC?

Name it

Definition 24.4.1 · Cevian

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

Definition 24.4.2 · Median

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

Definition 24.4.3 · Centroid

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

Theorem 24.4.4

The centroid of a triangle is on the median and it is $\frac{2}{3}$ of the way from one of the vertices to the midpoint of the opposite side.

Theorem 24.4.5 · Median in a Right Triangle

In a right triangle $ABC$ , let the median from point $B$ intersect $AC$ at a point $P$ . Then

AP = BP = CP.

Basically, in a right triangle $AC$ is the diameter of the circumcircle, and $PC$ , $PA$ , and $PB$ are radii of the circumcircle.

Contest cornerreal contest problems

Problem 24.6.13(additional)

BmMT

In right triangle $\triangle ABC$ with $AB = 5$ , $BC = 12$ , and $CA = 13$ , point $D$ lies on $CA$ such that $AD = BD$ . The length of $CD$ can then be expressed in the form $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$ . Compute $m + n$ .

Checkpoint for section 24.4

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

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Angle Bisector Theorem

In this section you will be able to use the Angle Bisector Theorem to split the opposite side in the ratio of the two other sides.

Cut the angle at A exactly in half and let the cut hit the far side at D. The longer side next to A gets the longer piece. But how much longer?

The bisector splits the opposite side in the same ratio as the two sides next to the angle. Want more? Watch More on the Angle Bisector Theorem.

Exploreplay first, formulas later

Try this: Drag B farther from A. Does D slide toward B or toward C?
Try this: Compare AB ÷ BD with AC ÷ CD in the readout. Can you make them different?
Try this: Make AB equal to AC. Where does D land now?

Name it

Theorem 24.5.1 · Angle Bisector Theorem

If the line $AD$ bisects angle $A$ , then

\frac{AB}{BD} = \frac{AC}{CD}

Watch itpredict, then reveal

Example 24.2

BmMT

In triangle $\triangle ABC$ , the angle trisector of $\angle BAC$ closer to $AC$ than $AB$ intersects $BC$ at $D$ . Given that triangle $\triangle ABD$ is equilateral with area 1, compute the area of triangle $\triangle ABC$ .

1
$\triangle ABD$ is equilateral, so $\angle BAD = 60^\circ$ . $AD$ is the trisector closer to $AC$ , so $\angle BAD$ is two of the three equal parts. How big is $\angle BAC$ ?

Predict first:
Your prediction

open on YouTube

Contest cornerreal contest problems

Problem 24.6.12(additional)

BmMT

Right triangle $\triangle ABC$ has $AB = 5$ , $BC = 12$ , and $CA = 13$ . Point $D$ lies on the angle bisector of $\angle BAC$ such that $CD$ is parallel to $AB$ . Compute the length of $BD$ .

Checkpoint for section 24.5

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

24.6

Practice Problems

In this section you will be able to mix area, Pythagoras and area ratios on contest problems.

One more challenge from the AMC 8! You’ll need area ratios: triangles with the same height have areas in the ratio of their bases. 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 24.6.11(additional)

AMC 8

In triangle $ABC$ , point $D$ divides side $\overline{AC}$ so that $AD : DC = 1 : 2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE$ . Given that the area of $\triangle ABC$ is $360$ , what is the area of $\triangle EBF$ ?