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Triangles: Area & Pythagorean

Triangle area, Pythagorean theorem, special right triangles, triangle inequality

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Triangles: Area & PythagoreanAMC 10 Step by Step

Name: Date: October 28, 2026

20 problems, easiest first

What you need to know

Area. Area $= \frac12 bh$ , with $h$ the height to base $b$ . With two sides $a$ , $b$ and the angle $C$ between them, area $= \frac12 ab\sin C$ . An equilateral triangle with side $s$ has area $\frac{\sqrt3}{4}s^2$ . Example: sides $6$ and $8$ with a $30^\circ$ angle between them give area $\frac12\cdot6\cdot8\cdot\frac12 = 12$ .
Pythagorean Theorem. In a right triangle with legs $a$ , $b$ and hypotenuse $c$ , $a^2+b^2=c^2$ . The converse is true too: if $a^2+b^2=c^2$ , the angle across from $c$ is a right angle. Know the triples $3$ - $4$ - $5$ , $5$ - $12$ - $13$ , $8$ - $15$ - $17$ , $7$ - $24$ - $25$ , and their multiples, such as $9$ - $12$ - $15$ .
Special right triangles. A $45$ - $45$ - $90$ triangle has sides in the ratio $1:1:\sqrt2$ . A $30$ - $60$ - $90$ triangle has sides $1:\sqrt3:2$ , with the short leg across from $30^\circ$ . Example: a $30$ - $60$ - $90$ triangle with hypotenuse $10$ has legs $5$ and $5\sqrt3$ .
Triangle Inequality. Any two sides add up to more than the third: $a+b>c$ , strictly. Example: $4$ , $5$ , $9$ is not a triangle, since $4+5 = 9$ .
Acute, right, or obtuse? Let $c$ be the longest side. If $a^2+b^2 > c^2$ , the triangle is acute; if equal, right; if less, obtuse. Example: $5$ , $6$ , $8$ is obtuse, since $25+36 = 61 < 64$ .
Heron's Formula. Area $=\sqrt{s(s-a)(s-b)(s-c)}$ , where $s$ is the semiperimeter (half the perimeter). Example: sides $5$ , $5$ , $6$ give $s = 8$ and area $\sqrt{8\cdot3\cdot3\cdot2} = 12$ .

How AMC 10 tests it

Problems 3–10: a figure of squares and right triangles; find a length or area by dropping one perpendicular.
"How many integers $x$ make a triangle with sides $a$ , $b$ , $x$ ?" or "...an obtuse triangle?"
Mid-paper: two sides and an altitude, or a $60^\circ$ or $120^\circ$ angle. The altitude splits the triangle into two right triangles that share a leg.
Shaded regions from overlapping triangles, or a triangle cut from a rectangle, found by subtraction.

Standard approaches

Drop an altitude and write a Pythagorean equation on each side of it; subtracting them removes the altitude.
Look for special right triangles or Pythagorean triples before working with square roots.
Find the area in two ways (two base–height pairs) to get an unknown height. Example: the $3$ - $4$ - $5$ triangle has area $6$ , so the height $h$ to the hypotenuse has $\frac12\cdot5\cdot h = 6$ , and $h = \frac{12}{5}$ .
For shaded regions, subtract simple pieces from a rectangle or triangle around the region.

Worked example

In triangle $ABC$ , $AB = 13$ , $AC = 15$ , and the altitude from $A$ to line $BC$ has length $12$ . What is the sum of all possible values of $BC$ ?

(A) $14$ (B) $16$ (C) $18$ (D) $20$ (E) $22$

Solution. Let $D$ be the foot of the altitude, so $AD = 12$ .

Step 1. Right triangle $ABD$ has hypotenuse $13$ and leg $12$ , so it is a $5$ - $12$ - $13$ triangle and $BD = 5$ .

Step 2. Right triangle $ACD$ has hypotenuse $15$ and leg $12$ , so it is a $9$ - $12$ - $15$ triangle and $CD = 9$ .

Step 3. If $D$ lies between $B$ and $C$ , then $BC = 5 + 9 = 14$ . If $D$ lies outside segment $BC$ , beyond $B$ (so angle $B$ is obtuse), then $BC = 9 - 5 = 4$ . Both triangles exist, so both values count.

Step 4. The sum is $14 + 4 = 18$ .

The answer is $\boxed{\textbf{(C)}\ 18}$ .

Pitfalls

Forgetting the obtuse picture, where the foot of an altitude falls outside the triangle. In the example, that picture gives $BC = 4$ .
Mixing up the legs of a $30$ - $60$ - $90$ triangle. The longer leg is across from $60^\circ$ and is $\sqrt3$ times the shorter leg.
Allowing $a+b=c$ (a flat triangle) when counting integer side lengths. Example: with sides $3$ and $7$ , the third side can be $5$ , $6$ , $7$ , $8$ , or $9$ , but not $4$ or $10$ .

Traps that recur

Counting 50 or 51 values per coordinate (including boundary points, which are not interior), or forgetting to subtract the points that are also 60-ray partitional. (2011 AMC 10A #25)
Reading 'the altitude to side AB' as a segment drawn from B, which would put P at B; the altitude meant is the one from the opposite vertex C, and its foot H is what fixes BP. (2025 AMC 10A #23)
Misreading the kite: its angles are 60°, 90°, 120°, 90° (the two √3 sides meet at 60°, the two unit sides at 120°), and its long diagonal is 2. (2024 AMC 10A #22)
Guessing 1:1 because AC = CD and DE = EB look symmetric, or setting AD:DB = AC:DE = 4:3 as if it were an angle bisector. (2019 AMC 10B #16)

1.2008 AMC 10A #14
Older television screens have an aspect ratio of $4: 3$ . That is, the ratio of the width to the height is $4: 3$ . The aspect ratio of many movies is not $4: 3$ , so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of $2: 1$ and is shown on an older television screen with a $27$ -inch diagonal. What is the height, in inches, of each darkened strip?
1. A) $2$
2. B) $2.25$
3. C) $2.5$
4. D) $2.7$
5. E) $3$
2.2007 AMC 10B #10
Two points $B$ and $C$ are in a plane. Let $S$ be the set of all points $A$ in the plane for which $\triangle ABC$ has area $1.$ Which of the following describes $S?$
1. A) $\text{two parallel lines}$
2. B) $\text{a parabola}$
3. C) $\text{a circle}$
4. D) $\text{a line segment}$
5. E) $\text{two points}$
3.2006 AMC 10B #10
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$ . What is the greatest possible perimeter of the triangle?
1. A) $43$
2. B) $44$
3. C) $45$
4. D) $46$
5. E) $47$
4.2005 AMC 10A #8
In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$ , $E$ is between $B$ and $H$ , and $BE = 1$ . What is the area of the inner square $EFGH$ ?
1. A) $25$
2. B) $32$
3. C) $36$
4. D) $40$
5. E) $42$
5.2005 AMC 10B #14
Equilateral $\triangle ABC$ has side length $2$ , $M$ is the midpoint of $\overline{AC}$ , and $C$ is the midpoint of $\overline{BD}$ . What is the area of $\triangle CDM$ ?
1. A) $\frac {\sqrt {2}}{2}$
2. B) $\frac {3}{4}$
3. C) $\frac {\sqrt {3}}{2}$
4. D) $1$
5. E) $\sqrt {2}$
6.2005 AMC 10B #10
In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$ ?
1. A) $3$
2. B) $2\sqrt{3}$
3. C) $4$
4. D) $5$
5. E) $4\sqrt{2}$
7.2004 AMC 10A #9
In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ , $\angle{EAB}$ and $\angle{ABC}$ are right angles, $AB=4$ , $BC=6$ , $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$ ?
1. A) $2$
2. B) $4$
3. C) $5$
4. D) $8$
5. E) $9$
8.2003 AMC 10A #7
How many non-congruent triangles with perimeter $7$ have integer side lengths?
1. A) $1$
2. B) $2$
3. C) $3$
4. D) $4$
5. E) $5$
9.2002 AMC 10A #13
Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
1. A) $6$
2. B) $12$
3. C) $12.5$
4. D) $13$
5. E) $15$
10.2000 AMC 10 #10
The sides of a triangle with positive area have lengths $4$ , $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ , $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$ ?
1. A) $2$
2. B) $4$
3. C) $6$
4. D) $8$
5. E) $10$
11.2000 AMC 10 #5
Points $M$ and $N$ are the midpoints of sides $PA$ and $PB$ of $\triangle PAB$ . As $P$ moves along a line that is parallel to side $AB$ , how many of the four quantities listed below change?
(a) the length of the segment $MN$
(b) the perimeter of $\triangle PAB$
(c) the area of $\triangle PAB$
(d) the area of trapezoid $ABNM$
1. A) $0$
2. B) $1$
3. C) $2$
4. D) $3$
5. E) $4$
12.2000 AMC 10 #7
In rectangle $ABCD$ , $AD=1$ , $P$ is on $\overline{AB}$ , and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$ . What is the perimeter of $\triangle BDP$ ?
1. A) $3+\frac{\sqrt{3}}{3}$
2. B) $2+\frac{4\sqrt{3}}{3}$
3. C) $2+2\sqrt{2}$
4. D) $\frac{3+3\sqrt{5}}{2}$
5. E) $2+\frac{5\sqrt{3}}{3}$
13.2023 AMC 10A #11
A square of area $2$ is inscribed in a square of area $3$ , creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?
1. A) $\frac15$
2. B) $\frac14$
3. C) $2-\sqrt3$
4. D) $\sqrt3-\sqrt2$
5. E) $\sqrt2-1$
14.2023 AMC 10A #17
Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$ . Point $P$ 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 $\triangle{APQ}$ ?
1. A) $84$
2. B) $86$
3. C) $88$
4. D) $90$
5. E) $92$
15.2020 AMC 10B #8
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?
1. A) $2$
2. B) $4$
3. C) $6$
4. D) $8$
5. E) $12$
16.2019 AMC 10B #15
Right triangles $T_1$ and $T_2$ have areas 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 other (third) sides of $T_1$ and $T_2$ ?
1. A) $\frac{28}{3}$
2. B) $10$
3. C) $\frac{32}{3}$
4. D) $\frac{34}{3}$
5. E) $12$
17.2018 AMC 10A #16
Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$ . Including $\overline{AB}$ and $\overline{BC}$ , how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$ ?
1. A) $5$
2. B) $8$
3. C) $12$
4. D) $13$
5. E) $15$
18.2014 AMC 10B #15
In rectangle $ABCD$ , $DC=2 \cdot CB$ and points $E$ and $F$ lie on $\overline{AB}$ so that $\overline{ED}$ and $\overline{FD}$ trisect $\angle ADC$ as shown. What is the ratio of the area of $\triangle DEF$ to the area of rectangle $ABCD$ ?
1. A) $\frac{\sqrt{3}}{6}$
2. B) $\frac{\sqrt{6}}{8}$
3. C) $\frac{3\sqrt{3}}{16}$
4. D) $\frac{1}{3}$
5. E) $\frac{\sqrt{2}}{4}$
19.2014 AMC 10B #13
Six regular hexagons surround a regular hexagon of side length $1$ as shown. What is the area of $\triangle ABC$ ?
1. A) $2\sqrt{3}$
2. B) $3\sqrt{3}$
3. C) $1+3\sqrt{2}$
4. D) $2+2\sqrt{3}$
5. E) $3+2\sqrt{3}$
20.2012 AMC 10A #15
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$ ?
1. A) $\frac16$
2. B) $\frac15$
3. C) $\frac29$
4. D) $\frac13$
5. E) $\frac{\sqrt{2}}{4}$

Triangles: Area & Pythagorean — answersAMC 10 Step by Step

Name: Date: October 28, 2026

Keep this page separate from the problems.

Answers

#	Answer	The idea it turns on
1	D	A 4 to 3 screen with a 27-inch diagonal is a 3-4-5 triangle scaled up: 21.6 wide and 16.2 tall. The movie is 10.8 tall, leaving 5.4 for two strips.
2	A	The base BC is fixed, so area 1 fixes the height. Points at that distance from line BC form two parallel lines, one on each side.
3	A	With sides x, 3x and 15, the two shorter sides must add to more than 3x. So x + 15 > 3x, and x is at most 7.
4	C	The corner triangle on side AB has legs 1 and (1 + inner side). The Pythagorean theorem with 50 gives the inner side 6.
5	C	Use CD = 2 as the base. M is halfway up side AC, so its height above line BD is half the triangle's height, or half of root 3.
6	A	Drop a perpendicular from C; it lands at the middle of AB. Its length squared is 48, so its foot is 4 from D, and BD = 4 - 1.
7	B	Both big right triangles contain the same piece, triangle ABD. Take it away from each and the difference stays the same: 16 - 12 = 4.
8	B	The longest side must be shorter than the other two together, so it is less than half of 7. That makes it 3, and the others add to 4.
9	B	15-20-25 is 3-4-5 made five times bigger, so it is a right triangle with area 150. The shortest height goes to the longest side: 300 divided by 25 is 12.
10	D	The third side must be strictly between 2 and 10. Two numbers in that range can differ by anything less than 8, but never by exactly 8.
11	B	P moves along a line parallel to AB, so its height above AB never changes. Since MN is always half of AB, only the perimeter can change.
12	B	Cutting the right angle at D into three parts gives 30 degree angles, so every triangle here is a 30-60-90 triangle and AD = 1 fixes all the other lengths.
13	C	The legs add to the big square's side and their squares add to 2. That gives their product, and sum plus product gives the ratio.
14	A	Go around the rectangle one right triangle at a time. Leg 30 forces 16-30-34; then leg 12 allows 5, 9 or 16, and only 9 makes the last triangle work.
15	D	The area forces R to be exactly 3 units from line PQ. Then check each place the right angle could be: at P, at Q, or at R.
16	A	The shared sides cannot play the same roles in both triangles, so the small triangle's hypotenuse must be a leg of the big one. That doubles a leg.
17	D	Sliding along the hypotenuse, the segment shrinks from 20 to the height (about 14.5), then grows to 21, hitting each whole number once per side.
18	A	Splitting the right angle at D into three equal parts gives 30 and 60 degree angles, so the 30-60-90 triangles fix where E and F land.
19	B	Measure from the center of the middle hexagon: each corner of triangle ABC is 2 units away, and the three corners are evenly spaced, so the triangle is equilateral.
20	B	AB has length 1, so only the height of C below AB matters. One segment drops 1 per 2 across, the other 2 per 1 across; they meet 2/5 down.

Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic note, tags and key insights are original to this site.