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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)

1.2012 AMC 10B #21
Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$ , $a$ , $a$ , $a$ , $2a$ , and $b$ . What is the ratio of $b$ to $a$ ?
1. A) $\sqrt{3}$
2. B) $2$
3. C) $\sqrt{5}$
4. D) $3$
5. E) $\pi$
2.2011 AMC 10B #18
Rectangle $ABCD$ has $AB = 6$ and $BC = 3$ . Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$ . What is the degree measure of $\angle AMD$ ?
1. A) $15$
2. B) $30$
3. C) $45$
4. D) $60$
5. E) $75$
3.2008 AMC 10A #18
A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?
1. A) $\frac{57}{4}$
2. B) $\frac{59}{4}$
3. C) $\frac{61}{4}$
4. D) $\frac{63}{4}$
5. E) $\frac{65}{4}$
4.2007 AMC 10B #17
Point $P$ is inside equilateral $\triangle ABC$ . Points $Q$ , $R$ , and $S$ are the feet of the perpendiculars from $P$ to $\overline{AB}$ , $\overline{BC}$ , and $\overline{CA}$ , respectively. Given that $PQ=1$ , $PR=2$ , and $PS=3$ , what is $AB$ ?
1. A) $4$
2. B) $3\sqrt{3}$
3. C) $6$
4. D) $4\sqrt{3}$
5. E) $9$
5.2005 AMC 10A #19
Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated $45^{\circ}$ , as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?
1. A) $1$
2. B) $\sqrt{2}$
3. C) $\frac{3}{2}$
4. D) $\sqrt{2}+\frac{1}{2}$
5. E) $2$
6.2004 AMC 10B #18
In the right triangle $\triangle ACE$ , we have $AC=12$ , $CE=16$ , and $EA=20$ . Points $B$ , $D$ , and $F$ are located on $AC$ , $CE$ , and $EA$ , respectively, so that $AB=3$ , $CD=4$ , and $EF=5$ . What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$ ?
1. A) $\frac{1}{4}$
2. B) $\frac{9}{25}$
3. C) $\frac{3}{8}$
4. D) $\frac{11}{25}$
5. E) $\frac{7}{16}$
7.2025 AMC 10A #23
Triangle $\triangle ABC$ has side lengths $AB = 80$ , $BC = 45$ , and $AC = 75$ . The bisector of $\angle B$ and the altitude to side $\overline{AB}$ intersect at point $P.$ What is $BP$ ?
1. A) $18$
2. B) $19$
3. C) $20$
4. D) $21$
5. E) $22$
8.2024 AMC 10A #22
Let $\mathcal K$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal K$ are used to form the polygon shown below. What is the area of triangle $\Delta ABC$ ?
1. A) $2+3\sqrt3$
2. B) $\dfrac92\sqrt3$
3. C) $\dfrac{10+8\sqrt3}{3}$
4. D) $8$
5. E) $5\sqrt3$
9.2019 AMC 10B #16
In $\triangle ABC$ with a right angle at $C,$ point $D$ lies in the interior of $\overline{AB}$ and point $E$ lies in the interior of $\overline{BC}$ so that $AC=CD,$ $DE=EB,$ and the ratio $AC:DE=4:3.$ What is the ratio $AD:DB?$
1. A) $2:3$
2. B) $2:\sqrt{5}$
3. C) $1:1$
4. D) $3:\sqrt{5}$
5. E) $3:2$
10.2018 AMC 10A #23
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
1. A) $\frac{25}{27}$
2. B) $\frac{26}{27}$
3. C) $\frac{73}{75}$
4. D) $\frac{145}{147}$
5. E) $\frac{74}{75}$
11.2018 AMC 10B #17
In rectangle $PQRS$ , $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ , $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$ ?
1. A) $1$
2. B) $7$
3. C) $21$
4. D) $92$
5. E) $106$
12.2017 AMC 10B #19
Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ ?
1. A) $9:1$
2. B) $16:1$
3. C) $25:1$
4. D) $36:1$
5. E) $37:1$
13.2015 AMC 10A #19
The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$ . The rays trisecting $\angle ACB$ intersect $AB$ at $D$ and $E$ . What is the area of $\bigtriangleup CDE$ ?
1. A) $\dfrac{5\sqrt{2}}{3}$
2. B) $\dfrac{50\sqrt{3}-75}{4}$
3. C) $\dfrac{15\sqrt{3}}{8}$
4. D) $\dfrac{50-25\sqrt{3}}{2}$
5. E) $\dfrac{25}{6}$
14.2014 AMC 10A #22
In rectangle $ABCD$ , $AB=20$ and $BC=10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$ . What is $AE$ ?
1. A) $\dfrac{20\sqrt3}3$
2. B) $10\sqrt3$
3. C) $18$
4. D) $11\sqrt3$
5. E) $20$
15.2005 AMC 10A #25
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 $BCED$ ?
1. A) $\frac{266}{1521}$
2. B) $\frac{19}{75}$
3. C) $\frac{1}{3}$
4. D) $\frac{19}{56}$
5. E) $1$
16.2004 AMC 10A #20
Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ ?
1. A) $\frac{4}{3}$
2. B) $\frac{3}{2}$
3. C) $\sqrt{3}$
4. D) $2$
5. E) $1+\sqrt{3}$
17.2002 AMC 10A #23
Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$ .
1. A) $15/2$
2. B) $8$
3. C) $17/2$
4. D) $9$
5. E) $19/2$
18.2002 AMC 10B #22
Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^{\circ}$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$ .
1. A) $24$
2. B) $26$
3. C) $28$
4. D) $30$
5. E) $32$
19.2001 AMC 10 #24
In trapezoid $ABCD$ , $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$ , with $AB+CD=BC$ , $AB<CD$ , and $AD=7$ . What is $AB\cdot CD$ ?
1. A) $12$
2. B) $12.25$
3. C) $12.5$
4. D) $12.75$
5. E) $13$
20.2011 AMC 10A #25
Let $R$ be a square region and $n\ge4$ an integer. A point $X$ in the interior of $R$ is called $n\text{-}ray$ partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?
1. A) $1500$
2. B) $1560$
3. C) $2320$
4. D) $2480$
5. E) $2500$

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	A	Two points 2a apart, each exactly a from a third point, force that third point to be the midpoint. The fourth point then makes an equilateral triangle.
2	E	Parallel sides make angle MDC equal angle AMD, so triangle MDC has two equal sides: CM = CD = 6. With CB = 3, angle CMB is 30.
3	B	Square the sum of the legs: it equals the hypotenuse squared plus 4 times the area. Since that sum is 32 minus the hypotenuse, the squares cancel.
4	D	Join P to the three corners. The three small triangles fill the big one, and adding their areas shows the triangle's height equals 1 + 2 + 3 = 6.
5	D	The tilted square rests on its neighbors' inner top corners, at height 1. Its bottom tip is half an inch lower, and B is one diagonal above the tip.
6	E	Find the middle triangle by cutting off the three corner triangles. Each corner triangle turns out to have area 18, and the big triangle has area 96.
7	D	Use Pythagoras twice to find where the altitude from C meets AB. The bisector from B then cuts that altitude in the ratio of the sides beside angle B.
8	B	Every kite corner is 60, 90 or 120 degrees, so all lengths come from 30-60-90 triangles: AB = 3 + 3 and C is 3√3/2 above AB.
9	A	The isosceles triangles copy the angles at A and B over to D, so angle CDE is a right angle. Then CDE is a 3-4-5 triangle and BC = 8.
10	D	Join the square's outer corner to the far corners. The pieces are the square, a height-2 triangle on the hypotenuse, and two thin triangles.
11	B	The octagon is the rectangle with four matching right triangles cut off its corners. Write both legs using the side length, then use the Pythagorean theorem.
12	E	Cut the big triangle into ABC plus three outside triangles. Each has sides 4s and 3s with a 120-degree angle, so area 12 times ABC.
13	D	Cut along the mirror line, the height from C. Each half is a right triangle with a 15-degree angle, and tan 15 degrees is 2 minus root 3.
14	E	Draw a 30-degree line from B; BE cuts that angle in half. The angle bisector rule gives CE, then DE is 10 root 3 and AE is 20.
15	D	Triangles ADE and ABC share the angle at A, so their areas compare as 19 times 14 against 25 times 42, which is 19 to 75. The rest is 56.
16	D	With side 1 and AE = CF = t, setting sides BE and EF equal gives (1 - t) squared = 2t, which makes the ratio exactly 2.
17	D	The line from E down to the middle of BC has length 8, so AE is the long side of a right triangle with legs 8 and AB + 6.
18	B	Call OM = x and ON = y. The two Pythagorean equations are mirror images; adding them gives x^2 + y^2 = 169, and XY is twice 13.
19	B	Draw a line from B straight across to CD. That right triangle has legs 7 and CD - AB, long side AB + CD, giving 4(AB)(CD) = 49.
20	C	The rays must pass through all four corners. That forces both coordinates of the point to be multiples of 2/n, so the question becomes counting grid points.

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.