Lesson 33 of 46 · Geometry

Triangles and the Pythagorean theorem

Bound a missing side with the triangle inequality, find areas with the right height, and use the Pythagorean theorem only where it applies.

About 20 minutes

In this lesson

  • Use the triangle inequality to find the possible lengths of a third side.
  • Find a triangle's area from any base and the height perpendicular to it.
  • Apply the Pythagorean theorem and its converse, and recognise the common whole-number right triangles.
  • Tell from three side lengths whether a triangle is acute, right or obtuse.

The idea

What every triangle obeys

Three facts hold for every triangle, whatever its shape.

  1. The angles add to 180∘180^\circ.
  2. Bigger sides face bigger angles. The longest side is opposite the largest angle, and the shortest side opposite the smallest. Equal sides face equal angles: in an isosceles triangle the two angles at the base are equal, and in an equilateral triangle every angle is 60∘60^\circ.
  3. The triangle inequality. Each side is shorter than the other two together. If two sides are aa and bb with a≥ba \ge b, the third side xx satisfies

a−b<x<a+ba - b < x < a + b

Why strict? If xx were exactly a+ba + b, the two shorter sides laid end to end would only just reach, and the "triangle" would flatten into a straight segment with no area. Questions that ask how many whole-number lengths are possible depend on getting those two endpoints excluded.

Area: base times height, with the right height

area=12×base×height\text{area} = \frac{1}{2} \times \text{base} \times \text{height}

Any side can be the base. The height is the perpendicular distance from the opposite vertex to the line containing that base. In a right triangle the two legs are a base and its height. In an obtuse triangle the height to one of the short sides falls outside the triangle, onto that side extended; the formula still works.

Because one triangle has one area, the three base–height pairs all give the same product. That is a useful tool: once you know the area, you can find the height to any other side by dividing twice the area by that side.

A slanted side is a height only if it is perpendicular to the base. Multiplying two sides that meet at an angle other than 90∘90^\circ gives too large an area.

The Pythagorean theorem

In a right triangle with legs aa and bb and hypotenuse cc (the side opposite the right angle, always the longest side):

a2+b2=c2a^2 + b^2 = c^2

The theorem works in both directions. If three sides satisfy a2+b2=c2a^2 + b^2 = c^2, the angle opposite cc is a right angle. Comparing c2c^2 with a2+b2a^2 + b^2 (where cc is the longest side) also sorts every triangle:

Longest side cc The angle opposite cc is The triangle is
c2<a2+b2c^2 < a^2 + b^2 less than 90∘90^\circ acute
c2=a2+b2c^2 = a^2 + b^2 exactly 90∘90^\circ right
c2>a2+b2c^2 > a^2 + b^2 more than 90∘90^\circ obtuse

The intuition: hold sides aa and bb fixed and open the angle between them. The wider the angle, the longer the third side; at exactly 90∘90^\circ it equals a2+b2\sqrt{a^2 + b^2}.

Whole-number right triangles worth knowing

Some side triples come up because they keep the arithmetic clean. Knowing them saves a square root.

Triple Check
3,4,53, 4, 5 9+16=259 + 16 = 25
5,12,135, 12, 13 25+144=16925 + 144 = 169
8,15,178, 15, 17 64+225=28964 + 225 = 289
7,24,257, 24, 25 49+576=62549 + 576 = 625

Any multiple of a triple is a triple too: 6,8,106, 8, 10 and 15,20,2515, 20, 25 are both 3,4,53, 4, 5 scaled up. Before you square anything, check whether the given sides are a multiple of a known triple. A leg of 1010 with a hypotenuse of 2626 is 5,12,135, 12, 13 doubled, so the other leg is 2424.

Make sure the triple is in the right order. Legs of 55 and 1313 do not give a hypotenuse of 1212: the hypotenuse is the longest side, so it would be 25+169=194\sqrt{25 + 169} = \sqrt{194}.

Where right triangles hide

Many questions do not show a right triangle; you build one. A rectangle's diagonal splits it into two right triangles. The height of an isosceles triangle drawn to its base cuts the base in half and makes two right triangles. A distance between two points on a grid is the hypotenuse of a triangle whose legs are the horizontal and vertical gaps. When a length is missing, look for a right angle you can create.

Worked examples

Example 1: How many whole-number third sides?

Two sides of a triangle measure 77 and 1212. If the third side is a whole number, how many lengths are possible?

  1. Apply the triangle inequality: 12−7<x<12+712 - 7 < x < 12 + 7, so 5<x<195 < x < 19.

  2. The whole numbers strictly between 5 and 19 run from 6 to 18.

  3. Count them: 18−6+1=1318 - 6 + 1 = 13.

Answer

There are 1313 possible lengths. Including the endpoints 5 and 19 gives the tempting wrong count of 15; both would flatten the triangle into a line.

Example 2: A missing leg, then the area

In right triangle DEFDEF, the right angle is at EE. The hypotenuse DFDF is 2626 and leg DEDE is 1010. What is the area of the triangle?

  1. The legs are DEDE and EFEF, because the right angle is at EE. So DE2+EF2=DF2DE^2 + EF^2 = DF^2.

  2. Spot the triple: 1010 and 2626 are 55 and 1313 doubled, so EF=12×2=24EF = 12 \times 2 = 24. Check: 100+576=676=262100 + 576 = 676 = 26^2.

  3. The two legs are a base and its height: area =12×10×24=120= \frac{1}{2} \times 10 \times 24 = 120.

Answer

The area is 120120. Using the hypotenuse as a height, 12×10×26=130\frac{1}{2} \times 10 \times 26 = 130, is the tempting wrong answer; the hypotenuse is not perpendicular to DEDE.

Example 3: An isosceles triangle and two heights

Triangle ABCABC has AB=AC=13AB = AC = 13 and BC=10BC = 10. Find its area, and then the height from BB to side ACAC.

  1. Drop the height from AA to BCBC at point MM. In an isosceles triangle this height meets the base at its midpoint, so BM=5BM = 5 and angle AMBAMB is a right angle.

  2. In right triangle ABMABM: AM2=132−52=169−25=144AM^2 = 13^2 - 5^2 = 169 - 25 = 144, so AM=12AM = 12 (a 5,12,135, 12, 13 triangle).

  3. Area =12×10×12=60= \frac{1}{2} \times 10 \times 12 = 60.

  4. Now use ACAC as the base. The area is still 60, so 12×13×h=60\frac{1}{2} \times 13 \times h = 60 and h=12013≈9.23h = \frac{120}{13} \approx 9.23.

Answer

The area is 6060 and the height from BB to ACAC is 12013\frac{120}{13}. Taking the slanted side 1313 as the height, 12×10×13=65\frac{1}{2} \times 10 \times 13 = 65, overstates the area.

Common traps

  • Pythagoras on a triangle with no right angle

    a2+b2=c2a^2 + b^2 = c^2 holds only when the angle opposite cc is 90∘90^\circ. If no right angle is stated or marked, the squares do not add up and you cannot use them to find a side. Look for a right angle you can create instead, such as an altitude.

  • Treating the missing side as the hypotenuse

    Given sides 6 and 10 of a right triangle, the third side is 88 if 10 is the hypotenuse, and 136\sqrt{136} only if 6 and 10 are both legs. Read where the right angle is before deciding which side is cc.

  • Including the endpoints of the triangle inequality

    The third side is strictly between the difference and the sum of the other two. With sides 4 and 9, a third side of 5 or 13 gives a flat figure with no area, not a triangle.

  • Using a slanted side as the height

    The height must be perpendicular to the base. A side that leans is longer than the height it stands in for, so the area comes out too large. If the height is not given, build a right triangle to find it.

Check yourself

Answer each question before you open its answer.

Question 1: A possible perimeter

Triangles

Two sides of a triangle have lengths 55 and 1111. Which of the following could be the perimeter of the triangle?

  1. A.2121
  2. B.2222
  3. C.2727
  4. D.3232
  5. E.3333
Show the answer

Answer: C. 2727

Working

The third side xx satisfies 11−5<x<11+511 - 5 < x < 11 + 5, so 6<x<166 < x < 16. The perimeter is 16+x16 + x, so it lies strictly between 2222 and 3232. Of the options only 2727 (third side 1111) is in that range.

Why the other options are wrong

  • A. Would make the third side 5, which is less than 11−5=611 - 5 = 6; the two short sides could not meet.
  • B. Would make the third side exactly 6, the difference of the other two. The figure would be flat, so the inequality is strict.
  • D. Would make the third side exactly 16, the sum of the other two: a flat figure, not a triangle.
  • E. Would make the third side 17, longer than 5+115 + 11; the other two sides could not reach.

What carries over

Turn a perimeter question into a third-side question, and keep both ends of the inequality strict.

Question 2: Area from a hypotenuse and a leg

The Pythagorean theorem

A right triangle has a hypotenuse of length 1717 and one leg of length 88. What is the area of the triangle?

  1. A.6060
  2. B.6868
  3. C.120120
  4. D.127.5127.5
  5. E.136136
Show the answer

Answer: A. 6060

Working

The other leg is 172−82=289−64=225=15\sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15 (an 8,15,178, 15, 17 triangle). The legs are perpendicular, so the area is 12×8×15=60\frac{1}{2} \times 8 \times 15 = 60.

Why the other options are wrong

  • B. Uses the hypotenuse as the height: 12×8×17\frac{1}{2} \times 8 \times 17. The hypotenuse is not perpendicular to the leg.
  • C. Finds the right legs, 8 and 15, but forgets the 12\frac{1}{2} in the area formula.
  • D. Multiplies the other leg by the hypotenuse, 12×15×17\frac{1}{2} \times 15 \times 17; those two sides are not perpendicular either.
  • E. Multiplies the two given sides with no 12\frac{1}{2}: 8×178 \times 17.

What carries over

In a right triangle the two legs are base and height; find the missing leg first.

Question 3: Acute, right or obtuse?

Quantitative Comparison

A triangle has sides of lengths 77, 88 and 1111.

Quantity A: the measure of the largest angle of the triangle, in degrees

Quantity B: 9090

  1. A.Quantity A is greater.
  2. B.Quantity B is greater.
  3. C.The two quantities are equal.
  4. D.The relationship cannot be determined from the information given.
Show the answer

Answer: A. Quantity A is greater.

Working

The largest angle is opposite the longest side, 11. Compare 112=12111^2 = 121 with 72+82=49+64=1137^2 + 8^2 = 49 + 64 = 113. Since 121>113121 > 113, the angle opposite 11 is greater than 90∘90^\circ.

Why the other options are wrong

  • B. Assumes a triangle with no stated right angle must be acute. Comparing squares shows the largest angle is obtuse.
  • C. Treats the triangle as right-angled. That would need 72+82=1127^2 + 8^2 = 11^2, but 113≠121113 \neq 121.
  • D. Three side lengths fix a triangle completely, so its angles are determined even though they are not given.

What carries over

Compare the square of the longest side with the sum of the other two squares to classify a triangle without finding any angle.

Key points

  • The third side lies strictly between the difference and the sum of the other two.

  • Area is half of base times the height perpendicular to that base; any side can be the base.

  • a2+b2=c2a^2 + b^2 = c^2 only in a right triangle, with cc the hypotenuse opposite the right angle.

  • Know 3,4,53, 4, 5; 5,12,135, 12, 13; 8,15,178, 15, 17; 7,24,257, 24, 25 and their multiples.

  • Longest side squared against the sum of the other two squares: less is acute, equal is right, more is obtuse.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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