One Triangle, Many, or None? Conditions and Errors | Grade 7 Math | LevelKite

You have drawn triangles from three side lengths and from angle measures. Now you will answer a bigger question with numbers alone: how many different triangles fit a set of conditions? A condition is a measurement the triangle must have, such as a side of 6 cm or an angle of $40^\circ$ .

First, decide what different means. Two triangles count as the same triangle if you can slide, turn or flip one so that it fits exactly on top of the other. They are different only if no slide, turn or flip makes them match.

Every set of conditions in this lesson gets one of three verdicts, or final answers:

Exactly one triangle. Mathematicians call this a unique triangle.
More than one triangle. At least two triangles fit, and they do not match.
No triangle. Nothing can meet all the conditions at once.

Start with three side lengths. Each one must be positive, because a side cannot have a length of 0 or less. All three must use the same unit so that you can add and compare them.

Picture the longest side lying flat, with one shorter side hinged to each end. To make a triangle, the two shorter sides must swing up and meet at a corner that is off the longest side.

If the two shorter lengths add to less than the longest, the sides cannot reach each other. They leave a gap.
If they add to exactly the longest, they touch only when both lie flat along it. You get a straight line segment with no inside, not a triangle.
If they add to more than the longest, they have length to spare, so they can swing up and meet at a corner.

So one comparison, the side test, decides it. Call the lengths $a$ , $b$ and $c$ , listed from least to greatest, so $c$ is the longest. If two lengths are equal, list them in either order. A triangle exists exactly when this is true:

a + b > c \quad \text{where } 0 < a \le b \le c

When three positive lengths pass the side test, they give exactly one triangle, counting slid, turned or flipped copies as the same triangle. Here is why. With all three side lengths fixed, no corner can open wider or close tighter unless a side changes length. Think back to drawing a triangle from three side lengths: your only choices were where to place it and which way to face it. Those choices are slides, turns and flips, so every result is the same triangle.

When the test fails, because $a + b$ is equal to $c$ or less than $c$ , the lengths give no triangle.

Worked example

Three side lengths: one triangle or none?

For each set of side lengths, decide whether it gives exactly one triangle, more than one triangle, or no triangle. Set A: 6 cm, 8 cm and 13 cm. Set B: 6 cm, 13 cm and 6 cm.

Start with Set A. All three lengths are positive and in centimeters, so the side test applies. Listed from least to greatest, they are 6 cm, 8 cm and 13 cm. The two shorter lengths are 6 cm and 8 cm, and the longest is 13 cm.
$a = 6\text{ cm},\quad b = 8\text{ cm},\quad c = 13\text{ cm}$
Add the two shorter lengths. The sum tells you how far those two sides reach when you lay them end to end.
$6\text{ cm} + 8\text{ cm} = 14\text{ cm}$
Compare the sum with the longest length. The sum is greater by 1 cm, so the two shorter sides have length to spare and can meet at a corner off the longest side. A triangle exists. Because all three side lengths are fixed, it is the only one: sliding, turning or flipping it gives the same triangle.
$14\text{ cm} > 13\text{ cm}$
Now Set B. All three lengths are positive and in centimeters. Listed from least to greatest, they are 6 cm, 6 cm and 13 cm. The two equal lengths are the shorter sides, and the longest is 13 cm.
$a = 6\text{ cm},\quad b = 6\text{ cm},\quad c = 13\text{ cm}$
Add the two shorter lengths and compare the sum with the longest length. The sum is 1 cm less, so the two shorter sides leave a gap and cannot meet. The side test fails.
$6\text{ cm} + 6\text{ cm} = 12\text{ cm},\quad 12\text{ cm} < 13\text{ cm}$

Answer

Set A gives exactly one triangle, because the two shorter sides total $14$ cm, which is greater than $13$ cm. Set B gives no triangle, because the two shorter sides total $12$ cm, which is less than $13$ cm.

Now take three angle measures. Angles tell you how wide each corner opens. They say nothing about how long any side is. So angles can fix a triangle's shape, but they leave its size free.

Think back to drawing a triangle from angle measures: the angles never told you how long to make the first side. Choose a longer first side and you get a larger triangle with the very same angles. A larger triangle cannot fit exactly on a smaller one, so the two are different triangles.

That gives the angle check. Each measure must be positive, which means greater than $0^\circ$ , and the three angles of any triangle add to exactly $180^\circ$ . So add the three measures.

If all three are positive and the sum is exactly $180^\circ$ , the measures give more than one triangle: one shape in many sizes.
If the sum is not $180^\circ$ , or if any measure is $0^\circ$ or less, they give no triangle.

Worked example

Three angle measures: many triangles or none?

For each set of angle measures, decide whether it gives exactly one triangle, more than one triangle, or no triangle. Set A: $35^\circ$ , $65^\circ$ and $80^\circ$ . Set B: $35^\circ$ , $65^\circ$ and $90^\circ$ .

Start with Set A. All three measures are greater than 0 degrees, so the angle check applies. Add the measures in two stages.
$35^\circ + 65^\circ = 100^\circ,\quad 100^\circ + 80^\circ = 180^\circ$
Compare the sum with 180 degrees. The sum equals 180 degrees, so a triangle with these angles exists. No condition sets a side length, so larger and smaller triangles with these same three angles all fit. The shape is fixed, but the size is not.
$180^\circ = 180^\circ$
Now Set B. All three measures are greater than 0 degrees. Add the measures in two stages.
$35^\circ + 65^\circ = 100^\circ,\quad 100^\circ + 90^\circ = 190^\circ$
Compare the sum with 180 degrees. The sum is 10 degrees too large. The angles of every triangle add to exactly 180 degrees, so no triangle can have these three angles.
$190^\circ \ne 180^\circ$

Answer

Set A gives more than one triangle, because its three positive angles add to $180^\circ$ , which fixes the shape but not the size. Set B gives no triangle, because its angles add to $190^\circ$ , not $180^\circ$ .

Key concept

How many triangles fit the conditions?

Count two triangles as the same when a slide, turn or flip makes one fit exactly on the other.

Three positive side lengths in the same unit: if the two shorter lengths add to more than the longest, they give exactly one triangle, a unique triangle. If the sum is equal to the longest or less, they give none.
Three angle measures: if all three are positive and add to exactly $180^\circ$ , they fix the shape but not the size, so they give more than one triangle. If the sum is not $180^\circ$ , or any measure is $0^\circ$ or less, they give none.

Seven sets of conditions. Compare rows 1 and 5: equal sides fix one triangle, but equal angles do not.

Conditions	Numeric check	What the check shows	Verdict
Sides 7 cm, 7 cm, 7 cm	Two shorter total $7 + 7 = 14$ cm. Longest is $7$ cm.	$14 > 7$	Exactly one triangle
Sides 3 cm, 9 cm, 7 cm	Two shorter total $3 + 7 = 10$ cm. Longest is $9$ cm.	$10 > 9$	Exactly one triangle
Sides 4 cm, 4 cm, 8 cm	Two shorter total $4 + 4 = 8$ cm. Longest is $8$ cm.	$8$ is not greater than $8$	No triangle
Sides 2 cm, 10 cm, 5 cm	Two shorter total $2 + 5 = 7$ cm. Longest is $10$ cm.	$7 < 10$	No triangle
Angles $60^\circ$ , $60^\circ$ , $60^\circ$	$60^\circ + 60^\circ + 60^\circ = 180^\circ$	All positive, and the sum is $180^\circ$	More than one triangle
Angles $50^\circ$ , $50^\circ$ , $70^\circ$	$50^\circ + 50^\circ + 70^\circ = 170^\circ$	All positive, but the sum is not $180^\circ$	No triangle
Angles $90^\circ$ , $90^\circ$ , $0^\circ$	$90^\circ + 90^\circ + 0^\circ = 180^\circ$	The sum is $180^\circ$ , but $0^\circ$ is not positive	No triangle

Now use these checks to judge someone else's reasoning. A claim about triangle conditions can go wrong in three places, so ask three questions in order:

Are the inputs valid? Side lengths must be positive and in the same unit. Angle measures must be greater than $0^\circ$ . A length or an angle measure of 0 or less gives no triangle.
Is it the right check? For sides, the sum of the two shorter lengths must be greater than the longest. For angles, the sum of all three must be exactly $180^\circ$ .
Does the verdict match the check? Three positive sides that pass give exactly one triangle, because slid, turned or flipped copies are the same triangle. Three positive angles that pass give more than one, because size is free. A failed check gives none.

When you find the error, say why it fails. Then redo that step correctly.

Worked example

Find the error in three claims

Jo made three claims. Find the error in each one and give the correct verdict. Claim 1: Sides of 4 cm, 5 cm and 10 cm give exactly one triangle, because $5 + 10 = 15$ and $15 > 4$ . Claim 2: Sides of 5 cm, 6 cm and 7 cm give more than one triangle, because $5 + 6 = 11$ , $11 > 7$ , and flipping the triangle makes a second one. Claim 3: Angles of $40^\circ$ , $60^\circ$ and $80^\circ$ give exactly one triangle, because they add to $180^\circ$ .

Claim 1, find the error. Jo added the two longer sides and compared the sum with the shortest side. For positive lengths that check can never fail: each longer side is already at least as long as the shortest, and adding a positive length makes the total larger still. A check that cannot fail proves nothing.
Claim 1, correct it. The check that can fail uses the two shorter sides. Add 4 cm and 5 cm, and compare the sum with the longest length, 10 cm. The sum is 1 cm short, so the two shorter sides cannot meet. The correct verdict is no triangle.
$4\text{ cm} + 5\text{ cm} = 9\text{ cm},\quad 9\text{ cm} < 10\text{ cm}$
Claim 2, find the error. Jo's check is correct: the two shorter lengths total 11 cm, which is greater than the longest, 7 cm, so a triangle exists. The error is in the verdict. A flipped copy fits exactly on the original when you flip it back, so it counts as the same triangle.
$5\text{ cm} + 6\text{ cm} = 11\text{ cm},\quad 11\text{ cm} > 7\text{ cm}$
Claim 2, correct it. Three positive side lengths that pass the side test leave nothing free except where the triangle sits and which way it faces. The correct verdict is exactly one triangle.
Claim 3, find the error. Jo's sum is correct and all three measures are greater than 0 degrees, so a triangle exists. The error is the jump from 'a triangle exists' to 'only one triangle exists'. Angles set the shape, but no condition here sets a side length.
$40^\circ + 60^\circ = 100^\circ,\quad 100^\circ + 80^\circ = 180^\circ$
Claim 3, correct it. Triangles with these angles can be any size. One can have a longest side of 5 cm and another a longest side of 10 cm. They share the same angles but cannot fit exactly on each other. The correct verdict is more than one triangle.

Answer

Claim 1 should say no triangle, because $4 + 5 = 9$ and $9$ is less than $10$ . Claim 2 should say exactly one triangle, because a flipped copy is the same triangle. Claim 3 should say more than one triangle, because three positive angles that add to $180^\circ$ fix shape but not size.

Here is the idea to keep: one numeric check tells you how many triangles fit a set of conditions.

Three positive side lengths: if the two shorter add to more than the longest, exactly one triangle fits, because fixed sides leave free only where the triangle sits and which way it faces. If the sum is equal or less, none fits.
Three angle measures: if all are positive and add to exactly $180^\circ$ , more than one triangle fits, because angles fix shape but not size. With any other sum, or a measure of $0^\circ$ or less, none fits.

When you check a claim, test the inputs, then the check, then the verdict, and explain why the faulty step fails before you fix it.