7th Grade · 7.G.A.2

Constructing triangles in 7th grade

Constructing triangles means drawing a triangle that fits three given facts, such as three side lengths or two angles and a side. In 7th grade, your child draws triangles freehand, with a ruler and protractor and on a computer. Each time, the question is whether the facts give exactly one triangle, more than one or none at all.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

7.G.A.2 Draw (freehand, with ruler and protractor, and with technology) geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle.

What it means, in plain words

Drawing triangles from given measures mixes tools and reasoning. A ruler measures the sides, and a protractor measures the angles in degrees. Your child usually starts with one side, then uses the other facts to find where the third corner goes.

Three sides do not always make a triangle. The triangle inequality says the two shorter sides together must be longer than the longest side. With sides of 3, 4 and 10 centimeters, the two short sides add to 7 and cannot meet across the 10 centimeter side, so there is no triangle. With sides of 5, 7 and 10 centimeters, they can meet, and there is exactly one triangle.

Angles have their own rule: the three angles of a triangle always add up to $180^\circ$ . When constructing triangles from angles alone, your child finds that three angles fix the shape but not the size. A small triangle and a large one can have the same three angles, so there are many triangles. Two angles and the side between them, or two sides and the angle between them, give exactly one triangle.

Your child should be able to:

Draw a triangle from given sides and angles with a ruler and protractor
Use the triangle inequality to decide whether three lengths make a triangle
Explain why three angles alone give many triangles of different sizes
Decide whether a set of facts gives one triangle, more than one or none

Example

Can sides of 6, 9 and 16 centimeters make a triangle? What about 6, 9 and 12 centimeters?

The two shorter sides add to $6 + 9 = 15$ centimeters in both cases.
First set: 15 is less than 16, so the short sides cannot meet. There is no triangle.
Second set: 15 is more than 12, so the short sides can meet. There is exactly one triangle with these sides.

Example

Draw a triangle with sides of 7 centimeters and 4 centimeters and a $110^\circ$ angle between them. How many triangles fit these facts?

Draw a 7 centimeter side with the ruler.
At its left end, use the protractor to draw a $110^\circ$ angle.
Along the new line, measure 4 centimeters and mark a point.
Join that point to the right end of the 7 centimeter side.
Every triangle drawn from these facts has the same size and shape, so exactly one triangle fits.

Where children usually get stuck

Your child says sides of 3, 5 and 8 inches make a triangle, because $3 + 5 = 8$ .
The rule was learned as "the short sides must reach". When they add up to exactly the longest side, they lie flat along it and the triangle has no height.
What helps. Cut paper strips 3, 5 and 8 inches long and try to build the triangle. The strips can only lie flat in a straight line.
Your child says that angles of $30^\circ$ , $60^\circ$ and $90^\circ$ give exactly one triangle.
Three facts usually fix a triangle, so three angles seem like enough. Angles fix the shape but not the size.
What helps. Draw two triangles with these angles together, one small and one large. Both fit, so there is more than one.
Asked for a $50^\circ$ angle, your child draws a $130^\circ$ angle.
A protractor has two rows of numbers that run in opposite directions, and your child read the wrong row.
What helps. Before measuring, ask whether the angle should be narrower or wider than a square corner. A $50^\circ$ angle is narrower, so a wide angle cannot be right.

Try four practice questions

Real 7th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up

You want to build a triangle from two sides, 5 units and 8 units long, and the angle between them. You can use only these angles: 0°, 40°, 90°, 135°, 180°. In each row, type the angle from the list that fits.

Condition	Angle between the two sides
The smallest angle in the list that makes exactly one triangle	°
The largest angle in the list that makes exactly one triangle	°
The angle in the list that is greater than 90° and makes no triangle	°

Show the answer and how to solve it

The answers are $40^{\circ}$ for Row 1, $135^{\circ}$ for Row 2 and $180^{\circ}$ for Row 3. Each row has only one correct answer.

1Draw the 5-unit and 8-unit sides from the same point, with the chosen angle between them. The two lengths and the angle fix where each free endpoint lands. For an angle between $0^{\circ}$ and $180^{\circ}$ , the starting point and the two free endpoints do not lie on one line. Only one segment joins the two free endpoints, so joining them closes exactly one triangle. At $0^{\circ}$ or $180^{\circ}$ , the starting point and both free endpoints lie on one line, so no triangle forms.
2Check each angle in the list. $40^{\circ}$ , $90^{\circ}$ and $135^{\circ}$ are between $0^{\circ}$ and $180^{\circ}$ , so each one makes exactly one triangle. $0^{\circ}$ and $180^{\circ}$ make no triangle.
3Row 1 asks for the smallest angle in the list that makes exactly one triangle. That is $40^{\circ}$ , because $0^{\circ}$ makes no triangle. Row 2 asks for the largest angle in the list that makes exactly one triangle. That is $135^{\circ}$ , because $180^{\circ}$ makes no triangle. Row 3 asks for the angle in the list that is greater than $90^{\circ}$ and makes no triangle. Of $0^{\circ}$ and $180^{\circ}$ , that is $180^{\circ}$ .

Common mistake. A common mistake is to pick $0^{\circ}$ as the smallest angle that makes a triangle, or $180^{\circ}$ as the largest. Each of these puts the two sides on one line, so it makes no triangle.

The idea underneath. Start both given sides at the same point. Set the angle between them and keep their lengths fixed. This fixes the two free endpoints. If the angle is between $0^{\circ}$ and $180^{\circ}$ , the starting point and the two free endpoints do not lie on one line, and the one segment that joins the free endpoints closes exactly one triangle. At $0^{\circ}$ or $180^{\circ}$ , they all lie on one line, so no triangle forms. The two side lengths do not change which angles work.

2On level
Look at the construction directions and Mia's triangle. Which statement best describes Mia's mistake?
$\displaystyle \text{Construct } \triangle ABC \text{ with } \angle A = 52^\circ, \angle C = 68^\circ, \overline{AC} = 6\text{ cm.}$
$\displaystyle \text{Side } \overline{AC} \text{ is between the two given angles.}$
Mia drew this triangle.
A
Mia put the 6 cm label on the wrong side. AC should be the side between angle A and angle C.
B
Mia should have put the 6 cm label on a side that is not between the two given angles.
C
Mia used the third angle, $60^{\circ}$ , instead of one of the given angles.
D
Mia placed the $68^{\circ}$ angle at the wrong vertex.
Show the answer and how to solve it
Mia's mistake is that the 6 cm label is on AB. The directions say AC must be the side between angle A and angle C, so the length should be on AC.
1. 1The directions give angle $A = 52^{\circ}$ , angle $C = 68^{\circ}$ , side AC = 6 cm, and state that AC is between the two given angles.
2. 2Because AC is between angles A and C, it must connect the left vertex A and the right vertex C directly. That is the bottom side of the triangle.
3. 3Mia placed $52^{\circ}$ at A and $68^{\circ}$ at C correctly. But the 6 cm label is on AB (the left side), not on AC (the bottom side).
4. 4So the best description is that Mia put the 6 cm label on the wrong side.
Common mistake. A common mistake is thinking that any side touching one of the given angles can carry the named length. When a side is between two given angles, it must directly connect those two vertices—here, A and C.
The idea underneath. A side is between two given angles only when it directly connects the vertices of those two angles.
3On level
You are constructing a triangle with side lengths 4, 6, and 9. Which drawing could be the finished construction?
A
B
C
D
Show the answer and how to solve it
The answer is the triangle with side labels 4, 6, and 9 because a correct finished construction must match all three given side lengths exactly.
1. 1The target side lengths are 4, 6, and 9.
2. 2Compare each drawing's three side labels as a set. Label order around the triangle does not matter, but all three lengths must match exactly.
3. 3The four drawings show the sets {4, 6, 9}, {4, 6, 8}, {3, 7, 9}, and {6, 6, 9}. The set {4, 6, 8} changes 9 to 8. The set {3, 7, 9} has the same perimeter of 19 but different individual lengths. The set {6, 6, 9} repeats 6 and omits 4.
4. 4Only the triangle labeled 4, 6, and 9 matches the target exactly, so that drawing is the correct finished construction.
Common mistake. A common mistake is choosing the triangle with the same perimeter. Here, $3 + 7 + 9 = 19$ matches $4 + 6 + 9 = 19$ , but the individual side lengths are completely different.
The idea underneath. A triangle described by three side lengths is identified by the whole set of lengths; orientation and label order do not matter, but each length and its repetition count do.
4Stretch
The drawing below is a rough sketch — the actual angle measures and side length are the labels, not the drawn shape. You are given two angle measurements and one side length; the third angle is marked with a question mark. Based on these two angles and this side length, is the triangle unique, is more than one triangle possible, or is no triangle possible?
A
More than one triangle possible
B
No triangle possible
C
Unique triangle possible
Show the answer and how to solve it
A triangle's three angles add to $180^{\circ}$ . These two already total $180^{\circ}$ , leaving $0^{\circ}$ for the third angle. A triangle needs three angles greater than $0^{\circ}$ . The answer is 'No triangle possible'.
1. 1Read the labels on the sketch: the two given angles are $100^{\circ}$ and $80^{\circ}$ , and the given side length is 7. The drawing is only a rough sketch — the decision comes from the labeled values, not the drawn shape.
2. 2A triangle's three angles add to $180^{\circ}$ , and every angle must be greater than $0^{\circ}$ . Add the two given angle measures.
3. 3A triangle's three angles add to $180^{\circ}$ . These two already total $180^{\circ}$ , leaving $0^{\circ}$ for the third angle. A triangle needs three angles greater than $0^{\circ}$ . The side length of 7 cannot change the angles. The answer is 'No triangle possible'. 'More than one triangle possible' is also wrong because no triangle meets these angle measures.
Common mistake. A common mistake is to accept $100^{\circ} + 80^{\circ} = 180^{\circ}$ as a valid triangle because the angles 'use up exactly $180^{\circ}$ .' But the third angle would be $0^{\circ}$ , which is not a positive angle, so no triangle can exist. Another common mistake is to pick 'More than one triangle possible'. Because no triangle fits these angles, there is not even one, so there cannot be more than one.
The idea underneath. A triangle's three angles add to $180^{\circ}$ , and every angle must be greater than $0^{\circ}$ . If a given angle is $0^{\circ}$ , or the two given angles total $180^{\circ}$ or more, no triangle is possible. Otherwise, subtract the total from $180^{\circ}$ to find the third angle. Every triangle with these three angles has the same shape and differs only in size, so the given side length fixes the size and exactly one triangle is possible.