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Isosceles and Equilateral Triangles: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Choose the base from equal sides

Triangle UVW has the following givens:
$\begin{aligned}UV&=VW.\\m\angle U&=46^\circ.\end{aligned}$
Name the base associated with the equal sides, then find angles V and W.

Schematic; only the stated givens are marked.

Text description of this figure

Triangle UVW has matching side ticks on UV and VW, meeting at V. Angle U has the given label 46 degrees. Angles V and W are not evaluated.
1. Hint 1
  Locate the common endpoint of the equal sides.
2. Hint 2
  The angles opposite the equal sides match; then use the triangle angle sum.
Answer

$\begin{aligned}\text{Base: }&UW.\\m\angle W&=46^\circ.\\m\angle V&=88^\circ.\end{aligned}$

Full solution

The equal sides meet at V, so UW is the base.
Angles U and W are opposite the equal sides and therefore match.

$\begin{aligned}m\angle W&=46^\circ.\end{aligned}$

Subtract both base angles from the triangle total:

$\begin{aligned}m\angle V&=180^\circ-46^\circ-46^\circ.\\m\angle V&=88^\circ.\end{aligned}$

After the theorem and angle sum, the two base angles match and the vertex uses the remaining angle measure.

Answer

$\begin{aligned}\text{Base: }&UW.\\m\angle W&=46^\circ.\\m\angle V&=88^\circ.\end{aligned}$

Key idea

The equal-side information identifies the vertex and the pair of opposite angles before any calculation.
Problem 2 A wide vertex angle

An isosceles triangle has a vertex angle of 116 degrees between its chosen equal sides. Find both base angles and classify the triangle by its angles.

Schematic; only the stated givens are marked.

Text description of this figure

An isosceles triangle has two equal-side ticks and the given obtuse vertex angle of 116 degrees. Its base angles have no numerical labels.
1. Hint 1
  Here the equal base angles cannot also be 116 degrees, since three such angles would exceed the triangle angle sum.
2. Hint 2
  Subtract the vertex angle from 180 degrees and share the remainder equally.
Answer

$\begin{aligned}\text{Each base angle: }&32^\circ.\\\text{Classification: }&\text{obtuse}.\end{aligned}$

Full solution

The two base angles together use the remaining angle measure:

$\begin{aligned}180^\circ-116^\circ&=64^\circ.\end{aligned}$

They are equal, so each is

$\begin{aligned}64^\circ/2&=32^\circ.\end{aligned}$

The 116-degree angle is obtuse, so the whole triangle is classified as obtuse.

The 116-degree vertex leaves 64 degrees for the two equal base angles.

Answer

$\begin{aligned}\text{Each base angle: }&32^\circ.\\\text{Classification: }&\text{obtuse}.\end{aligned}$

Key idea

An obtuse vertex angle is compatible with equal acute base angles.
Problem 3 Equal angles give a length equation

Triangle MNP has congruent angles at M and P. The lengths MN and NP in centimeters are
$\begin{aligned}MN&=3a+4.\\NP&=7a-20.\end{aligned}$
Find a and both lengths.

Schematic; only the stated givens are marked.

Text description of this figure

Triangle MNP has matching arcs at M and P. MN is labeled 3a plus 4 centimeters and NP 7a minus 20 centimeters. No equal-side ticks or solved values appear.
1. Hint 1
  Match each equal angle to its opposite side.
2. Hint 2
  The converse of the base-angle theorem gives the equation
  $3a+4=7a-20.$
Answer

$\begin{aligned}a&=6.\\MN&=22\text{ cm}.\\NP&=22\text{ cm}.\end{aligned}$

Full solution

The side opposite M is NP, and the side opposite P is MN.
The converse gives equal lengths:

$\begin{aligned}3a+4&=7a-20.\end{aligned}$

Rearrange:

$\begin{aligned}24&=4a.\\a&=6.\end{aligned}$

Check each expression separately:

$\begin{aligned}MN&=3(6)+4.\\MN&=22\text{ centimeters}.\\NP&=7(6)-20.\\NP&=22\text{ centimeters}.\end{aligned}$

Both lengths are positive.

Answer

$\begin{aligned}a&=6.\\MN&=22\text{ cm}.\\NP&=22\text{ cm}.\end{aligned}$

Key idea

Equal angles control the opposite sides, not the side joining their vertices.
Problem 4 Side length and angle measure

An equilateral triangle has perimeter 81 centimeters. Find each side length and each interior angle measure. Explain whether doubling the perimeter would change those angle measures.

Schematic; only the stated givens are marked.

Text description of this figure

An equilateral triangle has one matching tick on each of its three sides. Its given perimeter is 81 centimeters. No side length or angle measure is shown.
1. Hint 1
  Equal sides divide the perimeter into three equal lengths.
2. Hint 2
  The base-angle theorem makes all three angles equal, regardless of the side length.
Answer

Each side: 27 centimeters. Each angle: $60^\circ$ . With doubled perimeter, each angle remains $60^\circ$ .

Full solution

The equal sides share the total perimeter:

$\begin{aligned}81/3&=27.\end{aligned}$

Applying the base-angle theorem to two pairs of equal sides makes all three angles equal.
Their total is 180 degrees, so each is $60^\circ$ .

With doubled perimeter, each side is 54 centimeters.
The triangle remains equilateral, so its angles are still $60^\circ$ .

Answer

Each side: 27 centimeters. Each angle: $60^\circ$ . With doubled perimeter, each angle remains $60^\circ$ .

Key idea

Changing an equilateral triangle size changes its lengths, while its angles remain fixed.
Problem 5 One variable across three angles

An isosceles triangle has these angle measures:
$\begin{aligned}\text{Vertex angle}&=(5x-4)^\circ.\\\text{Each base angle}&=(2x+11)^\circ.\end{aligned}$
Find x and all three angle measures.

Schematic; only the stated givens are marked.

Text description of this figure

An isosceles triangle has equal-side ticks, vertex angle labeled 5x minus 4 degrees, and each base angle 2x plus 11 degrees. No value of x or numerical angle is shown.
1. Hint 1
  There is one vertex angle and two base angles.
2. Hint 2
  Use all three interior angles to form the equation
  $(5x-4)+2(2x+11)=180.$
Answer

$\begin{aligned}x&=18.\\\text{Vertex angle: }&86^\circ.\\\text{Each base angle: }&47^\circ.\end{aligned}$

Full solution

Use the full triangle angle sum:

$\begin{aligned}(5x-4)+2(2x+11)&=180.\end{aligned}$

Combine terms:

$\begin{aligned}9x+18&=180.\\9x&=162.\\x&=18.\end{aligned}$

Evaluate the vertex angle:

$\begin{aligned}5(18)-4&=86.\end{aligned}$

The vertex angle measures $86^\circ$ .

Evaluate each base angle:

$\begin{aligned}2(18)+11&=47.\end{aligned}$

Each base angle measures $47^\circ$ .

Check the total:

$\begin{aligned}86+47+47&=180.\end{aligned}$

Answer

$\begin{aligned}x&=18.\\\text{Vertex angle: }&86^\circ.\\\text{Each base angle: }&47^\circ.\end{aligned}$

Key idea

Count both copies of a repeated base-angle expression in the triangle sum.
Problem 6 A triangular fabric panel

A flat triangular fabric panel has two equal sloping edges and a base of 28 centimeters. A seam runs from the common vertex of the equal edges along the internal angle bisector to the base. One of the two angle parts at that vertex measures 36 degrees. Find each base piece cut by the seam, each base angle, and the angle the seam makes with the base. Assume the seam is straight and follows the stated bisector exactly.

Schematic; only the stated givens are marked.

Text description of this figure

A triangular panel has equal sloping sides and base 28 centimeters. Its seam is the internal vertex bisector, shown by equal arcs, and one vertex part has the given label 36 degrees. No midpoint or right-angle marks are drawn.
1. Hint 1
  The internal bisector from the vertex between equal sides is also a median and an altitude.
2. Hint 2
  The full vertex angle contains two 36-degree parts; then use the triangle angle sum for the equal base angles.
Answer

$\begin{aligned}\text{Each base piece: }&14\text{ cm}.\\\text{Each base angle: }&54^\circ.\\\text{Seam-base angle: }&90^\circ.\end{aligned}$

Full solution

The vertex bisector in an isosceles triangle is a median, so each base piece has length

$\begin{aligned}28/2&=14.\end{aligned}$

The full vertex angle has measure

$\begin{aligned}36^\circ+36^\circ&=72^\circ.\end{aligned}$

The equal base angles divide the remainder:

$\begin{aligned}(180^\circ-72^\circ)/2&=54^\circ.\end{aligned}$

The same bisector is also an altitude.
It is perpendicular to the base, so the angle between the seam and the base is $90^\circ$ .

After the vertex-segment theorem, the seam is a median and an altitude.

Answer

$\begin{aligned}\text{Each base piece: }&14\text{ cm}.\\\text{Each base angle: }&54^\circ.\\\text{Seam-base angle: }&90^\circ.\end{aligned}$

Key idea

The internal vertex bisector in an isosceles triangle supplies both equal base pieces and a right angle with the base.
Problem 7 Assess an altitude claim

In triangle ABC, the equal sides satisfy
$\begin{aligned}AB&=AC.\end{aligned}$
Point D lies inside BC and the altitude satisfies
$\begin{aligned}AD&\perp BC.\end{aligned}$
The given measurements are
$\begin{aligned}BD&=12\text{ centimeters}.\\m\angle BAD&=23^\circ.\end{aligned}$
Nia claims that DC and angle DAC cannot be found because an altitude need not bisect a side or an angle. Decide whether her claim applies to this triangle, and find DC and angle DAC if possible.

Schematic; only the stated givens are marked.

Text description of this figure

Triangle ABC has matching side ticks on AB and AC, AD perpendicular to BC, BD labeled 12 centimeters, and angle BAD labeled 23 degrees. There are no midpoint ticks or equal vertex-angle arcs.
1. Hint 1
  Her statement about a general triangle does not settle this special case.
2. Hint 2
  The altitude from the vertex between equal sides is the same segment as its median and internal bisector.
Answer

Her claim does not apply here.
$\begin{aligned}DC&=12\text{ cm}.\\m\angle DAC&=23^\circ.\end{aligned}$

Full solution

This is an isosceles triangle with A between the equal sides.
Its internal vertex bisector is perpendicular to BC and meets BC at its midpoint.

There is only one perpendicular to BC through A, so the given altitude AD is that same line.
Therefore

$\begin{aligned}DC&=BD.\\DC&=12\text{ centimeters}.\\m\angle DAC&=m\angle BAD.\\m\angle DAC&=23^\circ.\end{aligned}$

The midpoint and bisector marks are conclusions of the isosceles theorem.

Answer

Her claim does not apply here.
$\begin{aligned}DC&=12\text{ cm}.\\m\angle DAC&=23^\circ.\end{aligned}$

Key idea

A statement about general triangles can have a stronger conclusion under isosceles hypotheses.
Problem 8 Compare two triangular pieces

Triangle RST is equilateral with side length 6 centimeters. Triangle XYZ is equiangular with perimeter 45 centimeters. Decide whether the triangles are congruent and justify your answer.

Schematic: the first triangle supplies a side length; the second supplies equal angles and a perimeter.

Text description of this figure

Two schematic triangles are compared. Blue solid RST is equilateral with side length 6 centimeters. Red dashed XYZ has the given equal-angle arcs and perimeter 45 centimeters. No side length for XYZ or congruence conclusion is marked.
1. Hint 1
  Use the converse of the base-angle theorem to identify the sides of XYZ.
2. Hint 2
  A rigid motion preserves lengths as well as angles.
Answer

The triangles are not congruent. Triangle $XYZ$ has side length 15 centimeters; triangle $RST$ has side length 6 centimeters.

Full solution

The three equal angles of XYZ give three equal opposite sides.
Its perimeter therefore divides into equal thirds:

$\begin{aligned}45/3&=15.\end{aligned}$

RST has side lengths 6 centimeters, whereas XYZ has side lengths 15 centimeters.
Rigid motions cannot change 6 into 15, so there is no congruence.

Answer

The triangles are not congruent. Triangle $XYZ$ has side length 15 centimeters; triangle $RST$ has side length 6 centimeters.

Key idea

Equal angle measures do not by themselves establish congruence.
Problem 9 Prove the reverse direction

In nondegenerate triangle ABC, the given angle congruence is
$\begin{aligned}\angle ABC&\cong\angle ACB.\end{aligned}$
Prove the following without using the converse of the base-angle theorem as a reason:
$\begin{aligned}AB&=AC.\end{aligned}$

Schematic; only the stated givens are marked.

Text description of this figure

Triangle ABC has matching angle arcs at B and C. No equal-side ticks are drawn.
1. Hint 1
  Compare the triangle with itself under a different vertex correspondence, using the ordered names BCA and CBA.
2. Hint 2
  Their angles at the endpoints of $BC$ match in reversed order. The included side satisfies
  $BC=CB.$
Answer

By ASA,
$\triangle BCA\cong\triangle CBA.$
Thus
$BA=CA.$

Full solution

Compare the names BCA and CBA.
The first angle at B matches the second angle at C by the given.
The first angle at C matches the second angle at B for the same reason.

The side between those angles is BC in the first name and CB in the second, so its length equals itself.
ASA gives

$\begin{aligned}\triangle BCA&\cong\triangle CBA.\end{aligned}$

The correspondence exchanges B with C and keeps A fixed.
Thus side BA corresponds to CA, giving

$\begin{aligned}BA&=CA.\end{aligned}$

The side ticks record the conclusion after ASA, rather than a given.

Answer

By ASA,
$\triangle BCA\cong\triangle CBA.$
Thus
$BA=CA.$

Key idea

A converse needs a proof of its own; reordered names can expose a useful correspondence.
Problem 10 Start from the median

Triangle ABC has the equal sides
$\begin{aligned}AB&=AC.\end{aligned}$
Point D is the midpoint of BC. Prove that AD is both an internal angle bisector and an altitude. Then state why the line AD is a reflection axis of the triangle.

Schematic; only the stated givens are marked.

Text description of this figure

Triangle ABC has matching ticks on AB and AC. D is the midpoint of BC, marked by double ticks on BD and DC. AD is drawn without right-angle or equal vertex-angle markings.
1. Hint 1
  List three equal-side pairs for triangles ABD and ACD before making any angle claim.
2. Hint 2
  Their congruent angles at D form a linear pair. For the reflection, use the equal base halves on the perpendicular to AD.
Answer

Segment $AD$ bisects $\angle BAC$ and satisfies
$AD\perp BC.$
Reflection across line $AD$ fixes $A$ and exchanges $B$ with $C$ , so the triangle matches itself.

Full solution

The equal sides give

$\begin{aligned}AB&=AC.\end{aligned}$

The midpoint gives

$\begin{aligned}BD&=DC.\end{aligned}$

AD is shared.
Hence, by SSS,

$\begin{aligned}\triangle ABD&\cong\triangle ACD.\end{aligned}$

Corresponding angles at A are equal, so AD bisects $\angle BAC$ .
Corresponding angles at D are also equal.
Since B, D, C are collinear, those adjacent angles total 180 degrees and are each 90 degrees.
Therefore

$\begin{aligned}AD&\perp BC.\end{aligned}$

Reflection across AD fixes A and D.
Points B and C lie equal distances on opposite sides of D along the perpendicular line BC, so the reflection exchanges them.
The reflected triangle has the same three vertices and sides.

After proving the halves congruent, the bisector and altitude markings are justified.

Answer

Segment $AD$ bisects $\angle BAC$ and satisfies
$AD\perp BC.$
Reflection across line $AD$ fixes $A$ and exchanges $B$ with $C$ , so the triangle matches itself.

Key idea

Prove the two halves congruent before using their equal angles to establish the extra roles.

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