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Special Right 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 Two equal legs

A 45-45-90 triangle has hypotenuse $19\sqrt2$ mm.

Find each leg and the exact perimeter.

The given hypotenuse is opposite the marked right angle.

Text description of this figure

Right triangle CAB has its marked right angle at C. The two legs have matching ticks. Hypotenuse AB is labeled 19√2 mm.
1. Hint 1
  Identify the factor connecting a leg to the hypotenuse.
2. Hint 2
  Recover one leg by dividing by that factor, then add two equal legs and the hypotenuse.
Answer

Each leg is $19$ mm. The perimeter is $(38+19\sqrt2)$ mm.

Full solution

Each leg is the hypotenuse divided by $\sqrt2$ .
$s=\frac{19\sqrt2}{\sqrt2}.$
Therefore each leg is $19$ mm.

Add both equal legs and the hypotenuse.
$P=38+19\sqrt2\text{ mm}.$

The two leg squares total $722$ , which equals the square of $19\sqrt2$ , so the lengths check.

Answer

Each leg is $19$ mm. The perimeter is $(38+19\sqrt2)$ mm.

Key idea

Reverse the hypotenuse multiplier before adding the sides.
Problem 2 A longer leg

Triangle $UVW$ has a right angle at $V$ and a $30^\circ$ angle at $U$ .

The side $UV$ is $13\sqrt3$ cm long. Find $VW$ and $UW$ exactly.

The marked angles locate the given side.

Text description of this figure

Right triangle VUW has its marked right angle at V. The angle at U is 30 degrees and the angle at W is 60 degrees. Side VU is labeled 13√3 cm.
1. Hint 1
  Identify which of the legs is opposite the 30-degree angle.
2. Hint 2
  The given leg is opposite 60 degrees. Divide its length by the long-leg multiplier before finding the hypotenuse.
Answer

$\begin{aligned}VW&=13\text{ cm}.\\UW&=26\text{ cm}.\end{aligned}$

Full solution

The side $VW$ is opposite $30^\circ$ , so it is the short leg.
The given side $UV$ is the long leg.

Divide the long leg by $\sqrt3$ .
$VW=\frac{13\sqrt3}{\sqrt3}.$
Thus $VW$ is $13$ cm.

Double the short leg for the hypotenuse.
$UW=26\text{ cm}.$

The squared lengths check: $169+507$ equals $676$ , the square of $26$ .

The completed lengths follow the same side order.

Answer

$\begin{aligned}VW&=13\text{ cm}.\\UW&=26\text{ cm}.\end{aligned}$

Key idea

Assign side roles from their opposite angles before using a ratio.
Problem 3 An edge from a diagonal

A square has diagonal $7\sqrt6$ m. A square has four equal sides and four right angles.

Find its exact side length and perimeter.

The diagonal connects opposite vertices.

Text description of this figure

Square PQRS has all four equal sides and four right angles. Diagonal PR is dashed. Side PQ is labeled s. Diagonal PR is labeled 7√6 m.
1. Hint 1
  A diagonal and two adjacent sides form a right triangle with equal legs.
2. Hint 2
  Divide the diagonal by the equal-leg hypotenuse multiplier, then use four equal sides for the perimeter.
Answer

$\begin{gathered}\text{Side length}=7\sqrt3\text{ m}.\\\text{Perimeter}=28\sqrt3\text{ m}.\end{gathered}$

Full solution

The two equal sides are the legs of a 45-45-90 triangle.
$s=\frac{7\sqrt6}{\sqrt2}.$

Simplify the quotient of the positive roots.
$s=7\sqrt3\text{ m}.$

Four equal side lengths give the perimeter.
$P=28\sqrt3\text{ m}.$

For a check, each side square is $147$ and the diagonal square is $294$ , twice that value.

The completed square retains equal sides and right angles.

Answer

$\begin{gathered}\text{Side length}=7\sqrt3\text{ m}.\\\text{Perimeter}=28\sqrt3\text{ m}.\end{gathered}$

Key idea

A square diagonal is the hypotenuse when recovering the edge length.
Problem 4 A midpoint join

On paper, draw an equilateral triangle $ABC$ with each side labeled $22$ cm. Use a convenient drawing scale. Mark the midpoint $M$ of $BC$ and join $A$ to $M$ .

Add the equal-length and angle markings that can be justified. Explain why $AM$ is perpendicular to $BC$ and bisects angle $A$ . Label both half-base lengths and the exact altitude.
1. Hint 1
  Compare the two smaller triangles using three side pairs.
2. Hint 2
  Equal corresponding angles at M form a straight angle. The equal vertex parts divide the equilateral angle.
3. Hint 3
  Once the right angle is justified, the equilateral side is the hypotenuse of either half.
Answer

SSS justifies the matching halves. Both half-bases are $11$ cm, both vertex parts are $30^\circ$ , and both angles at $M$ are $90^\circ$ . The altitude is $11\sqrt3$ cm.

Full solution

Mark all three outer sides with one tick pattern.
Mark the two half-base segments with a different pattern.
The triangles $ABM$ and $ACM$ have equal outer sides, equal base halves, and shared $AM$ , so they are congruent by SSS.

Corresponding angles $AMB$ and $AMC$ are equal.
Since $B$ , $M$ , and $C$ are collinear, those angles sum to $180^\circ$ and each is $90^\circ$ .
Add right-angle marks only after this reasoning.

The two parts at $A$ correspond under the congruence.
They split the equilateral $60^\circ$ angle, so each is $30^\circ$ .
Add matching angle arcs or the two angle labels.

The half-base lengths are each $11$ cm.
In either 30-60-90 half, the altitude is the long leg.
$AM=11\sqrt3\text{ cm}.$

The perpendicular and angle-bisection markings are justified by SSS.

Answer

SSS justifies the matching halves. Both half-bases are $11$ cm, both vertex parts are $30^\circ$ , and both angles at $M$ are $90^\circ$ . The altitude is $11\sqrt3$ cm.

Key idea

Construction markings should record proved facts, not visual guesses.
Problem 5 A length relationship

A nondegenerate right triangle has two acute angles of $45^\circ$ and one leg of length $s>0$ .

Derive the hypotenuse length from the angle information and Pythagoras. Derive the reverse formula for $s$ when the hypotenuse is $h$ . Check your formulas for a leg of $13$ cm.

The positive leg and hypotenuse are labeled by role.

Text description of this figure

Right triangle CAB has its marked right angle at C. The two legs have matching ticks. Side CA is labeled s. Hypotenuse AB is labeled h.
1. Hint 1
  Equal angles have equal opposite sides.
2. Hint 2
  Apply Pythagoras after identifying both legs. Use the positive root because these quantities are lengths.
3. Hint 3
  To reverse the relationship, divide by the nonzero multiplier.
Answer

$\begin{aligned}h&=s\sqrt2.\\s&=\frac{h}{\sqrt2}.\end{aligned}$
For a $13$ cm leg, the hypotenuse is $13\sqrt2$ cm and the reverse formula returns $13$ cm.

Full solution

As a numerical check, equal acute angles force both legs to be $13$ cm.
Pythagoras gives
$h^2=13^2+13^2.$
Thus
$h^2=338.$
The positive root is $13\sqrt2$ cm.

For a general positive leg $s$ , equal opposite angles again force the other leg to be $s$ .
Pythagoras gives
$h^2=2s^2.$
Taking the positive root gives
$h=s\sqrt2.$

The multiplier is positive, so division recovers the leg.
$s=\frac{h}{\sqrt2}.$
With hypotenuse $13\sqrt2$ cm this returns $13$ cm.

Answer

$\begin{aligned}h&=s\sqrt2.\\s&=\frac{h}{\sqrt2}.\end{aligned}$
For a $13$ cm leg, the hypotenuse is $13\sqrt2$ cm and the reverse formula returns $13$ cm.

Key idea

The equal-angle theorem supplies the second leg before Pythagoras supplies the hypotenuse.
Problem 6 An altitude measurement

An equilateral triangular frame has altitude $15$ cm.

Find its exact side length and perimeter.

The stated altitude meets the opposite side at its midpoint.

Text description of this figure

Equilateral triangle ABC has midpoint M on BC and segment AM. Matching single ticks mark the equal outer sides AB and AC. The proved right angle at M and equal half-base double ticks are shown. The sloping sides are labeled e. AM is labeled 15 cm.
1. Hint 1
  The altitude is the long leg in either congruent half of the equilateral triangle.
2. Hint 2
  Recover half a side first, then double it. The frame has three equal sides.
Answer

Side length $10\sqrt3$ cm; perimeter $30\sqrt3$ cm.

Full solution

Let $s$ be a half-base length.
Since the altitude is the long leg,
$s\sqrt3=15.$

Divide and rationalize.
$s=5\sqrt3\text{ cm}.$

An equilateral side is twice the half-base.
$e=10\sqrt3\text{ cm}.$

The perimeter contains three equal sides.
$P=30\sqrt3\text{ cm}.$

A check in one half gives short-leg square $75$ plus altitude square $225$ , totaling $300$ , the square of the side length.

Each half has the checked side lengths.

Answer

Side length $10\sqrt3$ cm; perimeter $30\sqrt3$ cm.

Key idea

An equilateral altitude gives half a side before it gives a whole side.
Problem 7 The two projected pieces

Triangle $ABC$ is right at $C$ , angle $A$ is $30^\circ$ , and hypotenuse $AB$ is $34$ cm. The altitude from $C$ meets $AB$ at $H$ .

Find $AH$ and $HB$ exactly. Then explain why $AH:HB=3:1$ for every right triangle with these angle measures, whatever its size.

The marked right angles and labeled quantities are the givens.

Text description of this figure

Triangle ABC is right at C. H is the foot of the altitude on AB. AB is 34 cm; AH is AH; HB is HB. CA is AC. CB is BC. Angle A is 30°.
1. Hint 1
  Find both legs from the special side ratio before finding their projections.
2. Hint 2
  Use $AC^2=AB\cdot AH$ and $BC^2=AB\cdot HB$ .
3. Hint 3
  For the general explanation, let the hypotenuse be $c>0$ .
Answer

$AH=25.5$ cm and $HB=8.5$ cm. In general, $AH=3c/4$ and $HB=c/4$ , so $AH:HB=3:1$ .

Full solution

The short leg opposite $30^\circ$ is $BC=17$ cm.
The long leg is $AC=17\sqrt3$ cm.

Apply the leg projection relationships.
$\begin{aligned}AH&=\frac{AC^2}{AB}=\frac{867}{34}=25.5\text{ cm}.\\HB&=\frac{BC^2}{AB}=\frac{289}{34}=8.5\text{ cm}.\end{aligned}$
Their sum is the full $34$ cm hypotenuse.

Now let $AB=c>0$ .
The corresponding legs are $c\sqrt3/2$ and $c/2$ .
Hence
$\begin{aligned}AH&=\frac{(c\sqrt3/2)^2}{c}=\frac{3c}{4}.\\HB&=\frac{(c/2)^2}{c}=\frac{c}{4}.\end{aligned}$

Dividing the two projected lengths gives $AH:HB=3:1$ .
The leg ratio is $\sqrt3:1$ ; the projection ratio uses their squares, so it is $3:1$ .

The projected pieces follow from the corresponding leg squares.

Answer

$AH=25.5$ cm and $HB=8.5$ cm. In general, $AH=3c/4$ and $HB=c/4$ , so $AH:HB=3:1$ .

Key idea

A special leg ratio determines the hypotenuse-piece ratio through the leg projection identities.
Problem 8 A recorded calculation

Triangle $ABC$ has a right angle at $C$ and angle $A$ of $30^\circ$ . The side $AC$ measures $12\sqrt3$ cm.

Maya writes, "The hypotenuse is twice the given side, so AB is 24 square roots of 3 centimeters. Then BC is half of AC, so BC is 6 square roots of 3 centimeters."

Identify the side-role error. Find the correct lengths of $BC$ and $AB$ , and check them using squares.

The given side is located using the marked angles.

Text description of this figure

Right triangle CAB has its marked right angle at C. The angle at A is 30 degrees and the angle at B is 60 degrees. Side CA is labeled 12√3 cm.
1. Hint 1
  Which side is opposite the 30-degree angle? Which side is opposite the right angle?
2. Hint 2
  The hypotenuse is twice the short leg, not twice either chosen leg. Recover the short leg from AC first.
3. Hint 3
  Check whether the squares of your two legs add to the square of your hypotenuse.
Answer

Maya doubles the long leg and then halves that long leg. The correct lengths are
$\begin{aligned}BC&=12\text{ cm}.\\AB&=24\text{ cm}.\end{aligned}$
The squared lengths satisfy $144+432=576$ .

Full solution

Angle $A$ is $30^\circ$ , so its opposite side $BC$ is the short leg.
Side $AC$ is the long leg, and side $AB$ is the hypotenuse.
Maya applies the factor of two to the wrong side.

Recover the short leg from the given long leg.
$BC=\frac{12\sqrt3}{\sqrt3}.$
Thus
$BC=12\text{ cm}.$

Double the short leg for the hypotenuse.
$AB=24\text{ cm}.$

The leg squares are $144$ and $432$ .
Their sum is $576$ , which equals $24^2$ , so Pythagoras checks the repaired lengths.

The corrected lengths follow their angle-relative roles.

Answer

Maya doubles the long leg and then halves that long leg. The correct lengths are
$\begin{aligned}BC&=12\text{ cm}.\\AB&=24\text{ cm}.\end{aligned}$
The squared lengths satisfy $144+432=576$ .

Key idea

The factor of two relates the short leg and hypotenuse only.
Problem 9 Recognizing the special shape

Triangle $ABC$ is right at $C$ . Its hypotenuse $AB$ is $46$ cm, and leg $BC$ is $23$ cm.

Find $AC$ exactly. Prove that angle $A$ is $30^\circ$ and angle $B$ is $60^\circ$ by comparing with a known special triangle; do not assume a converse of the side-ratio rule.

Would knowing only that two sides of a triangle are $23$ cm and $46$ cm justify the same conclusion? Test the possible third side $30$ cm and explain.

The right angle and two lengths are given; the acute angles remain to be proved.

Text description of this figure

Triangle ABC is right at C. Hypotenuse AB is 46 cm, and BC is 23 cm. No acute angle measures are marked.
1. Hint 1
  Use Pythagoras to find the positive length of AC.
2. Hint 2
  A 30-60-90 triangle with short leg 23 cm supplies a comparison triangle with three known side lengths.
3. Hint 3
  Check triangle existence before comparing the largest-side square in the second case.
Answer

$AC=23\sqrt3$ cm. SSS with a 30-60-90 comparison triangle proves $A=30^\circ$ and $B=60^\circ$ . Two sides alone do not suffice: sides $23$ , $30$ , $46$ form an obtuse triangle.

Full solution

Pythagoras applies because the right angle is given.
$\begin{aligned}AC^2&=46^2-23^2=1587.\\AC&=23\sqrt3\text{ cm}.\end{aligned}$

Bisect an equilateral triangle of side $46$ cm by joining a vertex to the opposite midpoint.
The lesson proof gives a 30-60-90 half with sides $23$ , $23\sqrt3$ , and $46$ cm.

All three side lengths match triangle $ABC$ .
SSS proves congruence.
The angle opposite the $23$ cm side is $30^\circ$ , so $A=30^\circ$ ; the remaining acute angle $B$ is $60^\circ$ .

For sides $23$ , $30$ , and $46$ , the existence check passes because $23+30>46$ .
But
$\begin{aligned}46^2&=2116.\\23^2+30^2&=1429.\end{aligned}$
The longest-side square is greater, so that triangle is obtuse, not right.
Thus the two given side lengths alone do not force a 30-60-90 triangle.

Answer

$AC=23\sqrt3$ cm. SSS with a 30-60-90 comparison triangle proves $A=30^\circ$ and $B=60^\circ$ . Two sides alone do not suffice: sides $23$ , $30$ , $46$ form an obtuse triangle.

Key idea

Use all the stated hypotheses when recognizing a special triangle; a side ratio alone need not establish a right angle.
Problem 10 A shared length

A square has side length $a>0$ . An equilateral triangle has a side equal in length to a diagonal of the square.

The equilateral triangle's altitude is $10\sqrt6$ cm. Find the square's side length, its perimeter, and the equilateral triangle's perimeter. Explain the chain of side relationships and check the altitude exactly.
1. Hint 1
  The shared length has different roles: square diagonal and equilateral side.
2. Hint 2
  Recover the equilateral side from its altitude, then recover the square side from that diagonal.
3. Hint 3
  Use four equal sides for the square perimeter and three for the equilateral perimeter.
Answer

Square side $20$ cm; square perimeter $80$ cm; equilateral perimeter $60\sqrt2$ cm.

Full solution

Let $d$ be the common length of the square diagonal and the equilateral side.
The equilateral altitude is half its side times $\sqrt3$ , so
$\frac{d\sqrt3}{2}=10\sqrt6.$

Multiply by $2$ and divide by $\sqrt3$ .
$d=20\sqrt2\text{ cm}.$

A square diagonal is $\sqrt2$ times its side.
Therefore
$a=\frac{20\sqrt2}{\sqrt2}.$
Thus the square side is $20$ cm.

The square has four equal sides, so its perimeter is $80$ cm.
The equilateral triangle has three equal sides of length $20\sqrt2$ cm, so its perimeter is $60\sqrt2$ cm.

Check the equilateral altitude.
$h=\frac{20\sqrt2\cdot\sqrt3}{2}.$
This is exactly $10\sqrt6$ cm, the stated altitude.

The recovered equilateral side is the square diagonal.

Answer

Square side $20$ cm; square perimeter $80$ cm; equilateral perimeter $60\sqrt2$ cm.

Key idea

Follow a shared segment through its role in each figure before combining the two ratios.

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