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Similarity in 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 given pieces

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

The pieces are
$AH=5\text{ cm}.$
The other piece is
$HB=20\text{ cm}.$
Find the altitude and the whole hypotenuse.

The two measured pieces.

Text description of this figure

ABC is right at C. CH is perpendicular to AB at interior point H. AH is 5 centimeters and HB is 20 centimeters. Neither requested length is shown.
1. Hint 1
  Identify which lengths are pieces of the hypotenuse.
2. Hint 2
  The altitude uses the product of the pieces, while the whole uses their sum.
Answer

$\begin{aligned}CH&=10\text{ cm}\\AB&=25\text{ cm}.\end{aligned}$

Full solution

The altitude relation gives
$CH^2=5\cdot20.$

The positive root gives
$CH=10\text{ cm}.$

The whole consists of both pieces, so
$AB=5+20.$

Thus
$AB=25\text{ cm}.$

Answer

$\begin{aligned}CH&=10\text{ cm}\\AB&=25\text{ cm}.\end{aligned}$

Key idea

The altitude and the whole hypotenuse use the same pieces in different ways.
Problem 2 A leg and the whole

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

The measurements are
$AB=30\text{ cm}.$
Also,
$BC=18\text{ cm}.$
Find both pieces, the other leg, and the altitude. Give exact decimal values.

The supplied outer lengths.

Text description of this figure

The right triangle has hypotenuse AB 30 centimeters and leg BC 18 centimeters. CH is its hypotenuse altitude, with no unknown lengths labeled numerically.
1. Hint 1
  The known leg touches one particular piece of the hypotenuse.
2. Hint 2
  Find $HB$ from the leg relation before subtracting it from the whole.
3. Hint 3
  Use the recovered pieces to find the altitude, and use the whole with $AH$ to find $CA$ .
Answer

$\begin{aligned}AH&=19.2\text{ cm}\\HB&=10.8\text{ cm}\\CA&=24\text{ cm}\\CH&=14.4\text{ cm}.\end{aligned}$

Full solution

Leg $BC$ touches $HB$ , so
$18^2=30\cdot HB.$

Dividing by the whole gives
$HB=10.8\text{ cm}.$

Subtract from the whole to find
$AH=19.2\text{ cm}.$

The other leg uses $AH$ , giving
$CA^2=30\cdot19.2.$

Its positive root is
$CA=24\text{ cm}.$

The altitude uses both pieces, so
$CH^2=19.2\cdot10.8.$

The product is $207.36$ , whose positive square root is $14.4$ , so
$CH=14.4\text{ cm}.$

Completed lengths, in centimeters.

Answer

$\begin{aligned}AH&=19.2\text{ cm}\\HB&=10.8\text{ cm}\\CA&=24\text{ cm}\\CH&=14.4\text{ cm}.\end{aligned}$

Key idea

Recover the piece belonging to the given leg before finding the remaining lengths.
Problem 3 Exact lengths at the cut

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

The pieces are
$AH=7\text{ cm}.$
Also,
$HB=14\text{ cm}.$
Find $CH$ , $CA$ , and $BC$ in simplest radical form.

The given pieces.

Text description of this figure

ABC is right at C with interior hypotenuse altitude CH. AH is 7 centimeters and HB is 14 centimeters. No derived lengths are shown.
1. Hint 1
  Separate the altitude calculation from the two leg calculations.
2. Hint 2
  Add the pieces to get the whole hypotenuse before calculating a leg.
3. Hint 3
  Factor each radicand to take out its largest perfect-square factor.
Answer

$\begin{aligned}CH&=7\sqrt2\text{ cm}\\CA&=7\sqrt3\text{ cm}\\BC&=7\sqrt6\text{ cm}.\end{aligned}$

Full solution

The whole hypotenuse has length
$AB=21\text{ cm}.$

The altitude uses the two pieces:
$CH^2=7\cdot14.$

Taking the positive root and simplifying gives
$CH=7\sqrt2\text{ cm}.$

The leg at $A$ satisfies
$CA^2=21\cdot7.$

Its positive root is
$CA=7\sqrt3\text{ cm}.$

The leg at $B$ satisfies
$BC^2=21\cdot14.$

Its positive root is
$BC=7\sqrt6\text{ cm}.$

Completed exact lengths, in centimeters.

Answer

$\begin{aligned}CH&=7\sqrt2\text{ cm}\\CA&=7\sqrt3\text{ cm}\\BC&=7\sqrt6\text{ cm}.\end{aligned}$

Key idea

Keep exact radicals when the data are exact.
Problem 4 Three ordered names

Triangle $JKL$ is right at $K$ . The perpendicular $KM$ meets the interior of hypotenuse $JL$ at $M$ .

Prove by AA that all three triangles are similar. Write one ordered statement containing the whole triangle and both small triangles. Give the angle correspondences explicitly.

The named configuration.

Text description of this figure

JKL is right at K. KM is perpendicular to hypotenuse JL at interior point M. The diagram gives incidence and perpendicularity only.
1. Hint 1
  Locate the right angle in each of the three triangles.
2. Hint 2
  Use the shared endpoint angle for each small triangle.
3. Hint 3
  Choose the whole triangle's order so the right angle is last, and match that order in both small triangles.
Answer

$\triangle JLK\sim\triangle JKM\sim\triangle KLM.$
The corresponding angle measures are the whole angle at $J$ , the whole angle at $L$ , and $90^\circ$ , in that order.

Full solution

The whole triangle has acute endpoint angles at $J$ and $L$ , whose measures sum to $90^\circ$ .

Triangle $JKM$ has the same angle at $J$ as the whole and a right angle at $M$ .
Its angle $JKM$ therefore equals the whole angle at $L$ .

Triangle $KLM$ has the same angle at $L$ as the whole and a right angle at $M$ .
Its angle $LKM$ therefore equals the whole angle at $J$ .

Each comparison has two matching angle pairs, so AA gives
$\triangle JLK\sim\triangle JKM\sim\triangle KLM.$

Answer

$\triangle JLK\sim\triangle JKM\sim\triangle KLM.$
The corresponding angle measures are the whole angle at $J$ , the whole angle at $L$ , and $90^\circ$ , in that order.

Key idea

Give a shared acute angle as well as a right angle when applying AA.
Problem 5 Explain the two leg relations

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

Use ordered similarity and corresponding sides to prove the relation for leg $CA$ and then the relation for leg $BC$ . State each result using segment names, rather than remembered letter formulas.

Segments used in the argument.

Text description of this figure

The standard right triangle is labeled with segment names. CA touches AH at A, and BC touches HB at B. The altitude and both right angles are marked.
1. Hint 1
  For each leg, compare the whole triangle with the small triangle that has that leg as its hypotenuse.
2. Hint 2
  In $ABC$ to $ACH$ , $AB$ matches $AC$ and $AC$ matches $AH$ .
3. Hint 3
  For the other leg, use $ABC$ to $CBH$ and keep the same ratio direction.
Answer

$\begin{aligned}CA^2&=AB\cdot AH\\BC^2&=AB\cdot HB.\end{aligned}$

Full solution

AA gives $\triangle ABC\sim\triangle ACH$ .
The matching pairs $AB,AC$ and $AC,AH$ give
$\frac{AB}{AC}=\frac{AC}{AH}.$

All lengths are positive, so cross multiplication is valid and gives
$CA^2=AB\cdot AH.$

AA also gives $\triangle ABC\sim\triangle CBH$ .
The matching pairs $AB,CB$ and $BC,BH$ give
$\frac{AB}{BC}=\frac{BC}{HB}.$

Cross multiplication gives
$BC^2=AB\cdot HB.$

Answer

$\begin{aligned}CA^2&=AB\cdot AH\\BC^2&=AB\cdot HB.\end{aligned}$

Key idea

The piece in a leg relationship shares the leg's hypotenuse endpoint.
Problem 6 Check the recorded calculation

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

The measured pieces are
$AH=2\text{ cm}.$
Also,
$HB=18\text{ cm}.$
A student adds the pieces to obtain $AB=20$ cm, then writes the following altitude calculation:
$CH^2=20\cdot18.$
The student concludes
$CH=6\sqrt{10}\text{ cm}.$
Explain the error, find the correct altitude, and identify which actual segment has the student's computed length.

The measured pieces.

Text description of this figure

ABC is right at C, and CH is perpendicular to AB. The measured pieces AH and HB are 2 and 18 centimeters. No calculated lengths appear in the givens.
1. Hint 1
  Distinguish an altitude relationship from a leg relationship.
2. Hint 2
  The altitude uses both pieces, while the product of the whole and $HB$ belongs to a leg.
Answer

The student used the leg relationship for $BC$ .
$CH=6\text{ cm}.$
The computed length $6\sqrt{10}$ cm belongs to $BC$ .

Full solution

The whole is correctly found as $20$ cm, but the altitude uses $AH$ and $HB$ , not $AB$ and $HB$ .
Thus
$CH^2=2\cdot18.$

The squared altitude is $36$ , giving
$CH=6\text{ cm}.$

The student's product is the leg relationship
$BC^2=20\cdot18.$

Its positive root is indeed
$BC=6\sqrt{10}\text{ cm}.$

The two different calculated segments, in centimeters.

Answer

The student used the leg relationship for $BC$ .
$CH=6\text{ cm}.$
The computed length $6\sqrt{10}$ cm belongs to $BC$ .

Key idea

A correct computation can answer the wrong geometric question.
Problem 7 Draw and annotate

Draw a nondegenerate triangle $UVW$ with a right angle at $V$ . Place $V$ below hypotenuse $UW$ . The acute angle at $U$ measures $25^\circ$ .

Add the perpendicular $VD$ to hypotenuse $UW$ , with $D$ on the interior. Mark every right angle, label the other acute angles in the two small triangles, and write all three triangles in corresponding order.
1. Hint 1
  Rotation or reflection changes placement, but not the angle matching.
2. Hint 2
  The two small right triangles share the angles at the two endpoints of the hypotenuse.
3. Hint 3
  Read each triangle in the order $25^\circ$ , $65^\circ$ , $90^\circ$ .
Answer

Right angles: $\angle UVW,\angle UDV,\angle VDW$ . Angles $VWD,UVD$ are $65^\circ$ ; $DVW$ is $25^\circ$ .
$\triangle UWV\sim\triangle UVD\sim\triangle VWD.$

Full solution

Draw $UW$ as a segment and put $V$ below it so the angle at $V$ is right.
Since the angle at $U$ is $25^\circ$ , the angle at $W$ is $65^\circ$ .

The altitude meets the open hypotenuse segment at $D$ .
Mark the right angle of the whole triangle at $V$ and the right angle on each side of $VD$ at $D$ .

In $UVD$ , subtract $25^\circ$ and $90^\circ$ from $180^\circ$ , giving $65^\circ$ at $V$ .
The remaining part of the whole right angle at $V$ is $25^\circ$ .

Reading each triangle as $25^\circ$ , $65^\circ$ , $90^\circ$ gives
$\triangle UWV\sim\triangle UVD\sim\triangle VWD.$

One possible orientation of the completed construction.

Answer

Right angles: $\angle UVW,\angle UDV,\angle VDW$ . Angles $VWD,UVD$ are $65^\circ$ ; $DVW$ is $25^\circ$ .
$\triangle UWV\sim\triangle UVD\sim\triangle VWD.$

Key idea

Annotating angles first makes vertex order independent of orientation.
Problem 8 Test the missing hypothesis

A claim says, “Drawing an altitude to a side of any triangle makes two smaller triangles that are both similar to the whole triangle.”

Decide whether the claim is true. If it is false, give a nondegenerate triangle with an interior altitude foot that refutes it. Use angle measures to verify your counterexample, and state the hypothesis that repairs the three-triangle result.
1. Hint 1
  A right angle at the altitude foot does not force the original triangle to have a right angle.
2. Hint 2
  Try a triangle with two $40^\circ$ angles and one $100^\circ$ angle.
3. Hint 3
  Compare the full angle sets of the whole triangle and either small triangle.
Answer

False. A 40°, 40°, 100° triangle gives small triangles with angles 40°, 50°, 90°. Repair: start with a right triangle and draw the altitude from its right-angle vertex to its hypotenuse.

Full solution

Choose $ABC$ with angles $40^\circ$ at $A$ and $B$ and $100^\circ$ at $C$ .
Its acute endpoint angles ensure that the perpendicular from $C$ meets the interior of $AB$ .

Each small triangle has a $40^\circ$ endpoint angle and a $90^\circ$ angle at the foot, leaving $50^\circ$ as its third angle.

Neither small triangle can be similar to the whole: the whole contains a $100^\circ$ angle, whereas both small triangles contain a $90^\circ$ angle.

Requiring the original triangle to be right and choosing the altitude from that right-angle vertex to the hypotenuse supplies the three ordered similar triangles.

A possible counterexample configuration.

Answer

False. A 40°, 40°, 100° triangle gives small triangles with angles 40°, 50°, 90°. Repair: start with a right triangle and draw the altitude from its right-angle vertex to its hypotenuse.

Key idea

Check the whole triangle's angle hypothesis, as well as the perpendicular cut.
Problem 9 Equal legs and equal pieces

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

Prove that $CA=BC$ exactly when $AH=HB$ . Then determine the altitude as a fraction of the whole hypotenuse in this equal-leg case.
1. Hint 1
  The two leg squares share the same positive factor, the whole hypotenuse.
2. Hint 2
  Prove both directions of the stated equivalence.
3. Hint 3
  When the pieces are equal, their geometric mean equals either piece.
Answer

The two equalities are equivalent: $CA=BC$ exactly when $AH=HB$ . In this case,
$CH=\frac{AB}{2}.$

Full solution

The leg relations are
$CA^2=AB\cdot AH.$

The second leg relation is
$BC^2=AB\cdot HB.$

If $CA=BC$ , the squared lengths are equal.
Divide both products by positive $AB$ to obtain $AH=HB$ .

Conversely, if $AH=HB$ , the two products are equal, so the leg squares are equal.
Positive side lengths with equal squares are equal, giving $CA=BC$ .

In this case, both pieces have length $AB/2$ .
The altitude relation becomes
$CH^2=\left(\frac{AB}{2}\right)^2.$

Taking the positive root gives
$CH=\frac{AB}{2}.$

Answer

The two equalities are equivalent: $CA=BC$ exactly when $AH=HB$ . In this case,
$CH=\frac{AB}{2}.$

Key idea

An equivalence needs a separate justification in each direction.
Problem 10 Determine the internal measurements

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets hypotenuse $AB$ at interior point $H$ .

The whole hypotenuse has length
$AB=26\text{ cm}.$
The leg ratio is $BC:CA=3:2$ .

Find $AH$ , $HB$ , $CH$ , $CA$ , and $BC$ exactly. Explain why swapping the two hypotenuse pieces would not fit the specified leg ratio.

The supplied whole and leg ratio.

Text description of this figure

ABC is right at C, with interior hypotenuse altitude CH. AB is 26 centimeters. Leg BC is three ratio parts and leg CA is two ratio parts. No internal numerical answers are shown.
1. Hint 1
  The ratio of leg squares determines the ratio of the corresponding hypotenuse pieces.
2. Hint 2
  Square the given leg ratio, then use the whole length to determine each piece.
3. Hint 3
  After finding both pieces, use the altitude and leg relationships separately.
Answer

$\begin{aligned}AH&=8\text{ cm}\\HB&=18\text{ cm}\\CH&=12\text{ cm}\\CA&=4\sqrt{13}\text{ cm}\\BC&=6\sqrt{13}\text{ cm}.\end{aligned}$
Swapping the pieces would reverse the specified leg ratio to $2:3$ .

Full solution

The leg relations have the same positive factor $AB$ , so
$\frac{BC^2}{CA^2}=\frac{HB}{AH}.$

Squaring the supplied leg ratio gives
$HB:AH=9:4.$

The whole hypotenuse contains thirteen ratio parts totaling $26$ cm, so each part is $2$ cm.

Therefore
$AH=8\text{ cm}.$

The other piece is
$HB=18\text{ cm}.$

The altitude uses their product:
$CH^2=8\cdot18.$

Its positive root is
$CH=12\text{ cm}.$

The leg at $A$ satisfies
$CA^2=26\cdot8.$

Thus
$CA=4\sqrt{13}\text{ cm}.$

The leg at $B$ satisfies
$BC^2=26\cdot18.$

Thus
$BC=6\sqrt{13}\text{ cm}.$

The ratio fixes the longer piece at $B$ , where the longer leg $BC$ touches the hypotenuse.
Swapping the pieces would make $CA$ longer and give $BC:CA=2:3$ .

Completed lengths, in centimeters.

Answer

$\begin{aligned}AH&=8\text{ cm}\\HB&=18\text{ cm}\\CH&=12\text{ cm}\\CA&=4\sqrt{13}\text{ cm}\\BC&=6\sqrt{13}\text{ cm}.\end{aligned}$
Swapping the pieces would reverse the specified leg ratio to $2:3$ .

Key idea

Use a leg ratio to locate the pieces before recovering individual lengths.

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