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The Pythagorean Theorem and Converse: 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 Triangle RST

Triangle $RST$ has a right angle at $S$ . Side $RS$ is $5$ cm and side $ST$ is $3\sqrt7$ cm.

Identify the hypotenuse and find its exact length in simplest radical form.

The stated right angle is at S.

Text description of this figure

Triangle RTS has RS labeled 5 cm, ST labeled 3√7 cm, and RT labeled RT. The right angle is at S.
1. Hint 1
  Locate the side that does not touch the right-angle vertex.
2. Hint 2
  Square each leg, including the outside coefficient of the radical, and add.
3. Hint 3
  After taking the positive root, look for a perfect-square factor in the radicand.
Answer

The hypotenuse is $RT$ , with exact length $2\sqrt{22}$ cm, equivalently $\sqrt{88}$ cm.

Full solution

The right angle is at $S$ , so the side across from it is $RT$ .
The two given sides are its legs.

Apply the theorem.
$RT^2=5^2+(3\sqrt7)^2$

Squaring the radical length gives $63$ .
Thus
$RT^2=25+63$

Take the positive root and simplify.
$\begin{aligned}RT&=\sqrt{88},\\RT&=2\sqrt{22}.\end{aligned}$

Answer

The hypotenuse is $RT$ , with exact length $2\sqrt{22}$ cm, equivalently $\sqrt{88}$ cm.

Key idea

The right-angle vertex identifies the hypotenuse before any numbers enter the equation.
Problem 2 Another orientation

A right triangle has hypotenuse $3\sqrt{10}$ m and one leg $7$ m. Find the exact length of the other leg.

Only the stated measurements are labeled.

Text description of this figure

Triangle ABC has AC labeled 7 m, CB labeled x, and AB labeled 3√10 m. The right angle is at C.
1. Hint 1
  Decide whether the unknown is a leg or the hypotenuse.
2. Hint 2
  Subtract the known leg's square from the hypotenuse's square; square the coefficient $3$ too.
Answer

$\sqrt{41}$ m.

Full solution

The unknown side $x$ is a leg.
Rearrange the theorem to isolate its square.
$x^2=(3\sqrt{10})^2-7^2$

Evaluate the two squares.
$x^2=90-49$

The radicand is $41$ , which has no square factor greater than $1$ .
Take the positive root.
$x=\sqrt{41}$

The answer is shorter than the hypotenuse because its square is $41$ , less than $90$ .

Answer

$\sqrt{41}$ m.

Key idea

A missing leg is found by subtracting squares, even when a given side is a radical.
Problem 3 An asserted length

A right triangle has legs $8$ and $11$ units. Mara names the hypotenuse $t$ and writes the following lines.

Line 1: $t^2=8^2+11^2$ .

Line 2: $t=\sqrt{64+121}$ .

Line 3: $t=\sqrt{64}+\sqrt{121}$ .

Line 4: $t=19$ .

Explain the error in Line 3 and replace the incorrect conclusion with the exact hypotenuse. Give a direct squared-value check showing that $19$ cannot be the hypotenuse.
1. Hint 1
  A square root does not generally distribute over addition.
2. Hint 2
  Compare the square of the claimed length with the sum of the leg squares.
3. Hint 3
  Keep the sum under one radical before simplifying.
Answer

Line 3 distributes a square root over a sum, which is invalid. The correct hypotenuse is $\sqrt{185}$ units. The claimed $19$ units fails because $19^2=361$ , whereas the leg-square sum is $185$ .

Full solution

Lines 1 and 2 correctly identify and solve for the positive hypotenuse.
Line 3 treats a root of a sum as a sum of roots, a rule that does not hold here.

Add inside the one radical.
$t=\sqrt{185}$

The radicand is the product of $5$ and $37$ , so it has no perfect-square factor greater than $1$ .

Square the proposed answer $19$ .
$19^2=361$

Compare it with the required square.
$8^2+11^2=185$

The squares differ, so $19$ cannot be the hypotenuse.

The repaired length belongs across from the marked right angle.

Answer

Line 3 distributes a square root over a sum, which is invalid. The correct hypotenuse is $\sqrt{185}$ units. The claimed $19$ units fails because $19^2=361$ , whereas the leg-square sum is $185$ .

Key idea

Add the leg squares before taking one square root; a root of a sum does not split into two roots.
Problem 4 Two prior relations

Triangle $ABC$ is nondegenerate and right at $C$ . Let $a$ name $BC$ , $b$ name $CA$ , and $c$ name $AB$ . The altitude $CH$ meets the open segment $AB$ at $H$ , with $p$ naming $AH$ and $q$ naming $HB$ .

Use these prior results from similarity and segment addition.

$b^2=cp$ .

$a^2=cq$ .

$p+q=c$ .

Prove the Pythagorean theorem from these relations. Explain exactly where the fact that $H$ lies inside $AB$ is used.

The labels match the prior similarity relations in the statement.

Text description of this figure

ABC is right at C. The altitude CH meets the interior of AB at H. AH is p, HB is q, CA is b, BC is a, AB is c, and CH is h.
1. Hint 1
  Both leg-square products share the same whole hypotenuse.
2. Hint 2
  Add the two products and factor out that common length.
3. Hint 3
  The two interior pieces fill one whole hypotenuse.
Answer

The result is
$a^2+b^2=c^2$
The interior foot allows the two hypotenuse pieces to be replaced by the whole hypotenuse, by segment addition.

Full solution

Add the two leg-square relations.
$a^2+b^2=cq+cp$

Factor out the common hypotenuse.
$a^2+b^2=c(q+p)$

The interior position of $H$ means that the adjacent lengths $AH$ and $HB$ partition $AB$ .
This is the step that gives
$q+p=c$

Substitute the whole hypotenuse for the sum of its pieces.
$a^2+b^2=c^2$

Answer

The result is
$a^2+b^2=c^2$
The interior foot allows the two hypotenuse pieces to be replaced by the whole hypotenuse, by segment addition.

Key idea

The similarity proof adds two products with a common hypotenuse, then uses segment addition.
Problem 5 Four proposed frames

For each proposed set of side lengths, determine whether a nondegenerate triangle exists. Classify each existing triangle as acute, right, or obtuse. Show the existence check before the squared comparison.

Case A: $10$ , $12$ , and $17$ cm.

Case B: $4$ , $10$ , and $15$ cm.

Case C: $7$ , $10$ , and $12$ cm.

Case D: $14$ , $48$ , and $50$ cm.
1. Hint 1
  First compare the longest length with the sum of the other two.
2. Hint 2
  For each set that closes, compare the longest-side square with the sum of the other two squares.
3. Hint 3
  A greater longest-side square means an angle larger than a right angle, while equality gives a right angle.
Answer

Case A is obtuse. Case B does not form a nondegenerate triangle. Case C is acute. Case D is right.

Full solution

Case A exists because $10+12>17$ .
Its comparison is
$\begin{aligned}17^2&=289,\\10^2+12^2&=244.\end{aligned}$
The longest-side square is larger, so Case A is obtuse.

Case B fails the existence check.
$4+10<15$
It receives no angle classification.

Case C exists because $7+10>12$ .
Compare
$\begin{aligned}12^2&=144,\\7^2+10^2&=149.\end{aligned}$
The longest-side square is smaller, so Case C is acute.

Case D exists because $14+48>50$ .
Its comparison is
$\begin{aligned}14^2+48^2&=196+2304=2500.\\50^2&=2500.\end{aligned}$
Equality gives a right triangle by the converse.

Answer

Case A is obtuse. Case B does not form a nondegenerate triangle. Case C is acute. Case D is right.

Key idea

The square comparison classifies only a triangle that has already passed the existence check.
Problem 6 Three fixed lengths

Triangle $PQR$ has side lengths $PQ$ of $9$ units, $QR$ of $40$ units, and $PR$ of $41$ units. No angle is given.

Prove that angle $PQR$ is right by constructing a separate right triangle and using SSS. State how the Pythagorean theorem is used without assuming the original triangle is right.
1. Hint 1
  Make a comparison triangle whose right angle is known from its construction.
2. Hint 2
  Copy the $9$ -unit and $40$ -unit sides onto perpendicular rays.
3. Hint 3
  Find the constructed hypotenuse, then match all three sides to the given triangle.
Answer

Construct a right triangle with legs $9$ and $40$ units. Its hypotenuse is $41$ units by the theorem. SSS congruence with $PQR$ makes angle $PQR$ right; the theorem was applied only to the constructed triangle.

Full solution

Construct triangle $XYZ$ with a right angle at $Y$ , side $XY$ of $9$ units, and side $YZ$ of $40$ units.
Copying positive lengths onto perpendicular rays makes this a nondegenerate right triangle.

Call its hypotenuse $d$ .
The theorem applies here because the construction supplies the right angle.
$\begin{aligned}d^2&=9^2+40^2,\\d^2&=1681,\\d&=41.\end{aligned}$

The corresponding sides $PQ$ and $XY$ both have length $9$ .
Sides $QR$ and $YZ$ both have length $40$ .
Sides $PR$ and $XZ$ both have length $41$ .

SSS proves that triangles $PQR$ and $XYZ$ are congruent in that order.
Angle $PQR$ therefore matches the right angle $XYZ$ .

No right angle in the original triangle was assumed.
Its right angle follows after congruence.

The dashed triangle is constructed right. Its computed third side permits SSS.

Answer

Construct a right triangle with legs $9$ and $40$ units. Its hypotenuse is $41$ units by the theorem. SSS congruence with $PQR$ makes angle $PQR$ right; the theorem was applied only to the constructed triangle.

Key idea

A constructed right triangle lets the established theorem supply the third side needed for a noncircular converse proof.
Problem 7 A changing third length

Two side lengths are $5$ and $12$ units. The third is a real number $z$ of units, with $z\ge12$ .

Give the complete range of $z$ for which a nondegenerate triangle exists. Within that range, give the acute, right, and obtuse cases, including or excluding each boundary correctly.
1. Hint 1
  The given restriction makes $z$ a longest side, including the possibility of a tie.
2. Hint 2
  The triangle inequality gives an upper bound on $z$ .
3. Hint 3
  Compare $z^2$ with the sum of the two fixed squares, then intersect each case with the existence range.
Answer

Existence:
$12\le z<17$
Acute:
$12\le z<13$
Right:
$z=13$
Obtuse:
$13<z<17$

Full solution

The longest side is $z$ .
Existence requires
$5+12>z$
Combining this with the stated lower bound gives
$12\le z<17$

The fixed square sum is
$5^2+12^2=169$

The acute case requires $z^2<169$ , so the positive length is below $13$ .
Intersecting with existence gives
$12\le z<13$

The right case requires equality of squares.
Since $z$ is positive,
$z=13$

The obtuse case requires $z$ above $13$ , while still below the flat boundary.
$13<z<17$

The value $12$ is allowed and acute.
The value $17$ is excluded because it makes the arrangement flat.

Answer

Existence:
$12\le z<17$
Acute:
$12\le z<13$
Right:
$z=13$
Obtuse:
$13<z<17$

Key idea

A variable-length classification must satisfy both the triangle-inequality range and the squared-side comparison.
Problem 8 Four written lines

A triangle has side lengths $11$ , $13$ , and $20$ units. A student writes this argument.

Line 1: $11+13>20$ , so a nondegenerate triangle with these lengths exists.

Line 2: $20^2=11^2+13^2$ , because $20$ is the longest side.

Line 3: $20^2=290$ , by evaluating the right side of Line 2.

Line 4: The triangle is right by the converse of the Pythagorean theorem.

Identify the first invalid line, explain why its stated reason fails, and repair the argument to give the correct angle classification.
1. Hint 1
  The longest side tells you which square to compare; it does not force equality.
2. Hint 2
  Check the lines in order and distinguish an assumed equality from an arithmetic consequence.
3. Hint 3
  Evaluate the longest-side square independently of the other square sum.
Answer

Line 2 is the first invalid line. A longest side need not satisfy the right-triangle equality. The valid triangle is obtuse because its longest-side square is $400$ , greater than the other square sum of $290$ .

Full solution

Line 1 is valid: the positive lengths pass the strict triangle inequality.

Line 2 is the first invalid line.
Knowing the longest side selects the comparison side but gives no right-angle hypothesis.
The student has applied the theorem before knowing the triangle is right.

Compute the two quantities separately.
$\begin{aligned}20^2&=400,\\11^2+13^2&=290.\end{aligned}$

The longest-side square is greater than the sum of the other two squares.
The angle opposite the $20$ -unit side is obtuse, so the triangle is obtuse.

Answer

Line 2 is the first invalid line. A longest side need not satisfy the right-triangle equality. The valid triangle is obtuse because its longest-side square is $400$ , greater than the other square sum of $290$ .

Key idea

The first unsupported equality can invalidate an argument even when the arithmetic after it is performed correctly.
Problem 9 Three designs

Design three nondegenerate triangles. Each must have two sides of $9$ and $12$ units and a third side of a positive integer number of units strictly longer than $12$ .

Design A must be acute.

Design B must be right.

Design C must be obtuse.

Give one possible third length for each design and justify both existence and classification.
1. Hint 1
  Find the right-triangle reference length for the two fixed sides.
2. Hint 2
  Choose one integer below that reference and one above it, keeping both longer than $12$ .
3. Hint 3
  Every chosen third side must still be shorter than the sum of the fixed sides.
Answer

One choice, in units:
$\begin{aligned}A&:13,\\B&:15,\\C&:17.\end{aligned}$
Other correct designs are possible.

Full solution

The fixed square sum is
$9^2+12^2=225$
The right-triangle reference length is $15$ units.

All proposed third sides must be strictly between $12$ and $21$ , since $9+12$ is $21$ .

Design A can use $13$ .
It passes the length bounds, and its longest-side square satisfies
$13^2<225$
Thus it is acute.

Design B uses $15$ .
It also passes the length bounds.
$15^2=225$
The converse makes it right.

Design C can use $17$ .
It is less than $21$ , so the triangle exists.
$17^2>225$
Thus it is obtuse.

The same comparisons show all accepted integer choices: $13$ or $14$ for A, $15$ for B, and $16$ through $20$ for C.

Answer

One choice, in units:
$\begin{aligned}A&:13,\\B&:15,\\C&:17.\end{aligned}$
Other correct designs are possible.

Key idea

Designs can be chosen on either side of one right-triangle reference, while the triangle inequality limits the choices.
Problem 10 Replacing one side

Triangle $ABC$ is right at $C$ . Its altitude $CH$ meets the open segment $AB$ at $H$ . Length $AH$ is $4$ units and length $HB$ is $21$ units. As in the chapter, $a$ names $BC$ , $b$ names $CA$ , and $c$ names $AB$ .

Find $a$ and $b$ exactly.

A new side set consists of those same two lengths $a$ and $b$ , together with a third length of $24$ units. Decide whether it makes a nondegenerate triangle and classify it. Justify which side is longest before comparing squares.

The diagram shows the original right triangle, before the new side set is considered.

Text description of this figure

ABC is right at C. The altitude CH meets the interior of AB at H. AH is 4, HB is 21, CA is b, BC is a, AB is c, and CH is h.
1. Hint 1
  Use the whole hypotenuse and each adjacent piece to recover the original legs.
2. Hint 2
  For the new set, compare each known side square with $24^2$ to identify the longest side.
3. Hint 3
  The original right triangle supplies the sum of the two known side squares, but the new triangle need not be right.
Answer

The original lengths are
$\begin{aligned}a&=5\sqrt{21}\text{ units},\\b&=10\text{ units}.\end{aligned}$
The new lengths form an acute nondegenerate triangle, with the $24$ -unit side longest.

Full solution

Segment addition gives the original hypotenuse.
$c=4+21$
Thus $c$ is $25$ units.

The prior leg-square relations give
$\begin{aligned}a^2&=25(21),\\a&=5\sqrt{21}.\end{aligned}$

For the other leg,
$\begin{aligned}b^2&=25(4),\\b&=10.\end{aligned}$

In the new set, compare the squares of the three positive lengths.
$\begin{aligned}a^2&=525,\\b^2&=100,\\24^2&=576.\end{aligned}$
The $24$ -unit side is longest.

Check existence.
Since $\sqrt{21}>4$ , the length $a$ exceeds $20$ units.
Adding $b$ gives more than $30$ units, which is more than $24$ .
The two shorter lengths therefore add to more than the longest.

The original theorem also gives the two retained side squares as
$a^2+b^2=625$

This sum exceeds $576$ , the new longest-side square.
The new triangle is acute.
Replacing the original hypotenuse did not preserve the original right angle.

These are the derived lengths of the original right triangle.

Answer

The original lengths are
$\begin{aligned}a&=5\sqrt{21}\text{ units},\\b&=10\text{ units}.\end{aligned}$
The new lengths form an acute nondegenerate triangle, with the $24$ -unit side longest.

Key idea

Lengths derived in one right triangle can form a different triangle whose existence and angles need separate checks.

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