The Pythagorean Theorem

A right triangle has one angle of ${90}^{\circ }$. A small square often marks that corner. Its two sides are the legs. The third side, across from the right angle, is the hypotenuse. It is always the longest side.

Find the hypotenuse first

Imagine triangle PQR with its right angle at Q. The legs PQ and QR meet at Q. PR is opposite Q, so PR is the hypotenuse. Turning the page changes none of these names. A side's position on the page does not decide its role.

We often label the legs $a$ and $b$ and the hypotenuse $c$. These letters are a choice, not part of the shape. A problem can use other labels, so look for the right angle before writing an equation.

State the rule

The Pythagorean theorem says that the square of the hypotenuse equals the sum of the squares of the legs:

For triangle PQR, the same rule is $P{Q}^2+Q{R}^2=P{R}^2$. The hypotenuse's square stands alone. Each leg is squared separately before the two results are added. For example, $3^2+4^2=9+16=25=5^2$. Adding the lengths gives 7, not the hypotenuse length of 5.

See the rule as areas

Draw a square on each side of a right triangle. Each triangle side becomes one edge of its square. A side of 3 units gives a square of 9 square units; a side of 4 gives a square of 16. Together those areas equal 25 square units, the area of the square on the 5-unit hypotenuse.

You can try this on grid paper with legs of 5 and 12 units. Their squares have areas 25 and 144, totaling 169. The hypotenuse is 13 units because ${13}^2=169$. A drawing helps you see the pattern. A proof explains why the pattern always holds.

The equation concerns right triangles. Sides of 4, 5 and 6 give $4^2+5^2=41$ and $6^2=36$. These values differ, so that triangle has no right angle. To calculate a missing side with the theorem, first establish the right angle from the shape or the given information.

A rule with a long history

The theorem is named after Pythagoras, a Greek thinker born around 570 BCE. The number relationship was known much earlier. The Babylonian tablet Plimpton 322 dates from around 1800 BCE and contains numbers linked to right triangles. One example is 119, 120 and 169:

Checking a few triangles gives examples. A proof gives a reason that works for every right triangle. Start with any right triangle whose legs are $a$ and $b$ and whose hypotenuse is $c$.

Rearrange four copies

Cut out four identical copies of the triangle. Draw two large squares, each with sides of length $a+b$.

In the first square, put a triangle in each corner. Place one leg along each outside edge so every edge contains an $a$-length and a $b$-length. The four hypotenuses surround a tilted square in the middle. Its sides all have length $c$.

Why are its corners right angles? The two acute angles of a right triangle total ${90}^{\circ }$. At an outside meeting point they lie on a straight line beside the central corner. That corner is therefore ${180}^{\circ }-{90}^{\circ }={90}^{\circ }$. The middle shape really is a square, with area $c^2$.

In the second large square, split each outside edge into lengths $a$ and $b$. Use two opposite corner rectangles, each measuring $a$ by $b$. Each rectangle holds two of the triangles. The uncovered regions are squares of side $a$ and side $b$.

Both large squares have equal areas. Both contain the same four triangles. Their uncovered areas must be equal, giving $c^2=a^2+b^2$. Nothing in this argument depends on particular side lengths.

Write the area argument in algebra

In the first arrangement, the whole square contains four triangles and the middle square:

Expanding the left side means multiplying $(a+b)(a+b)$. This gives $a^2+2ab+b^2$, including two copies of $ab$. The four triangle areas total $2ab$, so

Subtract $2ab$ from both sides to get $a^2+b^2=c^2$. Forgetting the middle term in $(a+b{)}^2$ would break this area calculation.

Garfield's trapezoid

James A. Garfield published an area proof in 1876, before becoming US president. Picture two vertical parallel sides of lengths $a$ and $b$, separated horizontally by $a+b$. Join their tops and bottoms to form a trapezoid. Two copies of our triangle fill its lower corners; their hypotenuses meet at a point on the bottom edge.

The remaining upper triangle has two sides of length $c$. The angle between them is right: the other two angles along the bottom line total ${90}^{\circ }$. Its area is therefore $c^2\mathrm{/}2$. Equating the trapezoid's area with its three triangle areas gives

Double both sides, expand, and subtract $2ab$. Again, $a^2+b^2=c^2$.

Use similar triangles

Draw right triangle ABC with the right angle at C. Set $BC=a$, $AC=b$ and $AB=c$. Draw the perpendicular height from C to AB, meeting it at D. Write $AD=p$ and $DB=q$, so $p+q=c$.

Each small triangle has a right angle and shares an acute angle with ABC. Triangles with matching angles are similar: their matching sides have equal ratios. In ACD, AC matches AB, and AD matches AC. Thus $b\mathrm{/}c=p\mathrm{/}b$, giving $b^2=cp$. In BCD, BC matches AB, and BD matches BC. Thus $a\mathrm{/}c=q\mathrm{/}a$, giving $a^2=cq$.

Add these equations:

The ratios must match corresponding sides. A small triangle's hypotenuse matches the large triangle's hypotenuse even though the triangles face different ways.

When both legs are known, add their squares to find the square of the hypotenuse. Then take the positive square root. The formula is $c=\sqrt{a^2+b^2}$, where $a$ and $b$ are the legs.

Square, add, take the root

A right triangle has legs of 6 cm and 8 cm.

1. Square each leg: $6^2=36$ and $8^2=64$.
2. Add the squares: $c^2=36+64=100$.
3. Take the root: $c=\sqrt{100}=10$.

The hypotenuse is 10 cm. The value 100 is its square, not its length. Squaring means multiplying a number by itself; $6^2$ is 36, not 12.

Keep a root exact or round it

With legs of 5 cm and 7 cm, $c^2=25+49=74$. The exact answer is $\sqrt{74}$ cm. Since $64<74<81$, the length lies between 8 cm and 9 cm. A calculator gives about 8.602 cm, so to the nearest tenth it is 8.6 cm. Write $\sqrt{74}\approx 8.6$ because the rounded value is close to the exact value.

Some roots simplify. A right triangle with two 5-unit legs gives $c=\sqrt{50}$. Since $50=25\cdot 2$, this is $5\sqrt{2}$ units. A square's diagonal follows the same pattern: with side length $s$, its diagonal is $s\sqrt{2}$. The diagonal splits the square into two right triangles with equal legs.

The method also works with decimals. Legs of 2.5 m and 6 m give $c^2=6.25+36=42.25$, so $c=6.5$ m. A decimal input does not mean the answer must be rounded; 6.5 is exact here.

Convert units before squaring

Suppose the legs measure 1.2 m and 50 cm. Choose centimeters: 1.2 m is 120 cm. Now

The root is 130 cm, or 1.3 m. Adding ${1.2}^2$ and ${50}^2$ would mix square meters with square centimeters. They cannot be added as though they used one scale.

Check the size

The hypotenuse must be longer than either leg. It is also shorter than their sum: the straight side is shorter than the path along both legs. For the 5 cm and 7 cm legs, 8.6 cm is longer than 7 cm and shorter than 12 cm. A result of 74 cm fails this check and points to a missing square-root step. Keep full calculator values until the final rounding step.

A leg is shorter than the hypotenuse. Finding it uses subtraction, because the known leg's square is already part of the hypotenuse's square. Start with the full equation so each length goes in the right place.

Rearrange the equation

If $c$ is the hypotenuse and $a$ is the known leg, then $a^2+b^2=c^2$. Subtract $a^2$ from both sides:

Square the known lengths before subtracting. Then take the positive root, since the answer is a length.

A triangle has hypotenuse 13 cm and one leg 5 cm. Write $5^2+b^2={13}^2$. This gives $b^2=169-25=144$, so $b=12$ cm. Check in the original equation: $25+144=169$.

Let the right angle choose the side

In triangle XYZ, the right angle is at Z. Side XY is 25 cm and XZ is 7 cm. XY is opposite the right angle, so it is the hypotenuse. For the missing side YZ,

Thus YZ is 24 cm. The letters and the triangle's direction do not change the method. Treating XY as a leg would give a side longer than 25 cm, which cannot be the missing leg.

Work with non-whole-number answers

For hypotenuse 10 inches and leg 7 inches, $b^2=100-49=51$. The exact length is $\sqrt{51}$ inches. Its decimal value is about 7.141, so to the nearest tenth the leg is 7.1 inches.

A hypotenuse of 6 units and a leg of 4 units give $b^2=36-16=20$. The exact answer $\sqrt{20}=2\sqrt{5}$ units is useful when another calculation follows. Use the full root or calculator value during that next step.

Catch subtraction errors

Subtract the known leg's square from the hypotenuse's square. Reversing that order would give a negative square. Check your side labels and subtraction if this happens.

Subtracting the lengths themselves also fails. In the 13 cm example, $13-5=8$ cm, but a triangle with legs 5 and 8 would have $5^2+8^2=89$, not 169. The theorem joins squared lengths, so the subtraction must happen between the squares.

Finally, compare the result with the hypotenuse. In the 10-inch example, adding squares would give $\sqrt{149}$, about 12.2 inches. That is too long for a leg. The correct result, about 7.1 inches, is shorter than 10 inches and fits the original equation before rounding.

A ladder, a ramp and a rectangular screen all contain useful right triangles. A sketch connects the measurements in the story to the sides of the triangle. This is the part that decides whether you add or subtract squares.

Label the distances

Draw the main shape and mark the right angle. Label each known length with its unit, and use a letter for the unknown. A vertical wall and level ground meet at a right angle. Adjacent sides of a rectangle also meet at a right angle. The triangle's hypotenuse runs across from that corner.

A 13-foot ladder rests against a vertical wall, with its foot 5 feet away on level ground. The ladder crosses from the ground to the wall, so its length is the hypotenuse. The height reached is a leg:

The ladder reaches 12 feet up the wall. Moving the foot to 4 feet away gives $h=\sqrt{169-16}=\sqrt{153}\approx 12.4$ feet. The top rises as the foot moves closer because the same ladder has a smaller horizontal distance to span.

Choose the sloping distance

A ramp rises 2 feet over a horizontal distance of 24 feet. The ramp surface is the hypotenuse:

The ramp is about 24.1 feet long to the nearest tenth. The horizontal distance and the sloping distance are close here because the rise is small, but they are different lengths.

For a screen 48 inches wide and 27 inches tall, the diagonal is $\sqrt{{48}^2+{27}^2}=\sqrt{3033}\approx 55.1$ inches. Screen size refers to this corner-to-corner diagonal. Width alone would leave out the height of the screen.

Distinguish a route from a direct distance

A hiker walks 3 km north and then 4 km east. The two parts of the route meet at a right angle. The route is 7 km long; the direct distance from the finish back to the start is $\sqrt{3^2+4^2}=5$ km.

The same idea finds the diagonal across an 80 m by 60 m rectangular field: $\sqrt{6400+3600}=100$ m. For a square with 90-foot sides, such as a baseball diamond, the corner-to-opposite-corner distance is $90\sqrt{2}\approx 127.3$ feet.

Before calculating, make all units match. A length of 5 feet 6 inches is 66 inches or 5.5 feet; it is not 5.6 feet, because a foot contains 12 inches. Give the final answer as the distance requested, with its unit and the requested rounding. Compare it with the sketch: a ladder's vertical reach is shorter than the ladder, while a rectangle's diagonal is longer than either side.

Some shapes hide the right triangle you need. Draw a perpendicular height or use a shared side. Once you find a missing length, it may become an input for an area, a perimeter, or a second triangle.

Split an isosceles triangle

An isosceles triangle has two 10 cm sides and a 12 cm base. The height from the top meets the base at its midpoint. Each right triangle therefore has hypotenuse 10 cm and a horizontal leg of 6 cm.

The height is $\sqrt{{10}^2-6^2}=\sqrt{64}=8$ cm. Now use the whole base in the area formula:

Half the base belongs in the right triangle; the full base belongs in the area of the original triangle. The sloping 10 cm side is not its perpendicular height.

An equilateral triangle works the same way. For side length 6 cm, the height is $\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt{3}$ cm. Its area is $\frac{1}{2}\cdot 6\cdot 3\sqrt{3}=9\sqrt{3}$ square centimeters, or about 15.6 square centimeters.

Find a trapezoid's height

An isosceles trapezoid has parallel sides of 10 cm and 16 cm, and equal slanted sides of 5 cm. Drop a height from each end of the short parallel side. The 6 cm difference between the bases is split equally at the two ends, giving horizontal legs of 3 cm.

The height is $\sqrt{5^2-3^2}=4$ cm. Its area is $\frac{1}{2}(10+16)\cdot 4=52{\text{ cm}}^2$. Using the full 6 cm difference as one leg would ignore the equal pieces at the two ends.

Use half of each diagonal

The diagonals of a rhombus cut each other in half at right angles. With diagonals of 10 cm and 24 cm, each small right triangle has legs 5 cm and 12 cm. Its hypotenuse is a rhombus side, $\sqrt{25+144}=13$ cm. All four sides are equal, so the perimeter is $4\cdot 13=52$ cm.

Carry a side into the next triangle

Suppose two right triangles share a side. In the first, legs of 9 and 12 give the shared side $\sqrt{81+144}=15$. In the second, that 15-unit side is a leg and the hypotenuse is 17. Its other leg is $\sqrt{289-225}=8$. The shared length stays 15, but its role changes between triangles.

A shared length may be an awkward root. If the first triangle has legs 4 and 7, the shared side is $\sqrt{65}$. If it becomes a leg in a second triangle with hypotenuse 10, then the remaining side is

You can carry the squared value 65 directly into the second equation. Rounding the shared side to 8 would change that square to 64 and give 6 for the final length. Keep the root, the square, or all calculator digits until the requested final answer.

The theorem starts with a known right angle and gives a side-length equation. The converse starts with the side lengths: if $a^2+b^2=c^2$, the triangle has a right angle opposite side $c$.

Compare the squares

Put the longest side in the $c$ position. For sides of 9, 12 and 15 units, $9^2+{12}^2=81+144=225$, and ${15}^2=225$. The equality shows that the triangle is right. Its right angle is between the 9-unit and 12-unit sides, opposite the 15-unit side.

The order in which lengths are listed does not matter. Sides written as 5, 13 and 12 still use 13 as the longest side. Comparing $5^2+{13}^2$ with ${12}^2$ would test the wrong arrangement.

Decimal lengths work too: 1.5, 2 and 2.5 give $2.25+4=6.25={2.5}^2$. These sides form a right triangle.

Tell acute and obtuse triangles apart

First, the lengths must form a triangle: the two shorter lengths must add to more than the longest. Then compare their squares with the longest side's square.

For sides 6, 7 and 9, $36+49=85>81$, so the triangle is acute. For sides 5, 6 and 8, $25+36=61<64$, so it is obtuse. Choosing the longest side lets you test the largest angle.

Why the converse is true

Take a triangle with lengths $a$, $b$ and $c$ satisfying $a^2+b^2=c^2$. Build a second triangle with legs $a$ and $b$ meeting at a right angle. Call its hypotenuse $x$. The theorem gives $x^2=a^2+b^2=c^2$, so the positive lengths $x$ and $c$ are equal.

The two triangles now have all three side lengths equal. By SSS congruence, three matching side lengths make triangles the same size and shape. Their matching angles are equal, so the first triangle also has a right angle opposite $c$.

Equality and measurements

Exact side lengths of 2, 3 and 3.6 give $4+9=13$ and ${3.6}^2=12.96$. These are close but unequal, so the exact triangle is acute. If the numbers came from rounded measurements, they describe a corner that may be very nearly right. A measurement's precision limits what you can conclude.

Builders use the same converse test: measure 3 units along one edge and 4 along the other. A 5-unit distance between the marks makes the corner right. Scaling to 6, 8 and 10 checks the same shape over a larger span.

A Pythagorean triple is three positive whole numbers satisfying $a^2+b^2=c^2$. They give the side lengths of a right triangle. Common triples include 3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29 and 9-40-41. Some countries call them Pythagorean triads.

Check the relationship

For 11-60-61, ${11}^2+{60}^2=121+3600=3721={61}^2$. For 12-35-37, $144+1225=1369={37}^2$. Both are triples. The numbers 6, 8 and 11 fail because $36+64=100$, while ${11}^2=121$.

Consecutive numbers do not generally work. The set 4-5-6 gives $16+25=41$, not 36. In fact, 3-4-5 is the only triple of three consecutive positive whole numbers.

Scale all three sides together

Multiplying every side of 3-4-5 by 3 gives 9-12-15. A common scale factor preserves the shape and the right angle. In algebra, if the original triple satisfies the rule, then

A positive whole-number multiplier gives another whole-number triple. A triple is primitive when all three numbers have no common factor greater than 1. The triple 8-15-17 is primitive. The triple 6-8-10 is not, because all three share a factor of 2.

Scaling can save time. A right triangle with hypotenuse 26 and leg 10 matches twice 5-12-13. The missing leg is therefore 24. Check: ${10}^2+{24}^2=100+576=676={26}^2$. Match the same scale factor across all three sides.

Generate a pattern

Choose an odd whole number $a>1$. Set $b=(a^2-1)\mathrm{/}2$ and $c=b+1$. Choosing $a=9$ gives $b=40$ and $c=41$. It works because

The difference between the last two numbers is 1, while the first comes from the square relationship. Simply adding 1 to every member of a known triple does not preserve it.

A positive decimal scale factor also preserves a right triangle. Multiplying 3-4-5 by 1.5 gives 4.5, 6 and 7.5. This is not a Pythagorean triple because its sides are not all whole numbers. Nor does every right triangle have whole-number lengths: two 1-unit legs give hypotenuse $\sqrt{2}$.

The distance between two points is the length of the straight segment joining them. On a grid, moving across and then up makes a right-angle route. The straight segment cuts across that route as the hypotenuse.

Find horizontal and vertical lengths

Points on the same horizontal line have the same vertical coordinate. From $(-2,3)$ to $(5,3)$, the horizontal distance is $5-(-2)=7$ units. Crossing the vertical axis covers 2 units on one side and 5 on the other. For a vertical segment, use the difference between the vertical coordinates.

For slanted segments, draw a horizontal line from one point and a vertical line from the other until they meet. These lines make the legs. Their meeting point uses one endpoint's horizontal coordinate and the other's vertical coordinate.

Work through a grid triangle

Let $A=(1,2)$ and $B=(7,10)$. Draw from A across to $(7,2)$, then up to B. The horizontal leg is $7-1=6$ units and the vertical leg is $10-2=8$ units. Therefore

The direct distance is 10 units. Adding 6 and 8 would give 14 units, the length of the two-part route along the legs.

Handle negative coordinates

For $P=(-3,4)$ and $Q=(2,-8)$, the horizontal leg is $2-(-3)=5$ units. The vertical leg is $4-(-8)=12$ units. Both are positive lengths even though some coordinates are negative.

Now $PQ=\sqrt{5^2+{12}^2}=\sqrt{169}=13$ units. Keep each subtraction together: moving from negative 8 to positive 4 covers 12 units, not 4. The minus sign in a coordinate tells you where the point is, not that a side has negative length.

Read the scale

Count the spaces between grid lines. From horizontal coordinate 1 to coordinate 7 there are six unit spaces; counting both endpoint dots would give one too many. Read the axis labels as well. One drawn square may represent more than one unit.

If map squares represent 1 mile, points 3 squares east and 4 squares north are 5 miles apart directly. You can calculate in grid units and then multiply by the map scale. Alternatively, convert both legs to real distances before using the theorem. Use the same scale for both directions, or convert each direction with its own labeled scale.

Some distances stay as roots. From $(0,0)$ to $(2,3)$, the answer is $\sqrt{13}$ units, about 3.61 units. The root is the exact distance. Use the requested precision when turning it into a decimal.

The distance formula records the grid-triangle calculation in one line. For endpoints $(x_1,y_1)$ and $(x_2,y_2)$,

The leg lengths are the absolute differences of the coordinates. Squaring a signed difference gives the same result as squaring its positive length, so the formula works without absolute-value signs.

Keep matching coordinates together

For $(-1,5)$ and $(4,-7)$, the horizontal difference is $4-(-1)=5$. The vertical difference is $-7-5=-12$. Substitute:

Put a negative difference in parentheses before squaring. Many calculators read $-{12}^2$ as $-({12}^2)=-144$. The square you need is $(-12{)}^2=144$. Also match horizontal with horizontal and vertical with vertical; subtracting one point's horizontal coordinate from a vertical coordinate gives no side of this grid triangle.

Reversing the points changes 5 and −12 to −5 and 12. Their squares stay the same, so the distance stays 13 units.

Find a perimeter

Take $A=(0,0)$, $B=(4,0)$ and $C=(4,3)$. Two sides lie along grid lines: AB is 4 units and BC is 3 units. The slanted side is $AC=\sqrt{4^2+3^2}=5$ units. Add the actual side lengths to get the perimeter, $4+3+5=12$ units. Adding their squares would not give a perimeter.

Test a right angle without roots

For $P=(1,1)$, $Q=(5,3)$ and $R=(3,7)$, calculate the squared side lengths:

• $P{Q}^2=4^2+2^2=20$.
• $Q{R}^2=(-2{)}^2+4^2=20$.
• $P{R}^2=2^2+6^2=40$.

Since $20+20=40$, the converse shows a right angle. PR is the longest side, so the right angle is at Q, opposite PR. Equal squared lengths also show $PQ=QR$, making the triangle isosceles. No square roots were needed for either conclusion.

Compare distances efficiently

From the origin, point $(5,1)$ has squared distance 26, while $(4,3)$ has squared distance 25. The second point is closer. Positive numbers and their square roots have the same order.

Different points can also have the same distance. Points $(3,4)$, $(-3,4)$, $(4,-3)$ and $(0,-5)$ all have $x^2+y^2=25$. Each is exactly 5 units from the origin, despite being in a different position.

A rectangular prism is a box shape with rectangular faces. A face diagonal joins opposite corners of one face. A space diagonal joins opposite corners of the whole box and passes through its interior. They are different lengths.

Use two right triangles

Let the box have length $l$, width $w$ and height $h$. First draw the diagonal across its rectangular base. Call its length $f$. The base edges are perpendicular, so $f^2=l^2+w^2$.

Now draw the vertical edge at one end of that diagonal. It is perpendicular to the base, including the diagonal in the base. This edge, the base diagonal and the space diagonal make a second right triangle. If the space diagonal is $d$, then $d^2=f^2+h^2$.

For a box measuring 3 inches by 4 inches by 12 inches, the base diagonal is $\sqrt{9+16}=5$ inches. The space diagonal is $\sqrt{5^2+{12}^2}=\sqrt{169}=13$ inches. Stopping at 5 inches would give a face diagonal and leave out the box's height.

Combine the two steps

Substitute $f^2=l^2+w^2$ into the second equation:

This formula uses all three perpendicular box dimensions. You can keep the squared base diagonal in the calculation without finding its decimal value. That avoids early rounding.

For a cube, all three edges have length $s$, so $d=\sqrt{3s^2}=s\sqrt{3}$. A cube with 2 cm edges has space diagonal $2\sqrt{3}\approx 3.46$ cm. Its face diagonal is only $2\sqrt{2}$ cm.

Decide whether a straight object fits

The space diagonal is the longest straight segment inside a rectangular box. A 30-inch umbrella cannot fit in a box whose inside dimensions are 20 inches by 15 inches by 12 inches: the diagonal is $\sqrt{400+225+144}=\sqrt{769}\approx 27.7$ inches. Even this longest span is shorter than the umbrella.

Work backward for an edge

A box has a 6 cm by 8 cm base and space diagonal 26 cm. Its base diagonal has square $6^2+8^2=100$. The height is therefore

The space diagonal remains the hypotenuse, even when you are finding an edge. Adding instead of subtracting here would produce an edge longer than the diagonal.

The same box idea measures a segment between points in space. From $(0,0,0)$ to $(2,3,6)$, three perpendicular changes give distance $\sqrt{2^2+3^2+6^2}=7$ units. In a perspective sketch, the right angles may look slanted; use the box's geometry to identify them.

Name the endpoints of the length you want. Then draw a flat triangle containing those endpoints and the vertical height. The point where the height meets the base decides how much of the base belongs in the triangle.

Cross a cylinder from edge to opposite edge

In a right circular cylinder, the longest thin straight rod runs from the rim of the bottom to the opposite rim of the top. A vertical slice through the center makes a rectangle. Its width is the whole diameter $D$, and its height is $h$. The rod follows that rectangle's diagonal:

A cylinder 6 cm across and 8 cm tall has longest segment $\sqrt{36+64}=10$ cm. A 12 cm straw aligned along that diagonal extends 2 cm past the top rim. The diameter is needed because the line crosses the entire base width.

Go from a cone's tip to its rim

In a right circular cone, the tip is directly above the base's center. The height meets the base at a right angle. From the center to the rim is one radius $r$, so the slant height $\mathrm{\ell }$ satisfies

Radius 5 cm and height 12 cm give slant height $\sqrt{25+144}=13$ cm. If the base diameter is 10 cm, halve it first to get that 5 cm radius. Using the whole diameter would make the slant height too large.

You can work backward too. Slant height 10 cm and radius 6 cm give vertical height $h=\sqrt{100-36}=8$ cm. The slant height is the hypotenuse; the vertical height is a leg.

Find the height along a pyramid's face

Take a square pyramid with base side 10 cm and vertical height 12 cm. Its tip is directly above the base center. The face slant height runs from the tip to the midpoint of a base edge. From the center to that midpoint is half a base side, 5 cm.

Thus the face slant height is $\sqrt{{12}^2+5^2}=13$ cm. This is also the perpendicular height of the triangular face, measured along its surface.

Find the edge to a corner

A sloping edge ends at a base corner, farther from the center than an edge midpoint. The base diagonal is $\sqrt{{10}^2+{10}^2}=10\sqrt{2}$ cm, so center to corner is half of it, $5\sqrt{2}$ cm.

Use this longer horizontal leg with the same 12 cm height:

The edge is longer than the 13 cm face slant height. The two lengths start at the same tip but end at different points. A separate triangle sketch makes that difference visible before you calculate.