The Baudhayana-Pythagoras Theorem | CBSE Class 8 Maths Notes

This note covers doubling and halving squares, right-angled triangles, square roots and decimal bounds, the Baudhayana-Pythagoras theorem, finding missing sides, integer triples, Fermat’s Last Theorem, and applications to squares and the depth of a lake.

How can a square with double the area be constructed?

A square has four equal sides and four right angles. A right angle measures 90 degrees, written 90°. The area measures the surface enclosed by a shape. For a square, its area is its side length multiplied by itself.

A diagonal joins two opposite vertices, or corners, of a square. Drawing a new square with this diagonal as its side produces a square with twice the original area. This construction connects a length inside the first square with the area of a second square.

Result: The square on a diagonal has double the area

Doubling the original side does something different: the area becomes four times as large. Each of the two dimensions has doubled, so the area multiplier is 2 × 2 = 4. Here, × means multiplication, and = means equality.

The diagonal construction works because the two squares can be compared using congruent triangles, which have the same shape and size. The original square contains two of the small triangles. The square on its diagonal contains four such triangles.

A dotted, tilted square is drawn on a diagonal of the original square. Horizontal and vertical lines divide the construction into matching triangular regions.

Reference: NCERT Class 8, page 34, unnumbered diagram

Perpendicular lines meet at a right angle. The horizontal and vertical lines help expose the equal triangular pieces; counting those pieces explains the doubling without first calculating the diagonal’s length.

Repeating the construction produces a sequence of squares containing 2, 4 and 8 matching small triangles. Each square in this sequence has twice the area of the preceding square. The side length changes too, but it does not double at each step.

A paper activity uses two identical squares. Cut one into four equal triangular pieces and place them around the other square. The resulting larger square contains the area of both original squares, making the area relationship visible.

How can a square with half the area be constructed?

Halving a square’s area reverses the doubling construction. Draw a smaller, tilted square inside the original square, with its corners at the midpoints of the original sides. A midpoint divides a line segment into two equal lengths.

In the paper-folding version, the crease lines pass through these midpoints. Folding the corners inward leaves the required smaller square. The change concerns the enclosed area, so it must be checked by comparing regions rather than by looking only at one side.

The yellow paper square is folded inward. Its central tilted square has corners labelled P, Q, R and S, with R at the top, P below, S left and Q right.

Reference: NCERT Class 8, page 36, unnumbered diagram

The letters P, Q, R and S name the four corners of the inner square. Connecting opposite corners gives the segments PR and QS. These divide the inner square into triangular parts that can be compared with the regions outside it.

Why does halving the side give a different result?

A square whose side is half the original side occupies one quarter of the original area. Four such squares fill the original square. Therefore, taking half the side is not the construction needed when the aim is to take half the area.

The tilted construction uses equal triangular regions to show that the inner square and the parts outside it occupy equal areas. The inner square therefore occupies half the whole square. This is the same area comparison as doubling, read in the reverse direction.

Both constructions show why side length and area must be distinguished. A diagonal can be used as a new side, while congruent triangles establish how much area the resulting square encloses. An area comparison does not require the new side to be a whole number.

Why does an isosceles right triangle lead to a square root?

A right-angled triangle has a right angle. Its two sides meeting at that angle are its perpendicular sides. The side opposite that angle is the hypotenuse. An isosceles right triangle has two equal perpendicular sides.

Take an isosceles right triangle whose equal sides each measure 1 unit. Two copies form a square of side 1 unit. Its diagonal is the hypotenuse of either triangle. A square constructed on this diagonal has twice the area of the original square.

Let c represent the length of the hypotenuse. The area of the square on it is c × c, written c² and read “c squared”. Since the original square has area 1 square unit, the new square has area 2 square units.

Definition: For a positive number, its positive square root is the positive number whose square equals it. The symbol √ denotes this square root; √2 is the positive number whose square is 2.

Thus c² = 2, giving c = √2. The answer is a length of √2 units. By contrast, 2 square units describes the area of the square on that length. The squared length and the length itself answer different questions.

Result: Equal perpendicular sides give c² = 2a²

Let a be the common length of the two perpendicular sides, while c remains the hypotenuse length. The expression 2a² means twice the square of a. The same construction gives c² = 2a². Equivalently, c = a√2, where a√2 means a multiplied by √2.

Worked example 1. Find the hypotenuse when each equal perpendicular side is 12 units.

Answer: c² = 2 × 12² = 288, so c = √288 units. Since 16² = 256 and 17² = 289, the hypotenuse lies between 16 and 17 units.

Worked example 2. An isosceles right triangle has hypotenuse √72 units. Find its equal sides.

Answer: 72 = 2a², so a² = 36 and a = 6. Each equal perpendicular side measures 6 units.

The formula can therefore be used in either direction. When the equal side is known, double its square. When the hypotenuse is known, halve its square. In both cases, take the positive square root to obtain the required side length.

How can the value of √2 be bounded and understood?

A lower bound is a value below the number being considered, while an upper bound is a value above it. Here, the symbol < means “is less than”. Since 1² = 1 and 2² = 4, we have 1 < √2 < 2.

To obtain closer bounds, square decimal numbers between 1 and 2. For positive lengths, a larger square has a larger side. Consequently, squares just below and just above 2 identify lengths that enclose √2 more closely.

Worked example 3. Use 1.414² = 1.999396 and 1.415² = 2.002225 to bound √2.

Answer: The first square is below 2 and the second is above 2. Therefore, 1.414 < √2 < 1.415. Neither endpoint is exactly √2.

Why is √2 neither a terminating decimal nor a fraction?

A terminating decimal ends after a finite number of decimal places. The decimal expansion of √2 does not terminate: √2 = 1.41421356…, where … indicates that more digits follow. A finite decimal approximation must not be confused with this exact number.

If a terminating decimal between 1 and 2 is written without trailing zeros, it has a final non-zero decimal digit. Its square also has a non-zero final decimal digit. Its square therefore cannot equal the integer 2 exactly.

Positive integers are the counting numbers starting from 1. Now suppose √2 = m/n, where m and n are positive counting numbers and m/n means m divided by n. Squaring gives 2n² = m². In the prime factorisation of a number, that number is written as a product of prime numbers.

A prime number is an integer greater than 1 with no positive factors other than 1 and itself. A factor divides a number without leaving a remainder. Each prime occurs an even number of times in a square’s prime factorisation. Thus 2 occurs evenly in m² but oddly in 2n², an impossibility.

Therefore √2 cannot be expressed as a fraction of two positive integers. This conclusion is stronger than showing that its decimal does not terminate. The prime-factor argument rules out the fraction representation directly.

How do two different squares lead to the general theorem?

Doubling joins the areas of two equal squares. The more general problem asks for a square whose area equals the sum of the areas of two different-sized squares. To solve it, use their side lengths as the perpendicular sides of a right triangle.

For this general triangle, let a and b denote the perpendicular side lengths, and let c denote the hypotenuse length. All lengths use the same unit. The squares on these sides have areas a², b² and c² respectively.

Theorem: The Baudhayana-Pythagoras relationship

Definition: For a right-angled triangle with perpendicular sides a and b and hypotenuse c, the Baudhayana-Pythagoras theorem gives a² + b² = c². The symbol + means addition.

A theorem is a mathematical statement established by reasoning. The sum of the areas on the perpendicular sides equals the area on the hypotenuse. If a and b are equal, this becomes a² + a² = c², or 2a² = c², exactly the earlier isosceles-right-triangle relationship.

A right triangle has sides labelled a, b and c. A red square and a yellow square are drawn on the perpendicular sides; rearranged coloured pieces cover the square on the hypotenuse.

Reference: NCERT Class 8, page 46, unnumbered diagram

How does rearrangement explain the area equality?

The construction begins by joining the two smaller squares. A rectangular portion of the larger one is marked using the smaller square’s side. The rectangle’s diagonal gives the hypotenuse of a right triangle with perpendicular sides a and b.

Copies of this triangle create a four-sided figure on the hypotenuse. Equal hypotenuses make its sides equal. The angle relationships show that its corners are right angles, so the resulting figure is a square rather than merely a shape with four equal sides.

The small right triangles in the arrangement are congruent. Exchanging matching triangles preserves the total area. This rearrangement turns the regions of the original two squares into the square on the hypotenuse, establishing the theorem through areas.

The theorem is also known as the Pythagorean theorem. It is often called the Baudhayana-Pythagoras theorem. Its central condition is that the triangle is right-angled: the equation is a relationship between those perpendicular sides and their opposite side.

How are unknown sides calculated using the theorem?

Begin by locating the right angle. The side opposite it is c, the hypotenuse. Either perpendicular side can be named a or b. The labels are a way of recording the geometry; assigning c to the wrong side changes the equation incorrectly.

Which calculation should be used?

1. Identify the hypotenuse and the two perpendicular sides in the given right triangle.
2. Write a² + b² = c² before inserting the known lengths.
3. Add the two squares when finding c; subtract the known perpendicular side’s square when finding the other perpendicular side.
4. Take the positive square root and give the result in the stated length unit.

The symbol − means subtraction. The unit cm means centimetres. A length expressed in cm has its square expressed in square centimetres, written cm². These squared values must return to length units after the square root is taken.

Worked example 4. A right triangle has perpendicular sides 3 cm and 4 cm. Find its hypotenuse.

Answer: c² = 3² + 4² = 9 + 16 = 25. Thus c = 5 cm. A drawing should measure about 5 cm; the theorem gives the exact length.

Worked example 5. A right triangle has perpendicular sides 5 cm and 12 cm. Find its hypotenuse.

Answer: c² = 5² + 12² = 25 + 144 = 169. Therefore c = 13 cm.

Worked example 6. A right triangle has hypotenuse 17 cm and one perpendicular side 8 cm. Find the remaining side, b.

Answer: 8² + b² = 17², so b² = 289 − 64 = 225. Thus b = 15 cm.

The last example uses subtraction because the square on the hypotenuse already includes both smaller-square areas. Removing one leaves the other. Adding the squares of 17 and 8 would answer a different geometric question and would not find this triangle’s missing side.

Worked example 7. A right triangle has perpendicular sides 5 units and 7 units. Find c.

Answer: c² = 5² + 7² = 25 + 49 = 74. Hence c = √74 units. The square-root form gives the exact length.

The hypotenuse is the longest side: c² exceeds each of a² and b² because the other squared side is positive. Since all side lengths are positive, c is greater than either perpendicular side. This gives a useful check on a calculated answer.

What are Baudhayana triples and primitive triples?

A Baudhayana triple is a set of three positive integers, written (a, b, c), satisfying a² + b² = c². The brackets group the three numbers, with c in the hypotenuse position. Positive integers are the whole numbers starting from 1.

These are also called Baudhayana-Pythagoras triples, Pythagorean triples or right-angled triangle triples. They give right triangles whose three side lengths are integers. Examples include (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (12, 35, 37) and (15, 36, 39).

Do not judge a triple merely by the appearance of its numbers. Its defining test concerns their squares. The order should make clear which number represents the hypotenuse, and the square of that number must equal the sum of the other two squares.

How does scaling produce more triples?

Let k be any positive integer. Multiplying every entry of a triple by k produces its scaled version, written (ka, kb, kc). Here ka, kb and kc mean k multiplied by a, b and c respectively.

If a² + b² = c², then (ka)² + (kb)² = k²(a² + b²) = k²c² = (kc)². The common multiplier therefore preserves the relationship. Because k can keep increasing, there are infinitely many triples.

A common factor divides each number without a remainder. A primitive triple has no common factor greater than 1. Thus (3, 4, 5) is primitive, whereas (9, 12, 15) is not: all three entries in the latter share the factor 3.

Dividing (9, 12, 15) by its common factor 3 returns (3, 4, 5). More generally, reducing a non-primitive triple by the greatest factor common to all three entries gives a primitive triple. Scaling that primitive triple recovers the original one.

How can odd squares be used to generate triples?

Consecutive odd numbers are neighbouring terms in the sequence 1, 3, 5 and so on. Their sums connect directly with squares: 1 = 1², 1 + 3 = 2², and 1 + 3 + 5 = 3².

Let n denote the positive integer counting how many odd numbers are added. The nth odd number is 2n − 1, meaning twice n minus 1. The sum of the first n odd numbers is n².

Identity: (n − 1)² + (2n − 1) = n²

An identity is an equality that holds throughout the stated range of its variable, a symbol representing a value that can vary. Here the first n − 1 odd numbers total (n − 1)². Adding the next odd number, 2n − 1, gives n².

If that final odd number is itself a square greater than 1, the identity becomes a sum of two positive squares equal to a third square. Their positive square roots provide the entries of a Baudhayana triple. The method translates a number pattern into side lengths.

Worked example 8. Generate a triple using the odd square 9, the fifth odd number.

Answer: Set n = 5. Then (5 − 1)² + 9 = 5², giving 4² + 3² = 5². Thus (3, 4, 5) is a Baudhayana triple.

Worked example 9. Generate a triple using the odd square 25, the thirteenth odd number.

Answer: Set n = 13. Then (13 − 1)² + 25 = 13², giving 12² + 5² = 13². Thus (5, 12, 13) is a Baudhayana triple.

One perpendicular side obtained this way is one less than the hypotenuse. Those two consecutive integers cannot have a common factor greater than 1. Consequently, the three entries cannot share such a factor, and the triple is primitive.

This method does not produce every primitive triple. For example, (8, 15, 17) is primitive, but neither perpendicular side is one less than 17. Generating some primitive triples and generating all primitive triples are different claims.

How does Fermat’s Last Theorem extend the question?

Baudhayana triples show that a square of a positive integer can equal the sum of two other such squares. Fermat considered whether the same kind of equality could hold for higher powers, including cubes and fourth powers.

A power records repeated multiplication: the exponent tells how many copies of the base are multiplied. A cube is the third power, while a fourth power uses four copies. Changing the exponent changes the question about which integer sums are possible.

What does the higher-power statement say?

For this discussion, let x, y and z be positive integers, and let n be the integer exponent. The equation is written xⁿ + yⁿ = zⁿ, where the raised n means “to the power n”. The symbol > means “is greater than”.

Fermat’s Last Theorem states that this equation has no solution in positive integers when n > 2. The qualification “positive integers” is part of the statement, as is the condition on the exponent. Neither qualification should be omitted.

Fermat lived during the 17th century. He wrote that he had found a proof but lacked enough space in the book’s margin to include it. No one could ever find Fermat’s proof, and more than 300 years of failed attempts followed.

In 1963, the 10-year-old Andrew Wiles read about the problem and resolved to prove it. He eventually proved the theorem in 1994. The contrast with square powers remains central: there are infinitely many Baudhayana triples, but the higher-power equation has no positive-integer solutions.

This extension concerns arithmetic properties of positive integers. It does not change how the right-triangle theorem is used: when the geometry supplies a right angle, the relationship among the three side lengths still involves their squares.

How does the theorem solve square and lotus problems?

An application begins by finding the right triangle within the situation. Some shapes contain one immediately, such as the triangle formed by a square’s diagonal. In other problems, an assumption about the geometry must be stated before the equation can be used.

Worked example 10. Find the diagonal of a square whose side measures 5 cm.

Answer: Let d be the diagonal length. The square’s two adjacent sides meet at a right angle, so d² = 5² + 5² = 50. Therefore d = √50 cm, equivalently 5√2 cm.

The letter d in this example names the diagonal. It plays the role of the hypotenuse because it lies opposite the right angle between the adjacent sides. Identifying that role explains why the two equal squared side lengths are added.

What assumption is needed in the lotus problem?

In the problem from Bhaskaracharya’s Lilavati, a lotus stem extends 1 unit above the water. When swayed, its tip touches the water 3 units from its original position. We make the very reasonable assumption that the stem is initially perpendicular to the water’s surface.

Let x now denote the submerged stem length, which is the lake’s depth. The entire stem has length x + 1. Its original submerged position, the horizontal displacement and the slanted stem form a right triangle with sides x, 3 and x + 1.

One lotus is shown upright above the water and another touches the water to its left. Vertical and slanting segments meet at the same point below the surface; the exposed height is marked 1.

Reference: NCERT Class 8, page 52, unnumbered illustration

Worked example 11. A lotus extends 1 unit above water and touches the surface 3 units away when swayed. Assume its initial position is perpendicular to the surface and use its unchanged stem length. Find the depth.

Answer: With depth x, the hypotenuse is x + 1. Thus 3² + x² = (x + 1)². Expanding gives 9 + x² = x² + 2x + 1. Subtracting x² gives 9 = 2x + 1, so x = 4 units.

The length x + 1 is not an extra numerical measurement. It follows from the same stem being 1 unit longer than the submerged portion in the upright position. Recording this relationship supplies the link that makes the depth calculable.

The perpendicular assumption supplies the right angle. Keeping both that assumption and the relationship between depth and total length visible makes the solution complete: the equation follows from a clearly described triangle, and the answer represents the lake’s depth.

Glossary

• Diagonal — A line segment joining opposite vertices of a square or another quadrilateral.
• Congruent triangles — Triangles with the same shape and size, so their corresponding sides and angles match.
• Perpendicular lines — Lines meeting at a right angle, which measures ninety degrees.
• Hypotenuse — The side opposite the right angle in a right-angled triangle.
• Isosceles right triangle — A right-angled triangle whose two perpendicular sides have equal lengths.
• Positive square root — The positive number which, multiplied by itself, gives the specified positive number.
• Lower bound — A number below the value being studied, used to restrict its possible location.
• Upper bound — A number above the value being studied, used to restrict its possible location.
• Terminating decimal — A decimal representation that ends after a finite number of decimal places.
• Prime factorisation — The expression of a positive integer greater than one as a product of primes.
• Baudhayana triple — Three positive integers satisfying the square relationship for the sides of a right triangle.
• Scaled triple — A triple obtained by multiplying each entry of another triple by the same positive integer.
• Primitive triple — A Baudhayana triple whose three entries have no common factor greater than one.
• Identity — An equality that holds for every value in the stated range of its variable.

Common errors and misconceptions

• Misconception: Doubling a square’s side doubles its area. Correct: Doubling both dimensions makes its area four times as large. A square constructed on its diagonal has twice the area.
• Misconception: Halving a square’s side halves its area. Correct: It gives one quarter of the area. Joining the midpoints of the original sides gives the required square of half the area.
• Misconception: The hypotenuse is whichever side looks slanted. Correct: It is identified by its position opposite the right angle. The orientation of a drawing does not determine its name.
• Misconception: The equation c² = 25 means the side is 25 units. Correct: It means the squared length is 25. The positive side length is 5 units.
• Misconception: √2 equals 1.414 exactly. Correct: The square of 1.414 is 1.999396. The exact number lies between 1.414 and 1.415 and has a non-terminating decimal representation.
• Misconception: Every triple is primitive. Correct: A triple with a common factor greater than 1 is non-primitive. For example, (9, 12, 15) is obtained by scaling (3, 4, 5).
• Misconception: The two known squared lengths should be added in every problem. Correct: When the hypotenuse and one perpendicular side are known, subtract the latter’s square from the hypotenuse’s square.

Exam-style questions with model answers

Q1. State the Baudhayana-Pythagoras theorem and define the letters in your equation. [2 marks]

1. For a right-angled triangle, the theorem states that a² + b² = c².
2. Here a and b are the perpendicular side lengths, and c is the hypotenuse length, opposite the right angle.

Q2. An isosceles right triangle has hypotenuse √72 units. Find the length of each equal perpendicular side, showing your method. [3 marks]

1. Let a be the length of each equal perpendicular side. Because the triangle is right-angled and these sides are equal, the theorem gives a² + a² = (√72)².
2. Combining the equal terms gives 2a² = 72. Dividing both sides by 2 gives a² = 36.
3. Taking the positive square root gives a = 6. Therefore each of the two equal perpendicular sides measures 6 units.

Q3. A right triangle has perpendicular sides 5 cm and 12 cm. Calculate its hypotenuse, showing the equation, substitution and final length. [3 marks]

1. Let c be the hypotenuse length in centimetres. Since the given sides are perpendicular, their squared lengths add to the square of the hypotenuse: c² = 5² + 12².
2. Evaluating the squares gives c² = 25 + 144 = 169. This is the squared length, so one further step is needed.
3. The positive square root of 169 is 13. Thus the hypotenuse measures 13 cm.

Q4. A right triangle has hypotenuse 17 cm and one perpendicular side 8 cm. Find the other side and explain why subtraction is needed. [4 marks]

1. Let b denote the missing perpendicular side length in centimetres. The hypotenuse is the side of length 17 cm, so the theorem gives 8² + b² = 17².
2. The square on the hypotenuse equals the sum of the two smaller-square areas. Removing the known smaller-square area leaves b², so subtraction is needed.
3. Hence b² = 17² − 8² = 289 − 64 = 225.
4. Taking the positive square root gives b = 15. The missing side is therefore 15 cm, which is shorter than the hypotenuse.

Q5. Given 1.414² = 1.999396 and 1.415² = 2.002225, find bounds for √2. Explain why neither endpoint equals √2. [3 marks]

1. The number √2 is positive and has square 2. Since the given value 1.999396 is below 2, its positive square root, 1.414, is below √2.
2. The given value 2.002225 is above 2, so its positive square root, 1.415, is above √2.
3. Therefore 1.414 < √2 < 1.415. Neither endpoint equals √2 because neither of its stated squares is exactly 2.

Q6. Let (a, b, c) be a Baudhayana triple, so a² + b² = c², and let k be any positive integer. Prove that (ka, kb, kc) is also a triple. Use (3, 4, 5) to explain why infinitely many triples exist. [5 marks]

1. The entries a, b and c are positive integers. Multiplying them by the positive integer k gives positive integers ka, kb and kc, as required for a new triple.
2. Squaring the first two new entries gives (ka)² + (kb)² = k²a² + k²b².
3. Taking out the common factor k² gives k²(a² + b²). The given relationship a² + b² = c² makes this expression k²c².
4. Since k²c² = (kc)², the new entries satisfy (ka)² + (kb)² = (kc)². Thus they form a Baudhayana triple.
5. In particular, (3k, 4k, 5k) is a triple for every positive integer k. Increasing k gives distinct triples without end, proving that infinitely many triples exist.

Q7. A lotus stem initially stands perpendicular to the water’s surface, with its tip 1 unit above the water. Keeping the same base and length, its tip touches the surface 3 units horizontally from its original position. Find the lake’s depth. [5 marks]

1. Let x denote the depth of the lake in units. Because the upright stem extends 1 unit above the water, its total length is x + 1 units.
2. The vertical depth and horizontal displacement are perpendicular. Together with the slanted stem, they form a right triangle with perpendicular sides x and 3, and hypotenuse x + 1.
3. Apply the theorem to obtain x² + 3² = (x + 1)². Expanding the square gives x² + 9 = x² + 2x + 1.
4. Subtract x² from both sides, giving 9 = 2x + 1. Then subtract 1 to obtain 8 = 2x.
5. Dividing by 2 gives x = 4. Since x was defined as the submerged vertical length, the lake’s depth is 4 units.

Key takeaways

• A square constructed on the diagonal of another square has twice the area of the original square.
• The square joining the midpoints of a square’s sides occupies half the area of the original square.
• The hypotenuse lies opposite the right angle; identify it before substituting side lengths into the theorem.
• For perpendicular side lengths a and b and hypotenuse c, the central relationship is a² + b² = c².
• For an isosceles right triangle, the equal perpendicular sides give c² = 2a² and c = a√2.
• The number √2 lies between 1.414 and 1.415, and cannot be expressed as a terminating decimal or a fraction of positive integers.
• Multiplying every entry of a Baudhayana triple by the same positive integer produces another triple.
• A primitive triple has no common factor greater than 1; a non-primitive triple can be reduced to a primitive one.
• In application problems, identify the right angle and explain what each side length represents before forming the equation.

Test yourself

What happens to a square’s area if its side is doubled?

Its area becomes four times as large, because both dimensions are multiplied by 2.

What is the hypotenuse of an isosceles right triangle with equal sides of 1 unit?

It is √2 units, because its squared length is 1² + 1² = 2.

A right triangle has perpendicular sides 3 cm and 4 cm. What is its hypotenuse?

The hypotenuse is 5 cm, since 3² + 4² = 9 + 16 = 25.

Why is (9, 12, 15) a non-primitive triple?

All three entries share the factor 3. Dividing by it gives the primitive triple (3, 4, 5).

What triple comes from the odd square 25 using the odd-number-sum identity?

Since 25 is the thirteenth odd number, 12² + 5² = 13² gives (5, 12, 13).

A square has side length 5 cm. What is its diagonal length?

The diagonal is √50 cm, or 5√2 cm, because its square is 5² + 5².

What does Fermat’s Last Theorem say about positive integer powers greater than 2?

No positive integer raised to such a power equals the sum of two positive integers raised to that same power.

Why is the upright-stem assumption needed in the lotus problem?

Assuming the stem is perpendicular to the water supplies the right angle needed to apply the theorem.