The Baudhayana-Pythagoras Theorem Class 8: Doubling Squares, √2, Proof, Triples and Applications

What The Baudhayana-Pythagoras Theorem is about

Key concepts

Doubling and halving a square

NCERT section 2.1 Doubling a Square

Doubling the side of a square makes its area four times bigger, not two. Baudhayana's answer (Sulba-Sutra 1.9): the square built on the diagonal of a square has double its area.

*Why:* draw the diagonal; the original square is two congruent right isosceles triangles. Extending the sides of the original square (the east-west and north-south lines) cuts the square on the diagonal into four of the same triangles. Two triangles versus four: double the area. Repeating gives squares of 2, 4, 8, ... triangles, each double the last.

Halving a square reverses this: join the midpoints of the sides of a square to get a tilted square inside it. The outer square is 8 small triangles and the inner one is 4, so the inner square has half the area. With paper, fold the four corners to the centre: the creases form the half-area square.

Note: a square with half the side has a quarter of the area, not half.

Worked example

A square garden has side 11 m. What is the area of the square built on its diagonal? Between which two numbers with one decimal place does the diagonal lie?

Try it first, then show the solutionSolution

1. Area of the garden $=121$ m², so the square on the diagonal has area $242$ m².
2. The diagonal $d$ satisfies $d^2=242$.
3. $15.5^2=240.25<242<243.36=15.6^2$, so $15.5<d<15.6$.

Answer: 242 m²; the diagonal lies between 15.5 m and 15.6 m.

The isosceles right triangle and √2

NCERT section 2.3 Hypotenuse of an Isosceles Right Triangle

A unit square has area 1, so the square on its diagonal has area 2. If the diagonal (the hypotenuse of a right isosceles triangle with equal sides 1) is $c$, then $c^2=2$ and $c=\sqrt{2}$.

General rule: if the equal sides are $a$ and the hypotenuse is $c$, then $c^2=2a^2$, so $c=a\sqrt{2}$. For $a=12$, $c=\sqrt{288}$, between 16 and 17; for $c=\sqrt{72}$, $a^2=36$ and $a=6$.

**How big is $\sqrt{2}$?** Square trial numbers:

• $1^2=1<2<4=2^2$, so $1<\sqrt{2}<2$ (a lower and an upper bound);
• $1.4^2=1.96$ and $1.5^2=2.25$, so $1.4<\sqrt{2}<1.5$;
• $1.41^2=1.9881$, $1.42^2=2.0164$; then $1.414<\sqrt{2}<1.415$, and so on: $\sqrt{2}=1.41421356\dots$

It never ends: a terminating decimal with last digit $d\ne 0$ has a square whose last digit is the last digit of $d^2$, never 0, so the square cannot be exactly 2.000...

It is not a fraction: if $\sqrt{2}=\frac{m}{n}$, then $2n^2=m^2$. In a square every prime appears an even number of times, so the prime 2 appears an odd number of times on the left and an even number on the right: impossible. (Euclid's proof, about 300 BCE.)

Worked example

The equal sides of an isosceles right triangle are 7 cm. Find the hypotenuse as a square root and give bounds with one decimal place.

Try it first, then show the solutionSolution

1. $c^2=2\times 7^2=98$, so $c=\sqrt{98}$.
2. $9^2=81$ and $10^2=100$, so $9<c<10$.
3. $9.8^2=96.04$ and $9.9^2=98.01$, so $9.8<c<9.9$.

Answer: $c=\sqrt{98}$ cm, with $9.8<c<9.9$.

Baudhayana's Theorem: a² + b² = c²

NCERT section 2.4 Combining Two Different Squares

Theorem: if a right-angled triangle has shorter sides $a$, $b$ and hypotenuse $c$ (the side opposite the right angle), then $a^2+b^2=c^2$. Baudhayana (Sulba-Sutra 1.12) stated it about 800 BCE, a couple of centuries before Pythagoras (about 500 BCE), so it is called the Baudhayana-Pythagoras Theorem.

Baudhayana's proof idea (Sulba-Sutra 2.1): place the squares of sides $a$ and $b$ side by side. Mark off a rectangle of sides $a$ and $b$ inside the larger square and draw its diagonal: this gives a right triangle with legs $a$, $b$. Build three more copies of this triangle around the hypotenuse to form a four-sided figure. All four triangles are congruent, so its sides are equal; at each corner the two acute angles $x$ and $90°-x$ add to 90°, so it is a square of side $c$. Rearranging shows its area equals the two original squares: $c^2=a^2+b^2$. You can show this with paper by cutting the two joined squares into three pieces and rearranging them.

Using it: $c=\sqrt{a^2+b^2}$ for the hypotenuse, and $a=\sqrt{c^2-b^2}$ for a shorter side. Since $c^2$ is bigger than $a^2$ and $b^2$, the hypotenuse is always the longest side.

Constructing areas: to make a square of 3 times a given area, combine a square of area 2 (on the diagonal) with one of area 1.

Worked example

Find the hypotenuse of a right triangle with legs 20 cm and 21 cm, and the third side of a right triangle with one leg 33 cm and hypotenuse 65 cm.

Try it first, then show the solutionSolution

1. $c^2=20^2+21^2=400+441=841$, so $c=29$ cm.
2. $a^2=65^2-33^2=4225-1089=3136$.
3. $a=56$ cm.

Answer: 29 cm and 56 cm.

Baudhayana triples

NCERT section 2.5 Right–Triangles Having Integer Sidelengths

A Baudhayana (Pythagorean) triple is three positive integers $(a,b,c)$ with $a^2+b^2=c^2$. Baudhayana's list includes the smaller triples (3, 4, 5) and (5, 12, 13), and also (7, 24, 25), (8, 15, 17), (12, 35, 37) and (15, 36, 39).

Scaling: if $(a,b,c)$ is a triple, so is $(ka,kb,kc)$, because $(ka{)}^2+(kb{)}^2=k^2(a^2+b^2)=k^2{c}^2=(kc{)}^2$. So there are infinitely many triples: (3, 4, 5), (6, 8, 10), (9, 12, 15), ...

Primitive triples have no common factor greater than 1: (3, 4, 5) is primitive, (9, 12, 15) is not. Every triple is a primitive one or a scaled copy of one. The primitive triples with all numbers up to 30 are (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25) and (20, 21, 29).

Generating triples from odd squares: since $(n-1{)}^2+(2n-1)=n^2$, whenever the odd number $2n-1$ is itself a square, we get a triple. For the odd square 49: $2n-1=49$ gives $n=25$, so $24^2+7^2=25^2$. This method always gives $c-a=1$, so it misses triples like (8, 15, 17) and (20, 21, 29).

Worked example

Use the odd square 81 to generate a Baudhayana triple, and check it. Is it primitive?

Try it first, then show the solutionSolution

1. $2n-1=81$ gives $n=41$.
2. $(n-1{)}^2+81=n^2$ gives $40^2+9^2=41^2$.
3. Check: $1600+81=1681=41^2$. 9, 40 and 41 share no common factor, so it is primitive.

Answer: (9, 40, 41), a primitive triple.

Fermat's Last Theorem

NCERT section 2.6 A Long-Standing Open Problem

There are infinitely many squares that are sums of two squares. In the 17th century the French mathematician Pierre de Fermat asked: can a cube be a sum of two cubes, a fourth power a sum of two fourth powers, and so on? He claimed that

$x^n+y^n=z^n$ **has no solution in natural numbers when $n>2$**,

and wrote in a book's margin that he had a 'truly marvellous proof' that the margin was too small to hold. Nobody found his proof, and for more than 300 years mathematicians failed to prove the statement.

As a 10-year-old in 1963, Andrew Wiles read about the problem in a library book and decided to solve it. He proved it in 1994. The story shows how a simple-looking pattern can lead to very deep mathematics.

Compare: $3^2+4^2=5^2$ works, but for cubes the nearest misses, like $9^3+10^3=1729=12^3+1$, are always off by something.

Worked example

Show that $6^3+8^3$ is not a perfect cube, even though $6^2+8^2$ is a perfect square.

Try it first, then show the solutionSolution

1. $6^2+8^2=36+64=100=10^2$, a perfect square.
2. $6^3+8^3=216+512=728$.
3. $8^3=512<728<729=9^3$, so 728 is not a cube (it misses $9^3$ by just 1).

Answer: 728 lies between $8^3$ and $9^3$, as Fermat's statement predicts.

Applications: lotus, ladders, rhombuses and equilateral triangles

NCERT section 2.7 Further Applications of the Baudhāyana - Pythagoras Theorem

Look for a right angle, name the hypotenuse, write $a^2+b^2=c^2$ and solve.

• Bhaskaracharya's lotus problem (Lilavati): a lotus stands 1 unit above the water; pushed by the wind, its tip touches the water 3 units away. With depth $x$, the stem is $x+1$, and $3^2+x^2=(x+1{)}^2$ gives $9=2x+1$, so $x=4$.
• Diagonal of a square or rectangle: $d=\sqrt{l^2+w^2}$; a square of side $s$ has diagonal $s\sqrt{2}$.
• Rhombus: the diagonals bisect each other at right angles, so the side is $\sqrt{(\frac{d_1}{2}{)}^2+(\frac{d_2}{2}{)}^2}$.
• **Equilateral triangle of side $s$:** the altitude bisects the base, so height $h=\sqrt{s^2-(\frac{s}{2}{)}^2}$ and area $=\frac{1}{2}\times s\times h$.
• Grid squares: a tilted square with vertices on grid dots, whose side goes $a$ across and $b$ up, has area $a^2+b^2$. So areas 1, 2, 4, 5, 8, 9, 10, 13, ... are possible, but 3, 6, 7 are not.

Worked example

The diagonals of a rhombus are 16 cm and 30 cm. Find its side and perimeter.

Try it first, then show the solutionSolution

1. The diagonals bisect each other at 90°, giving right triangles with legs 8 cm and 15 cm.
2. Side $=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17$ cm.
3. Perimeter $=4\times 17=68$ cm.

Answer: Side 17 cm, perimeter 68 cm.

Maths in real life: The Baudhayana-Pythagoras Theorem

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Baudhayana-Pythagoras Theorem, Triples and Applications
• Doubling and Halving Squares, and the Isosceles Right Triangle

The Baudhayana-Pythagoras Theorem formula sheet

Baudhayana-Pythagoras Theorem

$a^2+b^2=c^2$

c is the hypotenuse.

Missing leg

$a=\sqrt{c^2-b^2}$

Isosceles right triangle

$c^2=2a^2,c=a\sqrt{2}$

Bounds for root 2

$1.414<\sqrt{2}<1.415$

Diagonal of a rectangle

$d=\sqrt{l^2+w^2}$

Rhombus side

$s=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$

Equilateral triangle height

$h=\sqrt{s^2-{\left(\frac{s}{2}\right)}^2}$

Scaled triples

$(a,b,c)\to (ka,kb,kc)$

Triples from odd squares

$(n-1{)}^2+(2n-1)=n^2$

Fermat's Last Theorem

$x^n+y^n=z^n\text{ has no solution in natural numbers for }n>2$

How the school exam asks The Baudhayana-Pythagoras Theorem

The question patterns to expect, and the method that works for each.

Tricks to remember The Baudhayana-Pythagoras Theorem

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Opposite the corner, longest of all

  The hypotenuse faces the right angle and is always the longest side.

  Use it when: Labelling c correctly.

• 3-4-5 family

  Remember 3-4-5, 5-12-13, 8-15-17, 7-24-25; multiply to get more.

  Use it when: Spotting answers quickly.

• Double the square, use the diagonal

  Diagonal square = double area; side doubled = four times area.

  Use it when: Sulba-Sutra construction questions.

• √2 ≈ 1.414

  'One four one four' is close enough for most estimates.

  Use it when: Diagonals of squares.

• Odd square gives a triple

  Odd square $q$: legs $\frac{q-1}{2}$ and $\sqrt{q}$, hypotenuse $\frac{q+1}{2}$.

  Use it when: Generating triples.

The Baudhayana-Pythagoras Theorem: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in The Baudhayana-Pythagoras Theorem

The slips examiners see most often, and what to write instead.

• ✗Mistake: Adding squares when a leg is missing

  The formula is applied without checking which side is the hypotenuse.

  Wrong working

  Leg 20, hypotenuse 29: third side $=\sqrt{20^2+29^2}$.

  Right working

  Third side $=\sqrt{29^2-20^2}=\sqrt{441}=21$.

  ✓Do this: Subtract when the hypotenuse is known.

  Usually costs: Whole answer.

• ✗Mistake: Taking the square root of each term

  Students think $\sqrt{a^2+b^2}=a+b$.

  Wrong working

  $\sqrt{20^2+21^2}=20+21=41$.

  Right working

  $\sqrt{400+441}=\sqrt{841}=29$.

  ✓Do this: Add the squares first, then take one square root.

  Usually costs: Answer mark.

• ✗Mistake: Doubling the side to double the area

  Area and side are confused.

  Wrong working

  To double a 6 m square, use a 12 m square.

  Right working

  A 12 m square has 4 times the area. Use the diagonal: side $\sqrt{72}\approx 8.49$ m.

  ✓Do this: Area scales by the square of the side factor.

  Usually costs: 1-2 marks.

• ✗Mistake: Writing $\sqrt{2}=1.414$ as exact

  A rounded value is treated as exact.

  Wrong working

  $\sqrt{2}=1.414$, so $\sqrt{2}$ is a terminating decimal.

  Right working

  $1.414^2=1.999396\ne 2$; $\sqrt{2}$ is non-terminating and not a fraction. Write $\sqrt{2}\approx 1.414$.

  ✓Do this: Use '≈' for rounded values and keep roots exact where asked.

  Usually costs: 1 mark in reasoning items.

• ✗Mistake: Testing a right angle with the wrong pair

  The largest side is not used as c.

  Wrong working

  Sides 10, 24, 26: $10^2+26^2\ne 24^2$, so not right-angled.

  Right working

  Use the largest side as c: $10^2+24^2=676=26^2$, so it is right-angled.

  ✓Do this: Always compare the sum of the two smaller squares with the largest square.

  Usually costs: Whole answer.

NCERT exercise guide: The Baudhayana-Pythagoras Theorem

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on The Baudhayana-Pythagoras Theorem

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The diagonal of a square of side 8 cm is about
   (a) 8 cm
   (b) 11.31 cm
   (c) 16 cm
   (d) 64 cm

   Hint

   $8\sqrt{2}$.

   Answer

   (b) 11.31 cm

2. Q2 · 1 mark · MCQExam level

   In a right triangle, the longest side is
   (a) always a leg
   (b) the hypotenuse
   (c) sometimes the hypotenuse
   (d) equal to the sum of the legs

   Hint

   $c^2=a^2+b^2$.

   Answer

   (b) the hypotenuse

3. Q3 · 1 mark · Very short answerExam level

   Find the hypotenuse when the legs are 8 cm and 15 cm.

   Hint

   $64+225$.

   Answer

   17 cm

4. Q4 · 2 marks · Very short answerExam level

   Do the lengths 10 cm, 24 cm and 26 cm form a right triangle? Show your check.

   Hint

   Use the longest as c.

   Answer

   $100+576=676=26^2$, so yes.

5. Q5 · 2 marks · Very short answerHarder exam level

   A rectangle is 12 cm by 35 cm. Find its diagonal.

   Hint

   A known triple.

   Answer

   $\sqrt{144+1225}=\sqrt{1369}=37$ cm

6. Q6 · 3 marks · Short answerHarder exam level

   A square has a diagonal of 16 cm. Find its area and bounds for its side with one decimal place.

   Hint

   $2s^2=256$.

   Answer

   $s^2=128$, so area 128 cm²; $11.3^2=127.69<128<129.96=11.4^2$, so $11.3<s<11.4$.

7. Q7 · 2 marks · Very short answerHarder exam level

   Show that $1.73<\sqrt{3}<1.74$.

   Hint

   Square both bounds.

   Answer

   $1.73^2=2.9929<3$ and $1.74^2=3.0276>3$.

8. Q8 · 3 marks · Short answerHarder exam level

   List all the primitive Baudhayana triples whose numbers are all at most 30.

   Hint

   There are five.

   Answer

   (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29).

9. Q9 · 3 marks · Short answerHarder exam level

   How would you construct a square whose area equals the difference of the areas of squares of sides 13 cm and 5 cm? Find its side.

   Hint

   Hypotenuse 13, leg 5.

   Answer

   Draw a right triangle with hypotenuse 13 cm and one leg 5 cm; the other leg is $\sqrt{169-25}=12$ cm, and the square on it has area $169-25=144$ cm².

10. Q10 · 5 marks · Long answerHarder exam level

    Two vertical poles 6 m and 11 m high stand 12 m apart on level ground. (a) Find the length of a wire stretched between their tops. (b) A bird flies straight from the top of the taller pole to the foot of the shorter pole. How far does it fly (give bounds to one decimal place)?

    Hint

    Height difference is 5 m.

    Answer

    (a) Right triangle with legs 12 and $11-6=5$: wire $=\sqrt{169}=13$ m. (b) Legs 12 and 11: $\sqrt{144+121}=\sqrt{265}$; $16.2^2=262.44$ and $16.3^2=265.69$, so between 16.2 m and 16.3 m.

11. Q11 · 3 marks · Short answerHOTS / competency

    On a dot grid, can you draw a square of area 10? Of area 7? Explain using the theorem.

    Hint

    A tilted side going $a$ across and $b$ up has length $\sqrt{a^2+b^2}$.

    Answer

    Area 10: yes, a side going 1 across and 3 up has length $\sqrt{1+9}=\sqrt{10}$. Area 7: no, because 7 is not a sum of two squares of whole numbers (the squares 0, 1, 4 never add to 7).

12. Q12 · 3 marks · Short answerHOTS / competency

    Show that no Baudhayana triple can have all three numbers odd.

    Hint

    Parity of squares.

    Answer

    The square of an odd number is odd. If $a$ and $b$ were odd, $a^2+b^2$ = odd + odd = even, so $c^2$ and hence $c$ would be even. So not all three can be odd.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What did Baudhayana say about the square on the diagonal of a square?Show answer

   It has double the area of the original square.

2. 2.State the Baudhayana-Pythagoras Theorem.Show answer

   In a right triangle with legs $a$, $b$ and hypotenuse $c$, $a^2+b^2=c^2$.

3. 3.In an isosceles right triangle with equal sides $a$, what is the hypotenuse?Show answer

   $a\sqrt{2}$, since $c^2=2a^2$.

4. 4.Give bounds for $\sqrt{2}$ to three decimal places.Show answer

   $1.414<\sqrt{2}<1.415$.

5. 5.Why can $\sqrt{2}$ not be a fraction $\frac{m}{n}$?Show answer

   Then $2n^2=m^2$, but 2 would appear an odd number of times on the left and an even number on the right.

6. 6.What is a primitive Baudhayana triple?Show answer

   A triple with no common factor greater than 1, like (3, 4, 5).

7. 7.If (a, b, c) is a triple, is (5a, 5b, 5c)?Show answer

   Yes, scaling by any positive integer gives a triple.

8. 8.Which identity generates triples from odd squares?Show answer

   $(n-1{)}^2+(2n-1)=n^2$.

9. 9.What does Fermat's Last Theorem say, and who proved it?Show answer

   $x^n+y^n=z^n$ has no natural-number solutions for $n>2$; Andrew Wiles proved it in 1994.

10. 10.What is the depth in Bhaskaracharya's lotus problem?Show answer

    4 units.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you explain why the square on the diagonal has double the area?
• ☐Can you halve a square by folding?
• ☐Can you find bounds for $\sqrt{98}$ to one decimal place?
• ☐Can you prove that $\sqrt{2}$ is not a fraction?
• ☐Can you describe Baudhayana's proof of $a^2+b^2=c^2$?
• ☐Can you find a missing leg given the hypotenuse?
• ☐Can you test whether three lengths make a right triangle?
• ☐Can you list primitive triples up to 30?
• ☐Can you generate a triple from an odd square?
• ☐Can you solve a lotus or ladder problem?

The Baudhayana-Pythagoras Theorem: frequently asked questions