● Ganita Prakash · Grade 8 · Chapter 2
The Baudhāyana–Pythagoras
The Baudhāyana–Pythagoras
Theorem — Full Solutions
Every Math Talk, Try This, and Figure it Out question from the chapter, solved step-by-step — with the same diagrams used in the textbook, redrawn clearly, and every formula shown the way a Grade 8 student would write it in their notebook.
2.1
Doubling a Square
From the textbook
Baudhāyana’s Śulba-Sūtra (c. 800 BCE) asks: how do you construct a square with double the area of a given square? The first guess — doubling the side length — actually gives 4 times the area, not 2 times, because area depends on side² .
“The diagonal of a square produces a square of double the area of the original square.”
Baudhāyana, Śulba-Sūtra, Verse 1.9
1
In-textWill doubling the side length of a square double its area?
Original square vs. a square with sides doubled
Answer
1
No. If the original square has side $a$, its area is $a^2$.
2
If we double the side to $2a$, the new area is $(2a)^2 = 4a^2$.
3
So the new square is 4 times the original area — not 2 times. Doubling the side multiplies the area by $2\times2=4$, because area is a two-dimensional quantity.
The new square has 4 times the area, not double.
2
In-textWhy does the new dotted square (built on the diagonal) have double the area of the original square?
Square built on the diagonal of the original square
Answer
1
Draw the diagonal of the original square. It cuts the square into 2 equal (congruent) triangles.
2
Now build a new square using this diagonal as its side. If you draw the horizontal and vertical lines through the centre (the “east-west” and “north-south” lines Baudhāyana describes), the new square gets divided into 4 triangles, each congruent to the 2 triangles of the original square.
3
Since the original square = 2 such triangles, and the new square = 4 such triangles, the new square’s area is exactly double.
The new square is made of 4 congruent triangles while the original is made of only 2 — so its area is exactly double.
3
In-text · Math TalkWhy should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted (new) square?
Answer
1
In the original square, the diagonal makes a $45°$ angle with each side (since the diagonal bisects the right angle at each vertex of a square).
2
The new (dotted) square is built using this diagonal as its side. Each angle of this new square is $90°$, and its own diagonals are the horizontal/vertical lines through the centre.
3
By the diagonal property of a square — the diagonal bisects the angle at each vertex — the horizontal and vertical lines, which bisect the $90°$ angles of the new square into two $45°$ parts, must pass through the opposite vertex of that square.
4
This is exactly why the extended sides of the original square (which are these horizontal/vertical lines) land precisely on the corners of the dotted square — they are acting as the bisecting diagonals of the new square.
Because the horizontal and vertical lines through the centre are the diagonals of the new square, and a square’s diagonal always bisects its vertex angle and runs straight into the opposite corner.
4
In-textAll the small triangles formed are congruent to each other. Can you explain why?
Answer
1
Each small triangle is a right-angled isosceles triangle: its two legs are halves of the sides of the original square (so they’re all equal in length), and the angle between them is $90°$.
2
By the SAS (Side-Angle-Side) congruence rule — two sides and the included angle are equal — every one of these small triangles is congruent to every other.
All small triangles have two equal legs (half the side of the square) and a right angle between them, so by SAS congruence, they are all congruent.
5
Try ThisDoubling a Square Using Paper — cut two identical squares as shown (pieces 1–4 and 5–8), then arrange pieces 5, 6, 7, 8 around Square 1. Why does this give double the area?
Square 1 cut into 4 triangles; identical Square 2 cut into 4 triangles (5,6,7,8)
Answer
1
Square 1 is cut by its two diagonals into 4 congruent triangles (1, 2, 3, 4).
2
The identical Square 2 is cut the same way into triangles 5, 6, 7, 8 — each congruent to triangles 1–4.
3
When pieces 5, 6, 7, 8 are placed around the outside of Square 1 (each one attached along an edge, pointing outward), they exactly form the four corner triangles needed to complete a bigger tilted square.
4
This bigger square is made up of all 8 small triangles total (4 from Square 1 + 4 from Square 2) — exactly twice as many triangles as one original square, so its area is double.
The new square contains all 8 triangles from both original squares, so it has exactly double the area of one original square.
2.2
Halving a Square
From the textbook
To make a square with half the area of a given square, we reverse the doubling idea: draw a tilted square inside the original one, joining the midpoints of its sides.
6
In-textWhy is the smaller inside (tilted) square half the area of the larger square?
Adding east-west / north-south lines splits each outer triangle to match an inner one
Answer
1
Join the midpoints of all 4 sides of the big square. This creates a tilted square inside, and 4 corner triangles outside it.
2
Draw the horizontal and vertical lines through the centre of the figure. Each of the 4 outer corner triangles gets split into 2 smaller triangles, and the inner tilted square also gets split into 4 small triangles.
3
All these small triangles (8 from outside + 4 from inside = wait, let’s recount) — actually, the 4 outer triangles together have the same total area as the 4 triangles that make up the inner square, because each pair of half-corner-triangles is congruent to one quarter-triangle of the inner square.
4
So: outer big square = inner square + 4 corner triangles, and those 4 corner triangles together have the same area as the inner square. This means: Big square = inner square + inner square = 2 × inner square.
Since Big Square = 2 × Inner Square, the inner tilted square has exactly half the area of the big square.
7
Try ThisWill the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?
Answer
1
No. If the original square has side $a$, a square with side $\dfrac{a}{2}$ has area $\left(\dfrac{a}{2}\right)^2 = \dfrac{a^2}{4}$.
2
This is one-fourth of the original area $a^2$, not one-half — because halving the side divides the area by $2\times 2 = 4$.
3
Since each small square is $\dfrac14$ of the original area, exactly 4 such squares (each with half the sidelength) will fill the original square.
No — it gives $\frac14$ of the area, not $\frac12$. You’d need 4 such squares to fill the original.
8
In-textFold the square paper inward so the crease lines pass through the midpoints of the sides, forming PQRS. Why is PQRS a square, and why is its area half that of the original paper?
PQRS formed by joining the midpoints of the original square
Answer
1
Why PQRS is a square: Connect QS and PR (the diagonals of PQRS). Since P, Q, R, S are midpoints of the sides of the original square, each of the 4 corner triangles cut off (like the one at each corner) is a right-angled isosceles triangle, because the two legs of each corner triangle are equal (each is half the side of the original square) and the angle between them is the original square’s $90°$ corner angle.
2
By SAS congruence, all 4 corner triangles are congruent to each other. This means PQ = QR = RS = SP — all four sides of PQRS are equal.
3
Since each corner triangle is isosceles right-angled, its base angles are each $45°$. At any vertex of PQRS (say Q), the angle of PQRS = $180° – 45° – 45° = 90°$. So all angles of PQRS are $90°$ too.
4
A 4-sided figure with all sides equal and all angles $90°$ is a square.
5
Why its area is half: The original square is made of PQRS plus 4 corner triangles. Each corner triangle has legs of length $\frac{a}{2}$ (half the side), so its area is $\frac12 \times \frac{a}{2} \times \frac{a}{2} = \frac{a^2}{8}$.
6
Total area of 4 corner triangles $= 4 \times \dfrac{a^2}{8} = \dfrac{a^2}{2}$, which is exactly half of the original square’s area $a^2$. So PQRS, the rest of the square, is also $\dfrac{a^2}{2}$ — half the original area.
PQRS has all 4 sides equal (congruent corner triangles) and all angles $90°$, so it’s a square — and its area works out to exactly $\dfrac{a^2}{2}$, half of the original.
2.3
Hypotenuse of an Isosceles Right Triangle
From the textbook
For a right triangle with both legs $= 1$ unit, the square built on its hypotenuse has area $2$ (double the area of the small unit square). So if $c$ is the hypotenuse, $c^2 = 2 \Rightarrow c = \sqrt2$.
9
In-textFind the hypotenuse of an isosceles right triangle with both legs = 1 unit.
Square PEAR (side 1) and square REST built on its diagonal
Answer
1
A unit square (side 1) is made of 2 such isosceles right triangles, each with legs 1, 1 and hypotenuse $c$.
2
The square built on the diagonal (hypotenuse) has double the area of the unit square: $\text{Area} = 2 \times 1 = 2$ sq. units.
3
If $c$ is the hypotenuse, then $\text{Area} = c \times c = c^2$. So $c^2 = 2$.
4
Taking square root: $c = \sqrt{2}$.
The hypotenuse is $\sqrt{2}$ units $\approx 1.414$ units.
10
In-textIs $\sqrt2$ less than or greater than 1? Less than or greater than 2? Can we find closer bounds?
Answer
1
Compare squares: $1^2 = 1$ and $(\sqrt2)^2 = 2$. Since $1 < 2$, we get $1 < \sqrt2$.
2
Also $(\sqrt2)^2 = 2$ and $2^2 = 4$. Since $2 < 4$, we get $\sqrt2 < 2$.
3
So $1 < \sqrt2 < 2$ — here 1 is a lower bound and 2 is an upper bound.
4
To get closer bounds, square numbers between 1 and 2 with finer decimals:
| Test number | Square | Conclusion |
|---|---|---|
| $1.4^2$ | $1.96$ | too small |
| $1.5^2$ | $2.25$ | too big |
| ⟹ $1.4 < \sqrt2 < 1.5$ | ||
| $1.41^2$ | $1.9881$ | too small |
| $1.42^2$ | $2.0164$ | too big |
| ⟹ $1.41 < \sqrt2 < 1.42$ | ||
| $1.414^2$ | $1.999396$ | too small |
| $1.415^2$ | $2.002225$ | too big |
| ⟹ $1.414 < \sqrt2 < 1.415$ | ||
$\sqrt2$ lies between $1$ and $2$, and more precisely between $1.414$ and $1.415$. Its actual value is the non-terminating decimal $\sqrt2 = 1.41421356\ldots$
1
ExerciseTwo identical square papers are cut into 4 right-triangle pieces (1,2,3,4). Can you arrange these pieces to create a square with double the area of either square?
Answer
1
Each square is cut along one diagonal into 2 right triangles, giving 4 right triangles total (pieces 1, 2, 3, 4) — all congruent to each other, since the two original squares are identical.
2
Arrange all 4 right-triangle pieces so that their hypotenuses face outward and their right-angle corners meet at the centre — like a pinwheel. Each triangle’s two legs align with the legs of the neighbouring triangle.
3
This arrangement forms a new square whose side is the hypotenuse of each small triangle (= the diagonal of the original square).
4
Since 4 triangles = 2 original squares’ worth of area (2 triangles per square × 2 squares), and they now form 1 new square, that new square’s area = sum of both original squares = double the area of either one.
Arrange the 4 triangles pinwheel-style around a centre point, hypotenuses facing out — this forms a square with exactly double the area of either original square (this is the same idea as Baudhāyana’s diagonal method, just built from cut pieces).
2
ExerciseGiven the equal sides of an isosceles right triangle, find the hypotenuse and bound it to at least one decimal place: (i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9
Answer
·
Using the rule: if both legs $= a$, then hypotenuse $c$ satisfies $c^2 = 2a^2$, so $c = \sqrt{2a^2}$. Here is the working for each:
| a | $c^2 = 2a^2$ | Bounds (decimal) |
|---|---|---|
| 3 | $2(9)=18$ | $4.2^2=17.64$, $4.3^2=18.49$ → $4.2 < c < 4.3$ |
| 4 | $2(16)=32$ | $5.6^2=31.36$, $5.7^2=32.49$ → $5.6 < c < 5.7$ |
| 6 | $2(36)=72$ | $8.4^2=70.56$, $8.5^2=72.25$ → $8.4 < c < 8.5$ |
| 8 | $2(64)=128$ | $11.3^2=127.69$, $11.4^2=129.96$ → $11.3 < c < 11.4$ |
| 9 | $2(81)=162$ | $12.7^2=161.29$, $12.8^2=163.84$ → $12.7 < c < 12.8$ |
(i) $4.2 < c < 4.3$ (ii) $5.6 < c < 5.7$ (iii) $8.4 < c < 8.5$ (iv) $11.3 < c < 11.4$ (v) $12.7 < c < 12.8$
Method: square two consecutive one-decimal numbers until one square is just below and the other just above $2a^2$ — that pair gives the bound.
General formula derived in the textbook
For an isosceles right triangle with equal sides $a$ and hypotenuse $c$: $\quad c^2 = 2a^2$.
This comes from: Area of the square on hypotenuse (SQVU) $= 2 \times$ Area of square PQRS (side $a$).
3
ExerciseThe hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths?
Answer
1
We use the formula $c^2 = 2a^2$, where $c = 10$.
2
$10^2 = 2a^2 \Rightarrow 100 = 2a^2 \Rightarrow a^2 = 50$.
3
So $a = \sqrt{50}$. Since $7^2=49$ and $8^2=64$, we know $\sqrt{50}$ is just a little more than 7 (approximately $7.07$).
Each of the two equal sides has length $\sqrt{50}$ units $\approx 7.07$ units.
2.4
Combining Two Different Squares
“The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.”
Baudhāyana, Śulba-Sūtra, Verse 1.12
Two squares of side $a$ and $b$ combine into one square of side $c$, the hypotenuse of a right triangle with legs $a, b$
11
In-textWhy does Baudhāyana’s method work? Does it agree with the method for combining two same-sized squares?
Answer
1
When $a = b$ (same-sized squares), Baudhāyana’s construction becomes: make a right triangle with both legs $= a$. The hypotenuse of this triangle is exactly the diagonal of a square of side $a$.
2
This matches exactly what we did in Section 2.1 — building a new square on the diagonal of the original square to double its area! So yes, it agrees perfectly: it’s the special case $a=b$ of the general method.
Yes — when both squares are equal ($a=b$), Baudhāyana’s general method reduces exactly to the “build on the diagonal” method we used earlier to double a square.
12
In-textThe 4-sided figure formed (using triangles T, U, V, W, X) is actually a square. Explain why all its angles are right angles.
Answer
1
Let one acute angle of the right triangle (legs $a, b$) be $x$. Since the triangle’s angles add to $180°$ and one angle is $90°$, the other acute angle is $90° – x$.
2
Triangles T, U, W, X are all congruent copies of this same right triangle, just rotated around the figure. At each vertex of the new 4-sided figure, one angle of $x$ from one triangle sits right next to an angle of $(90°-x)$ from the neighbouring triangle.
3
So each vertex angle of the new figure $= x + (90° – x) = 90°$.
4
Since T, U, W, X are all congruent, the 4 sides of the new figure (each being a hypotenuse of one of these triangles) are all equal in length, $= c$.
A 4-sided figure with all 4 sides equal ($=c$) and all 4 angles $= 90°$ is, by definition, a square — and its side is the hypotenuse $c$ of the right triangle with legs $a$ and $b$.
1
ExerciseIf a right-angled triangle has shorter sides 5 cm and 12 cm, what is the length of the hypotenuse? (Draw and measure, then check using Baudhāyana’s Theorem.)
Answer
1
By Baudhāyana’s (Pythagoras’) Theorem: $a^2 + b^2 = c^2$, with $a=5$, $b=12$.
2
$5^2 + 12^2 = c^2 \Rightarrow 25 + 144 = c^2 \Rightarrow 169 = c^2$.
3
$c = \sqrt{169} = 13$.
The hypotenuse is 13 cm. (When you draw this triangle and measure with a ruler, you should get approximately 13 cm too.)
2
ExerciseIf a right-angled triangle has a short side of 8 cm and hypotenuse of 17 cm, what is the length of the third side?
Answer
1
Using $a^2 + b^2 = c^2$ with $a = 8$, $c = 17$, we need to find $b$.
2
$8^2 + b^2 = 17^2 \Rightarrow 64 + b^2 = 289$.
3
$b^2 = 289 – 64 = 225$.
4
$b = \sqrt{225} = 15$.
The third side is 15 cm.
3
ExerciseHow would you construct a square whose area is triple the area of a given square? Five times the area of a given square?
Answer
1
For triple the area: Start with the original square of side $a$ (area $a^2$). First double it using the diagonal method (Section 2.1) to get a square of side $a\sqrt2$ and area $2a^2$.
2
Now combine this doubled square (side $a\sqrt2$) with the original square (side $a$) using Baudhāyana’s “combining two different squares” method (Section 2.4): make a right triangle with legs $a$ and $a\sqrt2$. Its hypotenuse $c$ satisfies $c^2 = a^2 + (a\sqrt2)^2 = a^2 + 2a^2 = 3a^2$.
3
A square built on this hypotenuse has area $3a^2$ — exactly triple the original.
4
For five times the area: Take the original square (side $a$, area $a^2$) and the doubled square from before (side $a\sqrt2$, area $2a^2$). Instead, build a square of area $4a^2$ first — that’s just a square of side $2a$ (double the side length, as in our earlier “first guess”).
5
Now combine the square of area $4a^2$ (side $2a$) with the original square of area $a^2$ (side $a$) using the same combining method: make a right triangle with legs $2a$ and $a$. Its hypotenuse $c$ satisfies $c^2 = (2a)^2 + a^2 = 4a^2 + a^2 = 5a^2$.
6
A square on this hypotenuse has area $5a^2$ — exactly 5 times the original.
Triple: Combine the original square (side $a$) with its doubled version (side $a\sqrt2$) using a right triangle of legs $a, a\sqrt2$ — the hypotenuse gives a square of area $3a^2$.
Five times: Combine the original square (side $a$) with a square of side $2a$ using a right triangle of legs $a, 2a$ — the hypotenuse gives a square of area $5a^2$.
Five times: Combine the original square (side $a$) with a square of side $2a$ using a right triangle of legs $a, 2a$ — the hypotenuse gives a square of area $5a^2$.
This is the same idea Baudhāyana describes in Verse 1.10 — repeatedly combining squares using the diagonal/hypotenuse trick lets you reach any whole-number multiple of the original area.
4
ExerciseFind the missing sidelength in each right triangle ($c$ = hypotenuse): (i) $a=5, b=7$ (ii) $a=8, b=12$ (iii) $a=9, c=15$ (iv) $a=7, b=12$ (v) $a=1.5, b=3.5$
Answer
·
Apply $a^2+b^2=c^2$ in each case:
| Case | Working | Missing side |
|---|---|---|
| (i) $a=5,b=7$ | $c^2 = 25+49=74$ | $c=\sqrt{74}\approx 8.60$ |
| (ii) $a=8,b=12$ | $c^2=64+144=208$ | $c=\sqrt{208}\approx 14.42$ |
| (iii) $a=9,c=15$ | $b^2=15^2-9^2=225-81=144$ | $b=\sqrt{144}=12$ |
| (iv) $a=7,b=12$ | $c^2=49+144=193$ | $c=\sqrt{193}\approx 13.89$ |
| (v) $a=1.5,b=3.5$ | $c^2=2.25+12.25=14.5$ | $c=\sqrt{14.5}\approx 3.81$ |
(i) $\sqrt{74}$ (ii) $\sqrt{208}$ (iii) $12$ (iv) $\sqrt{193}$ (v) $\sqrt{14.5}$
2.5
Right Triangles Having Integer Sidelengths
From the textbook
A Baudhāyana triple (or Pythagorean triple) is a set of 3 positive integers $(a,b,c)$ satisfying $a^2+b^2=c^2$. Baudhāyana listed: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$, $(12,35,37)$, $(15,36,39)$.
13
Math TalkList all Baudhāyana triples with numbers $\le 20$. Is (30, 40, 50) a triple? Is (300, 400, 500)? What pattern do you see?
Answer
1
Checking all integer triples with every number $\le 20$ against $a^2+b^2=c^2$, we get:
| Triple | Check |
|---|---|
| (3, 4, 5) | $9+16=25 ✓$ |
| (6, 8, 10) | $36+64=100 ✓$ |
| (9, 12, 15) | $81+144=225 ✓$ |
| (12, 16, 20) | $144+256=400 ✓$ |
2
Is $(30,40,50)$ a triple? Check: $30^2+40^2 = 900+1600=2500$, and $50^2=2500$. Yes ✓
3
Is $(300,400,500)$ a triple? Check: $300^2+400^2=90000+160000=250000$, and $500^2=250000$. Yes ✓
4
Pattern: Every triple in the table is just $(3,4,5)$ multiplied by some positive integer $k$: $(3,4,5)\times2=(6,8,10)$, $\times3=(9,12,15)$, $\times4=(12,16,20)$, $\times10=(30,40,50)$, $\times100=(300,400,500)$.
All Baudhāyana triples with numbers $\le 20$ are: $(3,4,5)$, $(6,8,10)$, $(9,12,15)$, $(12,16,20)$ — each a scaled copy of $(3,4,5)$. Both $(30,40,50)$ and $(300,400,500)$ are valid triples, confirming the pattern.
14
In-textIs $(5,12,13)$ a primitive Baudhāyana triple? What are the other primitive triples with numbers $\le 20$? Generate 5 scaled versions of each — are they primitive?
Answer
1
A triple is primitive if its three numbers share no common factor greater than 1. For $(5,12,13)$: $\gcd(5,12,13)=1$ — no common factor. So yes, $(5,12,13)$ is primitive.
2
Other primitive triples with all numbers $\le 20$: $(3,4,5)$ and $(8,15,17)$. (Checking: $8^2+15^2=64+225=289=17^2$ ✓, and $\gcd(8,15,17)=1$.)
3
5 scaled versions of $(3,4,5)$: $(6,8,10)$, $(9,12,15)$, $(12,16,20)$, $(15,20,25)$, $(18,24,30)$ — multiplying by $k=2,3,4,5,6$.
4
5 scaled versions of $(5,12,13)$: $(10,24,26)$, $(15,36,39)$, $(20,48,52)$, $(25,60,65)$, $(30,72,78)$.
5
5 scaled versions of $(8,15,17)$: $(16,30,34)$, $(24,45,51)$, $(32,60,68)$, $(40,75,85)$, $(48,90,102)$.
6
Are these scaled versions primitive? No — every one of them shares the scaling factor $k$ ($k=2,3,4,5,6$) as a common factor among all 3 numbers, so none of them are primitive.
$(5,12,13)$ is primitive. The full list of primitive triples $\le 20$ is $(3,4,5)$, $(5,12,13)$, $(8,15,17)$. Scaling any primitive triple by $k>1$ always produces a non-primitive triple.
15
In-textIf $(a,b,c)$ is non-primitive with common factor $f>1$, is $\left(\frac{a}{f},\frac{b}{f},\frac{c}{f}\right)$ a Baudhāyana triple? Check this for $(9,12,15)$.
Answer
1
Yes. Since $(a,b,c)$ is a triple, $a^2+b^2=c^2$. Dividing every term by $f^2$: $\dfrac{a^2}{f^2}+\dfrac{b^2}{f^2}=\dfrac{c^2}{f^2}$, i.e. $\left(\dfrac{a}{f}\right)^2+\left(\dfrac{b}{f}\right)^2=\left(\dfrac{c}{f}\right)^2$.
2
This is exactly the triple condition for $\left(\dfrac{a}{f},\dfrac{b}{f},\dfrac{c}{f}\right)$, so it is a Baudhāyana triple (as long as $\frac{a}{f}, \frac{b}{f}, \frac{c}{f}$ are still whole numbers, which they are since $f$ is a common factor).
3
Check with $(9,12,15)$: The common factor is $f=3$. Dividing: $\left(\dfrac{9}{3},\dfrac{12}{3},\dfrac{15}{3}\right) = (3,4,5)$.
4
Check: $3^2+4^2 = 9+16=25=5^2$ ✓. Yes, $(3,4,5)$ is indeed a valid (and primitive) Baudhāyana triple.
Yes — dividing a non-primitive triple by its common factor $f$ always gives a smaller valid Baudhāyana triple. For $(9,12,15)$, dividing by $f=3$ gives the primitive triple $(3,4,5)$.
Method used in these exercises
If the $n$th odd number ($2n-1$) is itself a perfect square, then $(n-1)^2 + (2n-1) = n^2$ gives us a Baudhāyana triple: $\big(n-1,\ \sqrt{2n-1},\ n\big)$.
1
ExerciseFind 5 more Baudhāyana triples using the odd-square idea.
Answer
1
We pick odd numbers that are themselves perfect squares: $1, 9, 25, 49, 81, 121, \ldots$ — i.e., squares of odd numbers $1,3,5,7,9,11,\ldots$
2
For an odd square $m^2$ (where $m$ is odd), it is the $n$th odd number where $2n-1=m^2$, so $n=\dfrac{m^2+1}{2}$. Then the triple is $\big(n-1,\ m,\ n\big)$.
| Odd square used | $n$ | Triple $(n-1, m, n)$ | Check |
|---|---|---|---|
| $49=7^2$ | $25$ | $(24, 7, 25)$ | $24^2+7^2=576+49=625=25^2$ ✓ |
| $81=9^2$ | $41$ | $(40, 9, 41)$ | $40^2+9^2=1600+81=1681=41^2$ ✓ |
| $121=11^2$ | $61$ | $(60, 11, 61)$ | $60^2+11^2=3600+121=3721=61^2$ ✓ |
| $169=13^2$ | $85$ | $(84, 13, 85)$ | $84^2+13^2=7056+169=7225=85^2$ ✓ |
| $225=15^2$ | $113$ | $(112,15,113)$ | $112^2+15^2=12544+225=12769=113^2$ ✓ |
Five new triples: $(24,7,25)$, $(40,9,41)$, $(60,11,61)$, $(84,13,85)$, $(112,15,113)$.
2
ExerciseDoes this method yield non-primitive Baudhāyana triples?
Answer
1
Look at the triples generated: $(4,3,5)$, $(12,5,13)$, $(24,7,25)$, $(40,9,41)$, $(60,11,61)$ — in each one, the larger leg is exactly one less than the hypotenuse ($n-1$ and $n$).
2
If a triple has any common factor $f>1$, then $f$ would have to divide both $n-1$ and $n$. But two consecutive integers ($n-1$ and $n$) can never share a common factor greater than 1 (consecutive integers are always coprime).
3
So every triple this method produces must be primitive — it cannot generate non-primitive ones.
No. This method only ever yields primitive Baudhāyana triples, because two of its three numbers are always consecutive integers, which can never share a common factor.
3
ExerciseAre there primitive triples that cannot be obtained through this method? If yes, give examples.
Answer
1
Yes. This method only produces triples where the two larger numbers differ by exactly 1 (consecutive). But there exist primitive triples where no two numbers are consecutive.
2
Example: $(8,15,17)$ is primitive ($\gcd=1$), but $15$ and $17$ differ by 2, not 1 — so this triple cannot come from the odd-square method described here.
3
Another example: $(20,21,29)$ — check: $400+441=841=29^2$ ✓, and it’s primitive, but again $21$ and $29$ are not consecutive.
Yes — for example, $(8,15,17)$ is a primitive triple that this method cannot generate, since it requires the two larger numbers to differ by exactly 1, but $15$ and $17$ differ by 2.
2.6
A Long-Standing Open Problem
Key idea — no exercise questions in this section
Since there are infinitely many whole-number triples satisfying $a^2+b^2=c^2$, the French mathematician Fermat (17th century) wondered: are there whole-number solutions to $x^n+y^n=z^n$ for powers greater than 2?
1
Fermat claimed that no such solution exists for any $n>2$ — no perfect cube is a sum of two perfect cubes, no fourth power is a sum of two fourth powers, and so on.
2
He wrote in the margin of a book that he had a “truly marvellous proof” — but it was never found, and it’s unlikely he actually had a complete one.
3
This became known as Fermat’s Last Theorem, and it stayed unproven for over 300 years, despite attempts by many brilliant mathematicians.
4
It was finally proven in 1994 by the British mathematician Andrew Wiles, who had been inspired by the problem as a 10-year-old in 1963 and worked on it (often in secret) for years as an adult.
$x^n + y^n = z^n$ has no positive whole-number solutions when $n>2$. This is one of the most famous theorems in the history of mathematics.
2.7
Further Applications of the Theorem
A classic problem from Bhāskarāchārya’s Līlāvatī
A lotus stem pokes 1 unit above a lake’s surface. A breeze blows the flower sideways until its tip touches the water, 3 units from where it started. How deep is the lake?
Right triangle formed by the stem: legs $3$ and $x$, hypotenuse $x+1$
Worked example from the textbook
1
Let $x$ = depth of the lake = length of stem underwater. The total stem length (underwater + 1 unit above) is $x+1$.
2
When the flower is pushed sideways, the stem (now slanted) is the hypotenuse of a right triangle with legs $3$ (horizontal distance) and $x$ (the original vertical depth, which stays fixed at the base).
3
By the theorem: $3^2 + x^2 = (x+1)^2 \Rightarrow 9 + x^2 = x^2+2x+1$.
4
Cancel $x^2$ from both sides: $9 = 2x+1 \Rightarrow x=4$.
The depth of the lake is 4 units.
1
ExerciseFind the diagonal of a square with sidelength 5 cm.
Answer
1
The diagonal of a square splits it into 2 right-angled isosceles triangles, with both legs $=5$ cm.
2
Using $c^2=2a^2$ (or directly $c^2 = 5^2+5^2$): $c^2 = 25+25=50$.
3
$c=\sqrt{50}=5\sqrt2 \approx 7.07$ cm.
The diagonal is $5\sqrt2 \approx 7.07$ cm.
2
ExerciseFind the missing sidelengths in the 6 right triangles shown in the textbook.
Answer
·
Apply $a^2+b^2=c^2$ to each triangle using the labelled sides (right angle marked at the vertex between the two known/unknown legs):
| Given sides | Working | Missing side |
|---|---|---|
| legs 7, 9 | $c^2=49+81=130$ | hyp $=\sqrt{130}\approx 11.40$ |
| leg 4, hyp 10 | $b^2=100-16=84$ | leg $=\sqrt{84}\approx 9.17$ |
| leg 40, hyp 41 | $b^2=1681-1600=81$ | leg $=\sqrt{81}=9$ |
| legs 10, $\sqrt{200}$ | $c^2=100+200=300$ | hyp $=\sqrt{300}=10\sqrt3\approx17.32$ |
| legs 10, $\sqrt{150}$ | $c^2=100+150=250$ | hyp $=\sqrt{250}=5\sqrt{10}\approx15.81$ |
| leg 27, hyp 45 | $b^2=2025-729=1296$ | leg $=\sqrt{1296}=36$ |
$\sqrt{130}$, $\sqrt{84}$, $9$, $10\sqrt3$, $5\sqrt{10}$, $36$ — in the order of the triangles above.
3
ExerciseFind the sidelength of a rhombus whose diagonals are 24 units and 70 units.
Answer
1
The diagonals of a rhombus bisect each other at $90°$. So each diagonal is split into halves: $\frac{24}{2}=12$ and $\frac{70}{2}=35$.
2
Each side of the rhombus is the hypotenuse of a right triangle with legs $12$ and $35$.
3
$\text{side}^2 = 12^2+35^2 = 144+1225=1369$.
4
$\text{side} = \sqrt{1369} = 37$.
Each side of the rhombus is 37 units. (Notice $(12,35,37)$ is one of Baudhāyana’s own listed triples!)
4
ExerciseIs the hypotenuse always the longest side of a right triangle? Justify your answer.
Answer
1
Yes, always. From $a^2+b^2=c^2$, since $a^2>0$ and $b^2>0$, we know $c^2 = a^2+b^2 > a^2$, which means $c>a$. Similarly $c^2 = a^2+b^2>b^2$, so $c>b$.
2
So $c$ (the hypotenuse) is strictly greater than both of the other two sides $a$ and $b$.
3
This also makes geometric sense: the hypotenuse is opposite the largest angle ($90°$, the right angle), and in any triangle, the longest side is always opposite the largest angle.
Yes — the hypotenuse is always the longest side, both algebraically (from $a^2+b^2=c^2$) and because it’s opposite the triangle’s largest angle ($90°$).
5
ExerciseTrue or False — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.
Answer
1
Take any Baudhāyana triple $(a,b,c)$. Let $f$ be the greatest common factor of $a, b, c$.
2
If $f=1$, the triple is already primitive by definition.
3
If $f>1$, then (as shown earlier) $\left(\frac{a}{f},\frac{b}{f},\frac{c}{f}\right)$ is also a valid Baudhāyana triple, and since we divided out the greatest common factor, this smaller triple has no common factor left — it is primitive. The original triple is just this primitive triple scaled up by $f$.
True. Every Baudhāyana triple is either primitive itself, or is a scaled-up version of some primitive triple.
6
ExerciseGive 5 examples of rectangles whose sidelengths and diagonal are all integers.
Answer
1
In a rectangle, the diagonal forms the hypotenuse of a right triangle with the two sides as legs. So we just need any Baudhāyana triple $(a,b,c)$ — use $a, b$ as the rectangle’s sides and $c$ as its diagonal.
| Sides | Diagonal |
|---|---|
| 3 × 4 | 5 |
| 5 × 12 | 13 |
| 8 × 15 | 17 |
| 7 × 24 | 25 |
| 20 × 21 | 29 |
Rectangles of size $3\times4$ (diagonal 5), $5\times12$ (diagonal 13), $8\times15$ (diagonal 17), $7\times24$ (diagonal 25), and $20\times21$ (diagonal 29) all have whole-number sides and diagonals.
7
ExerciseConstruct a square whose area equals the difference of the areas of squares with sidelengths 5 units and 7 units.
Answer
1
We need a square whose area $= 7^2 – 5^2 = 49-25=24$ sq. units.
2
Construction: Draw a right triangle where the hypotenuse is $7$ and one leg is $5$. By Baudhāyana’s Theorem, the other leg $b$ satisfies $5^2+b^2=7^2 \Rightarrow b^2 = 49-25=24$.
3
This leg $b=\sqrt{24}$ is exactly the side we need. Build a square on this leg — its area is $b^2=24$ sq. units, the required difference.
Construct a right triangle with hypotenuse $7$ and one leg $5$; the other leg, of length $\sqrt{24}=2\sqrt6$ units, is the side of the required square (area $=24$ sq. units).
8
Exercise · Math Talk + Try ThisUsing dots of a grid as vertices, can you create squares of area (a) 2, (b) 3, (c) 4, (d) 5 sq. units? If the grid extends indefinitely, what are the possible integer areas of squares you can make this way?
Tilted squares on a grid — area 2 and area 5 use tilted (diagonal) squares
Answer
1
(a) Area 2: Yes. Tilt a square so its side goes from one dot to a dot that is 1 step right and 1 step up — this side has length $\sqrt{1^2+1^2}=\sqrt2$, giving area $(\sqrt2)^2=2$.
2
(b) Area 3: Not possible with grid-dot vertices. Since any tilted square’s side connects two dots that are $p$ steps across and $q$ steps up (whole numbers $p,q$), the side length squared is $p^2+q^2$, which must be the area. We cannot write $3$ as a sum of two perfect squares $p^2+q^2$ (try all combos: $0+3$✗, $1+2$✗ — no perfect square pairs work). So no, a square of area 3 cannot be drawn this way.
3
(c) Area 4: Yes — simply use an upright square with side 2 (2 steps across, 0 steps up): $2^2+0^2=4$.
4
(d) Area 5: Yes. Tilt a square so its side goes 1 step across and 2 steps up (or 2 across, 1 up): side length squared $=1^2+2^2=5$.
5
(ii) General rule: A square of integer area $N$ can be drawn on the grid if and only if $N$ can be written as $p^2+q^2$ for some whole numbers $p,q$ (the horizontal and vertical “step” of one side).
6
Possible areas (checking small numbers): $1=1^2+0^2$ ✓, $2=1^2+1^2$ ✓, $3$ ✗, $4=2^2+0^2$ ✓, $5=2^2+1^2$ ✓, $6$ ✗, $7$ ✗, $8=2^2+2^2$ ✓, $9=3^2+0^2$ ✓, $10=3^2+1^2$ ✓…
(a) ✓ possible (tilted, side $\sqrt2$) (b) ✗ not possible (c) ✓ possible (upright, side 2) (d) ✓ possible (tilted, side $\sqrt5$).
In general, only areas $N$ expressible as a sum of two perfect squares ($N=p^2+q^2$) can be formed — numbers like $3, 6, 7, 11, 12, \ldots$ can never be grid-square areas.
In general, only areas $N$ expressible as a sum of two perfect squares ($N=p^2+q^2$) can be formed — numbers like $3, 6, 7, 11, 12, \ldots$ can never be grid-square areas.
9
ExerciseFind the area of an equilateral triangle with sidelength 6 units.
Answer
1
Drop an altitude from the top vertex to the base. In an equilateral triangle, this altitude bisects the base (because the two halves formed are congruent right triangles, sharing the altitude and having equal hypotenuses — the two equal original sides).
2
So the base of $6$ units is split into two halves of $3$ units each. Each half-triangle is right-angled, with one leg $=3$ (half the base) and hypotenuse $=6$ (the original side).
3
By Baudhāyana’s Theorem, the height $h$ satisfies: $3^2+h^2=6^2 \Rightarrow 9+h^2=36 \Rightarrow h^2=27$.
4
$h=\sqrt{27}=3\sqrt3$ units.
5
Area of triangle $=\dfrac12 \times \text{base}\times\text{height} = \dfrac12\times6\times3\sqrt3 = 9\sqrt3$ sq. units.
Area $=9\sqrt3 \approx 15.59$ sq. units.
★
Quick Index — All In-text Questions
These are the Math Talk and Try This questions that appear directly inside the chapter’s explanations (not the end-of-section “Figure It Out” exercises). Tap any one to jump straight to its full solution above.
Solutions crafted for Grade 8 learners · Diagrams redrawn for clarity · @EDUGROWN