Chapter 2: The Baudhāyana-Pythagoras Theorem (Geometry)

No. With side $2s$, the area is $(2s{)}^2=4s^2$, which is 4 times the original area. Four copies of the old square fit inside the new one.

No; it makes the area 4 times as large.

The diagonal of a square bisects its corner angles, making $45°$ with each side.

The dotted square is built on that diagonal, so at each end of the diagonal its side makes $90°-45°=45°$ with the original square's sides. So each extended side of the original square bisects a right angle of the dotted square.

In a square, the bisector of an angle is the diagonal, so it passes through the opposite vertex.

The small triangles are congruent:

• Each is a right triangle with two $45°$ angles (an isosceles right triangle).
• Each has the original square's diagonal or the dotted square's side, which are the same length, as its hypotenuse.

By ASA (or RHS), they are congruent. The original square has 2 of them and the dotted square has 4, so the area doubles.

The extended sides bisect the dotted square's right angles, and an angle bisector of a square is its diagonal. All the small triangles are isosceles right triangles on equal hypotenuses, so they are congruent; the dotted square holds 4 against 2.

The two diagonals cut Square 2 into 4 identical isosceles right triangles. Place one on each side of Square 1, with its long side (the hypotenuse) along that side and the right-angle corner pointing outwards.

The four outer corners make a tilted square, standing on its corner. Its area is Square 1 + Square 2 = twice the original area.

Put one diagonal triangle of Square 2 against each side of Square 1; the four tips form a tilted square of double the area.

The tilted square cuts the original into 4 corner triangles and itself. Draw its two diagonals, which are the lines through the midpoints. They divide the original into 4 small squares, and each small square is cut in half by a side of the tilted square.

So the tilted square holds 4 half-squares, while the original holds 8. Its area is half.

A square with half the side has ${\left(\frac{1}{2}\right)}^2=\frac{1}{4}$ of the area, so 4 such squares fill the original.

The midpoint square holds 4 of the 8 equal triangles, so its area is half. Half the side gives only a quarter of the area: 4 such squares fill the original.

P, Q, R, S are the midpoints of the sides. Join PR and QS; they meet at the centre O, and $OP=OQ=OR=OS$ (each is half a side). PR and QS are perpendicular, because they are parallel to the sides.

So the four triangles $OPQ,OQR,ORS,OSP$ are congruent (SAS: two equal sides with a $90°$ angle between them). Hence:

• Equal sides: $PQ=QR=RS=SP$.
• Right angles: each base angle of these triangles is $45°$, so each angle of PQRS is $45°+45°=90°$.

PQRS is therefore a square. Each triangle such as $OPQ$ is half of a quarter-square of the paper, so PQRS has half the area.

The 4 triangles round the centre are congruent (SAS), so PQRS has equal sides and 45° + 45° = 90° angles; it covers half of each quarter, hence half the paper.

No. If $\sqrt{2}=\frac{m}{n}$, then $2n^2=m^2$.

In the prime factorisation of a square, every prime occurs an even number of times. So 2 occurs an even number of times in $m^2$ and in $n^2$. But in $2n^2$ it occurs an odd number of times. Both sides cannot then be equal, so no such fraction exists.

No: 2n² = m² is impossible, since 2 would appear an odd number of times on one side and an even number of times on the other.

The four pieces are identical isosceles right triangles. Place them with their right-angle corners meeting at one point in the middle and their long sides (the diagonals) facing outwards.

The four long sides form the boundary of a square whose side is the diagonal of the original square. Its area is that of both squares together, which is double.

Meet the four right-angle corners at the centre; the four diagonals then form the sides of a square of double area.

Since $c^2=2a^2$, we have $c=\sqrt{2a^2}$:

√18, √32, √72, √128, √162, between 4.2–4.3, 5.6–5.7, 8.4–8.5, 11.3–11.4 and 12.7–12.8 respectively.

Two such triangles make a square of side $a$. The square on the hypotenuse has area $10^2=100$, which is double that square.

So $a^2=50$ and $a=\sqrt{50}\approx 7.07$ (since $7.07^2\approx 49.98$).

Each equal side is √50 ≈ 7.07.

Yes. With $a=b$, the right triangle is an isosceles right triangle, and its hypotenuse is the diagonal of the square. The rectangle of width $b-a$ shrinks to nothing.

The method says to build a square on the diagonal, which is exactly the earlier doubling method.

Yes: for equal squares the hypotenuse is the diagonal, so it is the same doubling construction.

Each of the congruent right triangles has acute angles $x$ and $90°-x$.

At each corner of the new figure, an angle $x$ of one triangle and an angle $90°-x$ of the neighbouring triangle lie along a straight line. So the figure's angle there is

$$180°-x-(90°-x)=90°$$

The four sides are equal (each is the hypotenuse $c$) and all four angles are $90°$, so it is a square with area $c^2$.

At every corner the two acute angles x and 90° − x leave exactly 90°, and all sides equal c, so the figure is a square of area c².

$c^2=25+144=169$, so $c=13$. The measured value should be close to 13 cm.

13 cm

$b^2=17^2-8^2=289-64=225$, so $b=15$.

15 cm

Let the given side be $a$.

Triple:

1. Draw the diagonal of the square; its length is $a\sqrt{2}$, so the square on it has area $2a^2$.
2. Make a right triangle with legs $a$ and the diagonal $a\sqrt{2}$.
3. Its hypotenuse $h$ has $h^2=a^2+2a^2=3a^2$. The square on $h$ is 3 times the original.

Five times: make a right triangle with legs $a$ and $2a$. Its hypotenuse has square $a^2+4a^2=5a^2$. (Or use legs $a$ and $\sqrt{3}a$, from the triple step, and then add $a$ again.)

Triple: right triangle with legs a and the diagonal a√2 (h² = 3a²). Five times: right triangle with legs a and 2a (h² = 5a²).

1. $c=\sqrt{25+49}=\sqrt{74}\approx 8.60$
2. $c=\sqrt{64+144}=\sqrt{208}\approx 14.42$
3. $b=\sqrt{225-81}=\sqrt{144}=12$
4. $c=\sqrt{49+144}=\sqrt{193}\approx 13.89$
5. $c=\sqrt{2.25+12.25}=\sqrt{14.5}\approx 3.81$

(i) √74 ≈ 8.60 (ii) √208 ≈ 14.42 (iii) b = 12 (iv) √193 ≈ 13.89 (v) √14.5 ≈ 3.81

Checking every pair with $c\le 20$ gives six triples:

$(3,4,5)$, $(6,8,10)$, $(5,12,13)$, $(9,12,15)$, $(8,15,17)$, $(12,16,20)$

(30, 40, 50): $900+1600=2500=50^2$ ✓

(300, 400, 500): $90000+160000=250000=500^2$ ✓

Six triples: (3,4,5), (6,8,10), (5,12,13), (9,12,15), (8,15,17), (12,16,20). Yes, both (30,40,50) and (300,400,500) are triples.

(5, 12, 13) is primitive: $25+144=169$, and 5, 12, 13 have no common factor.

The primitive triples with numbers ≤ 20 are (3, 4, 5), (5, 12, 13), (8, 15, 17).

Scaled versions (×2 to ×6):

• (6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), (18, 24, 30)
• (10, 24, 26), (15, 36, 39), (20, 48, 52), (25, 60, 65), (30, 72, 78)
• (16, 30, 34), (24, 45, 51), (32, 60, 68), (40, 75, 85), (48, 90, 102)

None of the scaled versions is primitive: each has the scale factor $k>1$ as a common factor.

Yes, (5, 12, 13) is primitive; the others are (3, 4, 5) and (8, 15, 17). Scaled versions such as (10, 24, 26) are never primitive.

Check: $(9,12,15)÷3=(3,4,5)$, and $9+16=25$ ✓

Justification: Write $a=f{a}^{\prime }$, $b=f{b}^{\prime }$, $c=f{c}^{\prime }$. Then $a^2+b^2=c^2$ becomes $f^2(a^{\prime 2}+b^{\prime 2})=f^2{c}^{\prime 2}$. Dividing by $f^2$ gives $a^{\prime 2}+b^{\prime 2}=c^{\prime 2}$.

Yes: dividing a² + b² = c² by f² keeps it true; (9, 12, 15) gives (3, 4, 5).

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

(n − 1)²

If the odd square $k^2$ is the $n$th odd number, then $2n-1=k^2$ and $(n-1{)}^2+k^2=n^2$:

For example, $9^2+40^2=81+1600=1681=41^2$ ✓

(7, 24, 25), (9, 40, 41), (11, 60, 61), (13, 84, 85), (15, 112, 113)

No. In every triple it makes, one side is $n-1$ and the hypotenuse is $n$. Two consecutive numbers have no common factor other than 1, so the triple has no common factor greater than 1.

No: the hypotenuse and one side are consecutive numbers, so every triple is primitive.

Yes: any primitive triple whose hypotenuse is not 1 more than a side. For example:

• $(8,15,17)$: $17-15=2$
• $(20,21,29)$: $400+441=841=29^2$
• $(12,35,37)$
• $(28,45,53)$

Yes, e.g. (8, 15, 17), (20, 21, 29), (12, 35, 37).

$d=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\approx 7.07$

5√2 ≈ 7.07 cm

1. Hypotenuse $=\sqrt{49+81}=\sqrt{130}\approx 11.40$
2. Hypotenuse $=\sqrt{16+100}=\sqrt{116}\approx 10.77$
3. Leg $=\sqrt{1681-1600}=\sqrt{81}=9$
4. Leg $=\sqrt{200-100}=\sqrt{100}=10$ (an isosceles right triangle)
5. Hypotenuse $=\sqrt{100+150}=\sqrt{250}\approx 15.81$
6. Leg $=\sqrt{2025-729}=\sqrt{1296}=36$

√130 ≈ 11.40; √116 ≈ 10.77; 9; 10; √250 ≈ 15.81; 36

The diagonals of a rhombus bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs 12 and 35:

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

37 units

Yes. $c^2=a^2+b^2$ is greater than $a^2$ (because $b^2>0$), so $c>a$. In the same way $c>b$.

Also, the right angle is the largest angle of the triangle, and the longest side is always opposite the largest angle.

Yes: c² = a² + b² exceeds both a² and b².

True. Let $f$ be the HCF of $a,b,c$. Then $(a/f,b/f,c/f)$ is a triple (see above) with no common factor, so it is primitive. The original triple is this primitive triple scaled by $f$.

True: divide by the HCF to get the primitive triple.

Use Baudhāyana triples: the sides are $a$ and $b$, and the diagonal is $c$.

3 × 4 (diagonal 5), 6 × 8 (10), 5 × 12 (13), 8 × 15 (17), 7 × 24 (25).

$7^2-5^2=24$, so we need a side of $\sqrt{24}$.

1. Draw $AB=5$.
2. Draw a perpendicular to AB at B.
3. With centre A and radius 7, cut the perpendicular at C.

Then $B{C}^2=A{C}^2-A{B}^2=49-25=24$. Build the square on BC.

Make a right triangle with hypotenuse 7 and one leg 5; the square on the other leg (√24) has area 49 − 25 = 24.

A tilted square whose side goes $p$ dots across and $q$ dots up has area $p^2+q^2$:

• (a) Area 2: $p=1,q=1$
• (b) Area 3: not possible, because 3 is not a sum of two squares
• (c) Area 4: the ordinary $2\times 2$ square ($p=2,q=0$)
• (d) Area 5: $p=2,q=1$

(ii) The possible areas are exactly the numbers that are a sum of two squares $p^2+q^2$: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, …

Areas such as 3, 6, 7, 11, 12, 14, 15, 19, 21 and 22 are impossible.

2, 4 and 5 are possible but 3 is not; in general, the possible areas are the sums of two squares (1, 2, 4, 5, 8, 9, 10, 13, …).

In equilateral triangle ABC, draw the altitude AD to BC. Triangles ABD and ACD are right triangles with $AB=AC$ and a common side AD, so they are congruent (RHS). Hence $BD=DC=3$.

The height is $AD=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt{3}$.

$$\text{Area}=\frac{1}{2}\times 6\times 3\sqrt{3}=9\sqrt{3}\approx 15.59$$

9√3 ≈ 15.59 sq. units (height √27 ≈ 5.2).

Open the box labelled RED. Its label is wrong, so it holds either blue or green balls. Say it holds blue:

• The box labelled GREEN cannot be green (its label is wrong), and blue is already taken, so it must be red.
• The box labelled BLUE is then green.

If the RED-labelled box holds green, then by the same reasoning the box labelled BLUE is red, and the box labelled GREEN is blue.

Opening any one box works, because only two arrangements have every label wrong. One look tells them apart.

Open any one box (say "RED"); what you see fixes the other two, since every label must be wrong.