?In-text Question — Why does the new dotted square have double the area of the original square?
Let PQRS be the original square and SQVU the square drawn on its diagonal SQ.
Draw the horizontal line SRV and the vertical line QRU through R.
These lines divide the dotted square SQVU into 4 small triangles: \(\triangle SRQ\), \(\triangle QRV\), \(\triangle VRU\) and \(\triangle URS\).
The diagonal SQ divides the original square PQRS into 2 triangles: \(\triangle SPQ\) and \(\triangle SRQ\).
All these small triangles are congruent to each other (see the next two questions).
Area of the dotted square SQVU \(=\) 4 small triangles
Area of the original square PQRS \(=\) 2 small triangles
So, area of SQVU \(=2\times\) area of PQRS.
MTMath Talk — Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square? [Hint: From the diagonal property of a square, the line that bisects an angle passes through the opposite vertex. Argue why the vertical and horizontal sides of the original square bisect the two angles of the dotted square.]
The diagonal SQ of the square PQRS bisects the right angle at S.
So \(\angle QSR=45^\circ\).
SQVU is a square, so \(\angle QSU=90^\circ\).
The side SR makes \(45^\circ\) with SQ, which is half of \(\angle QSU=90^\circ\).
So the horizontal side SR (extended) bisects the angle \(\angle QSU\) of the dotted square.
Similarly, the diagonal SQ bisects the right angle at Q, so \(\angle SQR=45^\circ\).
\(\angle SQV=90^\circ\), so the vertical side QR (extended) bisects the angle \(\angle SQV\) of the dotted square.
In a square, the line that bisects an angle passes through the opposite vertex.
So the line SR, which bisects the angle at S, passes through the opposite vertex V.
And the line QR, which bisects the angle at Q, passes through the opposite vertex U.
?In-text Question — Moreover, all these small triangles are congruent to each other. Can you explain why?
SV and QU are the diagonals of the square SQVU, and they meet at R (as shown in the previous question).
The diagonals of a square are equal, bisect each other and cut each other at right angles.
So R is the centre of the square SQVU, and \(RS=RQ=RV=RU\).
RS and RQ are sides of the original square.
Let each be \(a\), so \(RS=RQ=RV=RU=a\).
Also \(\angle SRQ=\angle QRV=\angle VRU=\angle URS=90^\circ\).
In each of the triangles \(SRQ\), \(QRV\), \(VRU\) and \(URS\), two sides are equal to \(a\) and the angle between them is \(90^\circ\).
By the SAS congruence rule, \(\triangle SRQ\cong\triangle QRV\cong\triangle VRU\cong\triangle URS\).
The triangle SPQ has \(PS=PQ=a\) and \(\angle SPQ=90^\circ\), so it is congruent to them too.
?In-text Question — Doubling a Square Using Paper: Cut out two identical squares of paper. Draw, label, and cut as follows. Now place the pieces 5, 6, 7, and 8 around Square 1 to get a square with double the area.
Let the side of each square be \(s\).
The pieces 5, 6, 7 and 8 are 4 right-angled isosceles triangles, each congruent to the triangles 1, 2, 3 and 4 of Square 1.
Place the longest side (hypotenuse) of each piece along a side of Square 1, with the right angle of the piece outside the square.
Square 1 is made up of 4 small triangles, and the 4 pieces added are 4 more triangles of the same size.
So the new figure is made up of 8 such triangles, which is double the 4 triangles of Square 1.
The new figure is a square: each side is made of the equal short sides of two pieces, and each corner is the right angle of a piece.
?In-text Question — Why is the smaller inside square half the area of the larger square?
Let ABCD be the larger square, and P, Q, R, S the midpoints of its sides.
PQRS is the smaller square drawn inside ABCD.
Draw the east-west line SQ and the north-south line RP.
They meet at the centre O.
These lines divide ABCD into 4 equal small squares.
In each small square, one side of PQRS is a diagonal.
So each small square is divided into 2 equal triangles, one inside PQRS and one outside it.
ABCD is made up of 8 equal triangles, and PQRS is made up of 4 of them.
Area of PQRS \(=\dfrac48\times\) area of ABCD \(=\dfrac12\times\) area of ABCD.
?In-text Question — Halving a Square Using Paper: Cut out a square from a piece of paper. Now make a square whose area is half the area of the first square.
Let ABCD be the paper square.
Mark the midpoints P, Q, R and S of its four sides.
Fold the paper inward so that the corner A goes inside, with the crease along SP.
In the same way fold B along PQ, C along QR and D along RS.
The crease lines pass through the midpoints of the sides, and the figure PQRS that remains is a square.
Its area is half the area of the paper square (this is explained in the question on PQRS).
?In-text Question — Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?
Let the sidelength of the original square be \(a\).
Area of the original square \(=a^2\).
A square with half the sidelength has sidelength \(\dfrac a2\).
Its area \(=\left(\dfrac a2\right)^2=\dfrac{a^2}{4}\).
This is \(\dfrac14\) of the area of the original square, not \(\dfrac12\).
Number of such squares needed to fill the original square \(=a^2\div\dfrac{a^2}{4}=4\).
?In-text Question — Why is PQRS a square? Why is its area half that of the original paper? Explain by connecting QS and PR, finding the different angles formed, and then using triangle congruence.
Let ABCD be the paper of side \(a\), and P, Q, R, S the midpoints of its sides.
Join QS and PR.
They meet at O.
QS and PR join the midpoints of opposite sides, so they are perpendicular to each other, each has length \(a\), and each is bisected at O.
So \(OP=OQ=OR=OS=\dfrac a2\), and \(\angle POQ=\angle QOR=\angle ROS=\angle SOP=90^\circ\).
In the triangles POQ, QOR, ROS and SOP, two sides are equal (each \(\dfrac a2\)) and the angle between them is \(90^\circ\).
So these 4 triangles are congruent by the SAS rule.
Hence \(PQ=QR=RS=SP\).
Each triangle is right-angled and isosceles, so its other two angles are \(45^\circ\).
\(\angle SPQ=\angle OPS+\angle OPQ=45^\circ+45^\circ=90^\circ\).
In the same way, \(\angle PQR=\angle QRS=\angle RSP=90^\circ\).
All the sides are equal and all the angles are \(90^\circ\), so PQRS is a square.
Area of each of the 4 triangles \(=\dfrac12\times\dfrac a2\times\dfrac a2=\dfrac{a^2}{8}\).
Area of PQRS \(=4\times\dfrac{a^2}{8}=\dfrac{a^2}{2}\).
Area of the paper ABCD \(=a^2\), so area of PQRS \(=\dfrac12\times\) area of ABCD.
Note: the textbook spells “triangle” as “tringle” in this question. It is a typing error, and we have written “triangle”.
TTTry This — Can \(\sqrt2\) be expressed as a fraction \(\frac mn\), where \(m\) and \(n\) are counting numbers?
Suppose \(\sqrt2=\dfrac mn\), where \(m\) and \(n\) are counting numbers.
Squaring both sides, \(2=\dfrac{m^2}{n^2}\).
So \(2n^2=m^2\).
In the prime factorisation of a square number, each prime occurs an even number of times.
\(m^2\) is a square number, so on the right side the prime 2 occurs an even number of times.
\(n^2\) is also a square number, so the prime 2 occurs an even number of times in \(n^2\).
In \(2n^2\) there is one more factor 2, so on the left side the prime 2 occurs an odd number of times.
A number cannot have the prime 2 an odd number of times and also an even number of times in its prime factorisation.
So \(2n^2=m^2\) is impossible, and our assumption is wrong.
Figure it Out — Set 1
Three questions from pages 39–40 of the textbook.
1Math Talk — Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way. Can you arrange these pieces to create a square with double the area of either square?
Let the side of each square be \(s\).
Each square is cut along a diagonal into 2 right-angled isosceles triangles with equal sides \(s\).
So the 4 pieces 1, 2, 3 and 4 are congruent right-angled isosceles triangles.
Put the right-angled corners of the 4 pieces together at one point, so that the equal sides of neighbouring pieces touch.
The outer boundary is made of the 4 hypotenuses, which are equal.
At each outer corner two angles of \(45^\circ\) meet, giving \(90^\circ\).
So the new figure is a square.
Area of the new square \(=4\times\dfrac12\times s\times s=2s^2\).
Area of either given square \(=s^2\).
2The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.
(i) 3
(ii) 4
(iii) 6
(iv) 8
(v) 9
Let the equal sides be \(a\) and the hypotenuse \(c\).
The square on the hypotenuse has double the area of the square on an equal side, so \(c^2=2a^2\).
(i) \(a=3\)
\(c^2=2\times3^2=18\), so \(c=\sqrt{18}\).
\(4^2=16\lt18\lt25=5^2\), so \(4\lt\sqrt{18}\lt5\).
\(4.2^2=17.64\) and \(4.3^2=18.49\).
So \(4.2\lt\sqrt{18}\lt4.3\).
(ii) \(a=4\)
\(c^2=2\times4^2=32\), so \(c=\sqrt{32}\).
\(5.6^2=31.36\) and \(5.7^2=32.49\).
So \(5.6\lt\sqrt{32}\lt5.7\).
(iii) \(a=6\)
\(c^2=2\times6^2=72\), so \(c=\sqrt{72}\).
\(8.4^2=70.56\) and \(8.5^2=72.25\).
So \(8.4\lt\sqrt{72}\lt8.5\).
(iv) \(a=8\)
\(c^2=2\times8^2=128\), so \(c=\sqrt{128}\).
\(11.3^2=127.69\) and \(11.4^2=129.96\).
So \(11.3\lt\sqrt{128}\lt11.4\).
(v) \(a=9\)
\(c^2=2\times9^2=162\), so \(c=\sqrt{162}\).
\(12.7^2=161.29\) and \(12.8^2=163.84\).
So \(12.7\lt\sqrt{162}\lt12.8\).
(ii) \(\sqrt{32}\), between 5.6 and 5.7
(iii) \(\sqrt{72}\), between 8.4 and 8.5
(iv) \(\sqrt{128}\), between 11.3 and 11.4
(v) \(\sqrt{162}\), between 12.7 and 12.8
3The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]
Let each of the two equal sides be \(a\).
Two such right triangles joined along the hypotenuse make a square of side \(a\).
The hypotenuse is a diagonal of this square.
The square on the diagonal (hypotenuse 10) has double the area of this square.
So \(10^2=2a^2\).
\(100=2a^2\)
\(a^2=50\)
\(a=\sqrt{50}=5\sqrt2\).
\(7^2=49\) and \(7.1^2=50.41\), so \(7\lt a\lt7.1\).
?In-text Question — Why does Baudhāyana’s method work?
Let the two squares have sidelengths \(a\) and \(b\), with \(b\gt a\).
Join them side by side as shown.
Mark a rectangular portion of the larger square using a side of the smaller square, and draw its diagonal.
This gives a right triangle X with perpendicular sides \(a\) and \(b\).
Let its hypotenuse be \(c\).
Draw 3 more such right triangles W, T and U so that a 4-sided figure is formed over the hypotenuse.
T, U, X and W are congruent right triangles, so the sides of the 4-sided figure are all equal to \(c\).
The angles of this figure are right angles (explained in the next question).
So it is a square of side \(c\).
The two given squares together make up the pieces V, W and X.
The square on the hypotenuse is made up of the pieces T, U and V.
\(T\cong W\) and \(U\cong X\), so area of \(T+U+V\) \(=\) area of \(W+X+V\).
So area of the square on the hypotenuse \(=\) area of the two given squares.
That is, \(c^2=a^2+b^2\).
?In-text Question — Can you see why the method works in the case where the two squares are the same size? Does it agree with the method we used earlier to combine two same sized squares into a bigger square?
Let both squares have sidelength \(a\), so \(b=a\).
The right triangle has perpendicular sides \(a\) and \(a\), so it is an isosceles right triangle.
Its hypotenuse is the diagonal of a square of side \(a\).
The square on this hypotenuse has area \(c^2=2a^2=a^2+a^2\).
This is the sum of the areas of the two given squares, so the method works.
Earlier, we combined two same sized squares by drawing a square on the diagonal of one of them.
That is exactly the square on the hypotenuse of the isosceles right triangle above.
?In-text Question — Explain why all the angles of this new 4-sided figure are right angles and so it is a square.
T, U, X and W are congruent right triangles with acute angles \(x\) and \(90^\circ-x\).
Consider the bottom corner of the 4-sided figure, where the hypotenuses of W and X meet on the straight base line.
The angle of W at this corner is \(x\) (marked in the figure).
The angle of X at this corner is \(90^\circ-x\) (marked in the figure).
The angles on a straight line add up to \(180^\circ\).
So the angle of the 4-sided figure at this corner \(=180^\circ-x-(90^\circ-x)=90^\circ\).
All 4 sides of the figure are equal (each is the hypotenuse \(c\)), so it is a rhombus.
In a rhombus, opposite angles are equal and adjacent angles add up to \(180^\circ\).
So one right angle makes all the 4 angles right angles.
A figure with 4 equal sides and 4 right angles is a square.
?In-text Question — Combining Two Squares Using Paper: Cut out and join two different sized squares as follows. Now make two cuts to make three pieces. Rearrange the three pieces into a larger square. You have now combined two smaller squares into a larger square! Now make a right triangle using the two smaller squares. Draw a square on the hypotenuse. Cover the square on the hypotenuse using your pieces.
Cut out two squares of sidelengths \(a\) and \(b\) (\(b\gt a\)) and join them side by side.
Cut 1: join the top left corner of the smaller square to the point on the base at distance \(b\) from the left end.
Cut 2: join that point to the top right corner of the larger square.
Each cut has length \(c=\sqrt{a^2+b^2}\), and the two cuts meet at a right angle.
Rearrange: move W to the position T and X to the position U.
The 3 pieces now form a larger square of side \(c\).
Make a right triangle with perpendicular sides \(a\) and \(b\), and draw a square on its hypotenuse \(c\).
The 3 pieces, which come from the two smaller squares, cover this square exactly.
So area of the square on the hypotenuse \(=\) area of the two smaller squares, that is \(c^2=a^2+b^2\).
Figure it Out — Set 2
Four questions from page 47 of the textbook.
1If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana’s Theorem.
Draw a right angle, mark 5 cm on one arm and 12 cm on the other, and join the two end points.
On measuring the hypotenuse with a ruler, we get 13 cm.
Check using Baudhāyana’s Theorem:
Let \(a=5\), \(b=12\) and the hypotenuse be \(c\).
\(a^2+b^2=c^2\)
\(5^2+12^2=c^2\)
\(25+144=c^2\)
\(169=c^2\)
\(c=13\) cm
2If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana’s Theorem.
Draw a right angle and mark 8 cm on one arm.
With the compass centred at the end of this 8 cm side, and radius 17 cm, mark a point on the other arm.
Join them.
On measuring the third side with a ruler, we get 15 cm.
Check using Baudhāyana’s Theorem:
Let \(a=8\), \(c=17\) and the third side be \(b\).
\(a^2+b^2=c^2\)
\(8^2+b^2=17^2\)
\(64+b^2=289\)
\(b^2=289-64=225\)
\(b=15\) cm
3Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhāyana’s Śulba-Sūtra, Verse 1.10)
Let the given square have sidelength \(s\), so its area is \(s^2\).
Triple the area: \(3s^2=s^2+2s^2\).
The square on the diagonal of the given square has area \(2s^2\), and its side is the diagonal \(\sqrt2\,s\).
Draw a right triangle with perpendicular sides \(s\) and \(\sqrt2\,s\) (the side and the diagonal of the given square).
The square on its hypotenuse has area \(s^2+(\sqrt2\,s)^2=s^2+2s^2=3s^2\).
Construction for triple:
Draw the given square ABCD of side \(s\) and its diagonal AC.
Draw a line segment PQ of length \(s\).
At P draw PR perpendicular to PQ, with PR equal to AC (measure AC with a compass).
Join QR.
Then \(QR^2=s^2+(\sqrt2\,s)^2=3s^2\).
The square drawn on QR has triple the area of the given square.
Five times the area: \(5s^2=s^2+4s^2\).
A square of sidelength \(2s\) has area \(4s^2\).
Draw a right triangle with perpendicular sides \(s\) and \(2s\).
The square on its hypotenuse has area \(s^2+(2s)^2=s^2+4s^2=5s^2\).
Construction for five times:
Draw a line segment PQ of length \(s\).
At P draw PR perpendicular to PQ, with PR \(=2s\) (mark off the side \(s\) twice with a compass).
Join QR.
Then \(QR^2=s^2+(2s)^2=5s^2\).
The square drawn on QR has five times the area of the given square.
Five times: make the square on the hypotenuse of a right triangle with perpendicular sides \(s\) and \(2s\).
4Let \(a\), \(b\) and \(c\) denote the length of the sides of a right triangle, with \(c\) being the length of the hypotenuse. Find the missing sidelength in each of the following cases:
(i) \(a=5\), \(b=7\)
(ii) \(a=8\), \(b=12\)
(iii) \(a=9\), \(c=15\)
(iv) \(a=7\), \(b=12\)
(v) \(a=1.5\), \(b=3.5\)
By Baudhāyana’s Theorem, \(a^2+b^2=c^2\).
(i) \(a=5\), \(b=7\)
\(c^2=5^2+7^2=25+49=74\)
\(c=\sqrt{74}\approx8.60\)
(ii) \(a=8\), \(b=12\)
\(c^2=8^2+12^2=64+144=208\)
\(c=\sqrt{208}=\sqrt{16\times13}=4\sqrt{13}\approx14.42\)
(iii) \(a=9\), \(c=15\)
\(b^2=c^2-a^2=15^2-9^2=225-81=144\)
\(b=12\)
(iv) \(a=7\), \(b=12\)
\(c^2=7^2+12^2=49+144=193\)
\(c=\sqrt{193}\approx13.89\)
(v) \(a=1.5\), \(b=3.5\)
\(c^2=1.5^2+3.5^2=2.25+12.25=14.5\)
\(c=\sqrt{14.5}\approx3.81\)
(ii) \(c=\sqrt{208}=4\sqrt{13}\approx14.42\)
(iii) \(b=12\)
(iv) \(c=\sqrt{193}\approx13.89\)
(v) \(c=\sqrt{14.5}\approx3.81\)
MTMath Talk — List down all the Baudhāyana triples with numbers less than or equal to 20.
We need positive integers \(a,b,c\) with \(a^2+b^2=c^2\) and all three numbers at most 20.
The hypotenuse \(c\) is the largest number, so we check the values of \(c\) up to 20.
For each \(c\) from 5 to 20, try \(a=1,2,3,\dots\) and check whether \(c^2-a^2\) is a perfect square \(b^2\) with \(b\ge a\).
| c | \(c^2\) | Pairs \((a,b)\) with \(a\le b\) and \(a^2+b^2=c^2\) |
|---|---|---|
| 5 | 25 | \((3,4)\) |
| 6 | 36 | none |
| 7 | 49 | none |
| 8 | 64 | none |
| 9 | 81 | none |
| 10 | 100 | \((6,8)\) |
| 11 | 121 | none |
| 12 | 144 | none |
| 13 | 169 | \((5,12)\) |
| 14 | 196 | none |
| 15 | 225 | \((9,12)\) |
| 16 | 256 | none |
| 17 | 289 | \((8,15)\) |
| 18 | 324 | none |
| 19 | 361 | none |
| 20 | 400 | \((12,16)\) |
The values that work are \(c=5,10,13,15,17,20\):
\(3^2+4^2=9+16=25=5^2\), so \((3,4,5)\)
\(6^2+8^2=36+64=100=10^2\), so \((6,8,10)\)
\(5^2+12^2=25+144=169=13^2\), so \((5,12,13)\)
\(9^2+12^2=81+144=225=15^2\), so \((9,12,15)\)
\(8^2+15^2=64+225=289=17^2\), so \((8,15,17)\)
\(12^2+16^2=144+256=400=20^2\), so \((12,16,20)\)
Note: the textbook’s printed list on this page shows only (3, 4, 5), (6, 8, 10), (9, 12, 15) and (12, 16, 20). The triples (5, 12, 13) and (8, 15, 17) also have all three numbers at most 20, and the textbook itself uses them on the next page.
?In-text Question — Is (30, 40, 50) a Baudhāyana triple? Is (300, 400, 500) a Baudhāyana triple?
(30, 40, 50):
\(30^2+40^2=900+1600=2500\)
\(50^2=2500\)
So \(30^2+40^2=50^2\).
(300, 400, 500):
\(300^2+400^2=90000+160000=250000\)
\(500^2=250000\)
So \(300^2+400^2=500^2\).
?In-text Question — Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive Baudhāyana triples with numbers less than or equal to 20?
\(5^2+12^2=25+144=169=13^2\), so (5, 12, 13) is a Baudhāyana triple.
The only common factor of 5, 12 and 13 is 1 (13 is a prime number and does not divide 5 or 12).
So (5, 12, 13) is a primitive Baudhāyana triple.
The Baudhāyana triples with numbers at most 20 are (3, 4, 5), (5, 12, 13), (6, 8, 10), (8, 15, 17), (9, 12, 15), (12, 16, 20).
(6, 8, 10) has common factor 2, (9, 12, 15) has common factor 3 and (12, 16, 20) has common factor 4.
These are not primitive.
(3, 4, 5) and (8, 15, 17) have no common factor greater than 1.
These are primitive.
?In-text Question — Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?
The scaled version of \((a,b,c)\) is \((ka,kb,kc)\), where \(k\) is a positive integer.
We take \(k=2,3,4,5,6\).
| \(k\) | (3, 4, 5) | (5, 12, 13) | (8, 15, 17) |
|---|---|---|---|
| 2 | (6, 8, 10) | (10, 24, 26) | (16, 30, 34) |
| 3 | (9, 12, 15) | (15, 36, 39) | (24, 45, 51) |
| 4 | (12, 16, 20) | (20, 48, 52) | (32, 60, 68) |
| 5 | (15, 20, 25) | (25, 60, 65) | (40, 75, 85) |
| 6 | (18, 24, 30) | (30, 72, 78) | (48, 90, 102) |
In each scaled version \((ka,kb,kc)\), the number \(k\) is a common factor of all three numbers.
Since \(k\gt1\), the triple has a common factor greater than 1.
?In-text Question — If \((a, b, c)\) is non-primitive, and the integers have \(f\) — greater than 1 — as a common factor, then is \(\left(\frac af,\frac bf,\frac cf\right)\) a Baudhāyana triple? Check this statement for (9, 12, 15). Justify this statement.
Check for (9, 12, 15):
The common factor is \(f=3\).
\(\left(\dfrac93,\dfrac{12}{3},\dfrac{15}{3}\right)=(3,4,5)\)
\(3^2+4^2=9+16=25=5^2\), so \((3,4,5)\) is a Baudhāyana triple.
Justification:
\((a,b,c)\) is a Baudhāyana triple, so \(a^2+b^2=c^2\).
\(\left(\dfrac af\right)^2+\left(\dfrac bf\right)^2=\dfrac{a^2}{f^2}+\dfrac{b^2}{f^2}\)
\(=\dfrac{a^2+b^2}{f^2}\)
\(=\dfrac{c^2}{f^2}\)
\(=\left(\dfrac cf\right)^2\)
Since \(f\) is a common factor, \(\dfrac af,\dfrac bf,\dfrac cf\) are integers.
Figure it Out — Set 3
Three questions from page 50 of the textbook.
1Find 5 more Baudhāyana triples using this idea.
The idea: if the \(n\)th odd number \(2n-1\) is a square, say \(2n-1=m^2\), then \((n-1)^2+(2n-1)=n^2\), that is \((n-1)^2+m^2=n^2\).
So \((n-1,\ m,\ n)\) is a Baudhāyana triple.
We take the odd squares \(49,81,121,169,225\) (the squares of 7, 9, 11, 13, 15).
| Odd square \(2n-1\) | \(n\) | \((n-1)^2+(2n-1)=n^2\) | Triple |
|---|---|---|---|
| 49 | 25 | \(24^2+7^2=25^2\) | (7, 24, 25) |
| 81 | 41 | \(40^2+9^2=41^2\) | (9, 40, 41) |
| 121 | 61 | \(60^2+11^2=61^2\) | (11, 60, 61) |
| 169 | 85 | \(84^2+13^2=85^2\) | (13, 84, 85) |
| 225 | 113 | \(112^2+15^2=113^2\) | (15, 112, 113) |
Check: \(24^2+7^2=576+49=625=25^2\).
In the same way, \(40^2+9^2=1600+81=1681=41^2\).
2Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]
The triples from this method are \((n-1,\ m,\ n)\).
The hypotenuse is \(n\) and one of the smaller sides is \(n-1\), which is one less than \(n\).
Suppose a number \(f\) divides both \(n-1\) and \(n\).
Then \(f\) divides their difference \(n-(n-1)=1\).
So \(f=1\).
Therefore the triple has no common factor greater than 1.
3Math Talk — Are there primitive triples that cannot be obtained through this method? If yes, give examples.
This method gives only triples in which one leg is exactly 1 less than the hypotenuse.
Consider (8, 15, 17).
\(8^2+15^2=64+225=289=17^2\), so it is a Baudhāyana triple.
It has no common factor greater than 1, so it is primitive.
The hypotenuse is 17, but the legs are 8 and 15, which are 9 and 2 less than 17.
Neither is 1 less.
So (8, 15, 17) cannot be obtained by this method.
Similarly, \(12^2+35^2=144+1225=1369=37^2\), and (12, 35, 37) is primitive, but \(37-35=2\) and \(37-12=25\).
Figure it Out — Set 4
Nine questions from pages 52–54 of the textbook.
1Find the diagonal of a square with sidelength 5 cm.
The diagonal \(d\) is the hypotenuse of a right triangle whose perpendicular sides are the sides of the square, each 5 cm.
\(d^2=5^2+5^2\)
\(d^2=25+25=50\)
\(d=\sqrt{50}=5\sqrt2\)
\(7^2=49\) and \(7.1^2=50.41\), so \(d\) is between 7 and 7.1 cm.
2Find the missing sidelengths in the following right triangles:
In each triangle the right angle is marked, so the hypotenuse is the side opposite to it.
Use \(a^2+b^2=c^2\).
(i) perpendicular sides 7 and 9
\(c^2=7^2+9^2=49+81=130\)
\(c=\sqrt{130}\approx11.40\)
(ii) perpendicular sides 4 and 10
\(c^2=4^2+10^2=16+100=116\)
\(c=\sqrt{116}=2\sqrt{29}\approx10.77\)
(iii) one side 40 and hypotenuse 41
\(b^2=41^2-40^2=1681-1600=81\)
\(b=9\)
(iv) one side 10 and hypotenuse \(\sqrt{200}\)
\(b^2=(\sqrt{200})^2-10^2=200-100=100\)
\(b=10\)
(v) perpendicular sides 10 and \(\sqrt{150}\)
\(c^2=10^2+(\sqrt{150})^2=100+150=250\)
\(c=\sqrt{250}=5\sqrt{10}\approx15.81\)
(vi) one side 27 and hypotenuse 45
\(b^2=45^2-27^2=2025-729=1296\)
\(b=36\)
(ii) \(\sqrt{116}=2\sqrt{29}\approx10.77\)
(iii) 9
(iv) 10
(v) \(\sqrt{250}=5\sqrt{10}\approx15.81\)
(vi) 36
3Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
The diagonals of a rhombus bisect each other at right angles.
So the rhombus is divided into 4 congruent right triangles.
The perpendicular sides of each triangle are \(\dfrac{24}{2}=12\) and \(\dfrac{70}{2}=35\).
The side \(s\) of the rhombus is the hypotenuse of this triangle.
\(s^2=12^2+35^2\)
\(s^2=144+1225=1369\)
\(s=\sqrt{1369}=37\)
4Is the hypotenuse the longest side of a right triangle? Justify your answer.
Let the perpendicular sides be \(a\) and \(b\), and the hypotenuse \(c\).
By Baudhāyana’s Theorem, \(c^2=a^2+b^2\).
\(b^2\gt0\), so \(c^2=a^2+b^2\gt a^2\).
So \(c\gt a\), as the lengths are positive.
In the same way, \(a^2\gt0\), so \(c^2\gt b^2\) and \(c\gt b\).
5True or False — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.
Let \((a,b,c)\) be any Baudhāyana triple and let \(f\) be the greatest common factor of \(a\), \(b\) and \(c\).
Case 1: \(f=1\).
Then the triple has no common factor greater than 1, so it is primitive.
Case 2: \(f\gt1\).
Then \(\left(\dfrac af,\dfrac bf,\dfrac cf\right)\) is also a Baudhāyana triple.
It has no common factor greater than 1, because \(f\) was the greatest common factor.
So it is primitive.
And \((a,b,c)=\left(f\times\dfrac af,\ f\times\dfrac bf,\ f\times\dfrac cf\right)\), which is a scaled version of this primitive triple.
For example, \((9,12,15)=3\times(3,4,5)\).
6Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
The diagonal \(d\) of a rectangle with sides \(a\) and \(b\) is the hypotenuse of a right triangle with perpendicular sides \(a\) and \(b\).
So \(d^2=a^2+b^2\), and \((a,b,d)\) is a Baudhāyana triple.
Take 5 Baudhāyana triples:
\(3^2+4^2=25=5^2\): rectangle \(3\times4\), diagonal 5
\(6^2+8^2=100=10^2\): rectangle \(6\times8\), diagonal 10
\(5^2+12^2=169=13^2\): rectangle \(5\times12\), diagonal 13
\(8^2+15^2=289=17^2\): rectangle \(8\times15\), diagonal 17
\(7^2+24^2=625=25^2\): rectangle \(7\times24\), diagonal 25
7Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
Difference of the areas \(=7^2-5^2=49-25=24\) square units.
We need a square of area 24, that is, of sidelength \(\sqrt{24}\).
Take a right triangle with hypotenuse 7 and one perpendicular side 5.
By Baudhāyana’s Theorem, \(5^2+b^2=7^2\), so \(b^2=49-25=24\).
So the other perpendicular side is \(b=\sqrt{24}\).
Construction:
Draw a line segment AB of length 5 units.
At B draw a line BX perpendicular to AB.
With A as centre and radius 7 units, draw an arc cutting BX at C.
Then \(BC=\sqrt{7^2-5^2}=\sqrt{24}\).
At B draw a perpendicular to BC, and with centre B and radius BC cut it at D.
With centres C and D and the same radius BC, draw arcs that meet at E.
Join DE and CE.
BCED is the required square, of side \(\sqrt{24}\) and area 24.
8Math Talk / Try This — (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq.units, and (d) 5 sq. unit?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
(i) A side of the square joins two dots.
Suppose it goes \(p\) units across and \(q\) units up (\(p\) and \(q\) are whole numbers).
By Baudhāyana’s Theorem, (side)\(^2=p^2+q^2\), and the area of the square \(=p^2+q^2\).
(a) \(p=1,\ q=1\): area \(=1+1=2\).
Yes, a square of area 2 is possible.
(b) Area \(=3\) needs \(p^2+q^2=3\).
\(p,q\) can only be 0 or 1 (since \(2^2=4\gt3\)), giving \(0,1,2\).
So 3 is not possible.
No square of area 3.
(c) \(p=2,\ q=0\): area \(=4+0=4\).
Yes, a square of area 4 is possible.
(d) \(p=1,\ q=2\): area \(=1+4=5\).
Yes, a square of area 5 is possible.
(ii) By the same reasoning, the area of any square on the grid is \(p^2+q^2\), where \(p\) and \(q\) are whole numbers, not both 0.
Conversely, for any such \(p\) and \(q\), the dots \((0,0)\), \((p,q)\), \((p-q,\ p+q)\), \((-q,\ p)\) are the vertices of a square of area \(p^2+q^2\).
So the possible areas are the numbers that are the sum of two squares.
\(1=1^2+0^2,\ 2=1^2+1^2,\ 4=2^2+0^2,\ 5=1^2+2^2,\ 8=2^2+2^2,\ 9=3^2+0^2,\ 10=1^2+3^2,\ 13=2^2+3^2,\ 16,\ 17,\ 18,\ 20,\ 25,\ \ldots\)
Numbers such as 3, 6, 7, 11, 12, 14, 15, 19 are not the sum of two squares, so no grid square has these areas.
(ii) The possible areas are the numbers of the form \(p^2+q^2\) (the sum of two squares): 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, …
9Try This — Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]
Let ABC be the equilateral triangle with \(AB=BC=CA=6\), and let AD be the altitude from A to BC.
AD bisects BC:
In \(\triangle ABD\) and \(\triangle ACD\):
\(\angle ADB=\angle ADC=90^\circ\)
\(AB=AC=6\) (hypotenuses)
\(AD=AD\) (common side)
So \(\triangle ABD\cong\triangle ACD\) (RHS rule), hence \(BD=DC=\dfrac62=3\).
Height: in the right triangle ABD, \(AD^2+BD^2=AB^2\).
\(AD^2=6^2-3^2=36-9=27\)
\(AD=\sqrt{27}=3\sqrt3\)
Area: \(\dfrac12\times BC\times AD=\dfrac12\times6\times3\sqrt3=9\sqrt3\).
Since \(\sqrt3\approx1.732\), the area \(\approx9\times1.732\approx15.59\).
Puzzle Time — Find the Colours!
Find the Colours!
There are 3 closed boxes — one containing only red balls, the second containing only blue balls and the third containing only green balls. The boxes are labelled RED, BLUE and GREEN such that ‘no’ box has the correct label. We need to find which label goes with which box. How can this be done if we are allowed to open only one box?
Solution: open the box labelled GREEN.
No box has the correct label, so the box labelled GREEN does not contain green balls.
So it contains either red balls or blue balls.
Case 1: the GREEN-labelled box contains red balls.
The BLUE-labelled box cannot contain blue balls (wrong label), and the red balls are already taken. So it contains green balls.
The remaining box, labelled RED, contains blue balls.
Case 2: the GREEN-labelled box contains blue balls.
The RED-labelled box cannot contain red balls (wrong label), and the blue balls are already taken. So it contains green balls.
The remaining box, labelled BLUE, contains red balls.
| The GREEN-labelled box holds | RED-labelled box holds | BLUE-labelled box holds |
|---|---|---|
| red | blue | green |
| blue | green | red |
Continue with Part 2, Chapter 3
Move on to the next Part 2 chapter next.