Figure it Out
Q1. 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. [Diagram shows a square cut into 4 pieces: 1, 2, 3, 4 along the diagonals] Can you arrange these pieces to create a square with double the area of either square?
Answer:
Yes, arrange the 4 pieces (which are right-angled isosceles triangles) around a central empty space. The central space will be a square of the same size as the original. The outer square formed by the outer boundaries of the 4 pieces will have double the area of the original square. Specifically, place the triangles so that their hypotenuses form the outer boundary of a larger square, and their legs form the sides of the inner square.
Q2. The 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
Answer:
For an isosceles right triangle with equal sides a, the hypotenuse c is given by c = a√2. Using √2 ≈ 1.4142...
(i) a = 3: c = 3√2 ≈ 4.2426... Bounds: 4.2 < c < 4.3 (since 4.2² = 17.64, 4.3² = 18.49, and 18 = (3√2)² = 18, so 4.24² = 17.9776, 4.25² = 18.0625, so 4.24 < c < 4.25)
(ii) a = 4: c = 4√2 ≈ 5.65685... Bounds: 5.6 < c < 5.7 (5.65² = 31.9225, 5.66² = 32.0356, and (4√2)² = 32, so 5.65 < c < 5.66)
(iii) a = 6: c = 6√2 ≈ 8.48528... Bounds: 8.4 < c < 8.5 (8.48² = 71.9104, 8.49² = 72.0801, and (6√2)² = 72, so 8.48 < c < 8.49)
(iv) a = 8: c = 8√2 ≈ 11.3137... Bounds: 11.3 < c < 11.4 (11.31² = 127.9161, 11.32² = 128.1424, and (8√2)² = 128, so 11.31 < c < 11.32)
(v) a = 9: c = 9√2 ≈ 12.7279... Bounds: 12.7 < c < 12.8 (12.72² = 161.7984, 12.73² = 162.0529, and (9√2)² = 162, so 12.72 < c < 12.73)
Q3. The 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.]
Answer:
Let the equal sides have length a. The square formed by two such right triangles has area = a² (since two triangles form a square of side a). Alternatively, the square on the hypotenuse has area c² = 10² = 100. This square has double the area of the square on the equal sides (as per Baudhayana's method for isosceles right triangles). So, 2 × (area of square on side a) = 100, so area of square on side a = 50. Thus a² = 50, so a = √50 = 5√2 ≈ 7.07. So the other two sides are each of length 5√2 units.
Q1 (second set). If 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 Baudh?yana's Theorem.
Answer:
Draw a right-angled triangle with legs 5 cm and 12 cm. Measure the hypotenuse; it should be about 13 cm. Using Baudhayana's Theorem: a² + b² = c², so 5² + 12² = 25 + 144 = 169 = c². Therefore c = √169 = 13 cm.
Q2 (second set). If 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 Baudh?yana's Theorem.
Answer:
Draw a right-angled triangle with one leg 8 cm and hypotenuse 17 cm. Measure the other leg; it should be about 15 cm. Using Baudhayana's Theorem: a² + b² = c². Let the unknown side be b. Then 8² + b² = 17², so 64 + b² = 289, so b² = 289 - 64 = 225, so b = √225 = 15 cm.
Q3 (second set). Using 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? (Baudh?yana's ?ulb... [text truncated])
Answer:
For triple area: Start with the given square of side a. Construct a square with area double (using diagonal method) to get side a√2. Now combine the original square (area a²) with this double square (area 2a²) using Baudhayana's method for combining different squares. Make a right triangle with legs a and a√2. The hypotenuse will be √(a² + 2a²) = √(3a²) = a√3. The square on this hypotenuse has area 3a², which is triple the original area.
For five times area: Combine the double square (side a√2, area 2a²) with the triple square (side a√3, area 3a²). Make a right triangle with legs a√2 and a√3. The hypotenuse will be √(2a² + 3a²) = √(5a²) = a√5. The square on this hypotenuse has area 5a², which is five times the original area.
Alternatively, for five times, combine the original square (area a²) with the square of side 2a (area 4a²) using a right triangle with legs a and 2a, giving hypotenuse a√5.
Q4. Let 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
Answer:
(i) a=5, b=7: c² = 5²+7² = 25+49=74 → c=√74 ≈ 8.6
(ii) a=8, b=12: c² = 64+144=208 → c=√208 = 4√13 ≈ 14.42
(iii) a=9, c=15: b² = c² - a² = 225-81=144 → b=12
(iv) a=7, b=12: c² = 49+144=193 → c=√193 ≈ 13.89
(v) a=1.5, b=3.5: c² = (1.5)²+(3.5)² = 2.25+12.25=14.5 → c=√14.5 ≈ 3.81
Q1. Find 5 more Baudhāyana triples using this idea.
Answer:
Using the method of odd squares: If the nth odd number (2n-1) is a perfect square, we get (n-1)² + (2n-1)²? Actually from (n-1)² + (2n-1) = n² where (2n-1) is a square. Let (2n-1)=k². Examples:
1) 49 is 7², 49=2×25-1 → n=25: (24)² + 7² = 25² → (24,7,25)
2) 81 is 9², 81=2×41-1 → n=41: (40)² + 9² = 41² → (40,9,41)
3) 121 is 11², 121=2×61-1 → n=61: (60)² + 11² = 61² → (60,11,61)
4) 169 is 13², 169=2×85-1 → n=85: (84)² + 13² = 85² → (84,13,85)
5) 225 is 15², 225=2×113-1 → n=113: (112)² + 15² = 113² → (112,15,113)
Q2. Does 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.]
Answer:
Yes, this method can yield non-primitive triples. For example, when the odd square is 9 (which is 3²), we get (4,3,5). But if we take a scaled odd square like 225 = 15² = 2×113-1, we get (112,15,113). The triple (112,15,113) has gcd(112,15,113)=1, so it's primitive. However, if the odd square chosen is a multiple of a smaller odd square squared? Actually, the method always gives a triple where the smaller sides are (n-1) and √(2n-1), and hypotenuse = n, so one leg is one less than hypotenuse. Non-primitive triples occur when n-1 and √(2n-1) share a common factor. Example: For odd square 25 = 5², n=13 gives (12,5,13) - primitive. For odd square 49 = 7², n=25 gives (24,7,25) - primitive. Odd square 225 = 15², n=113 gives (112,15,113) - gcd is 1. Actually all triples from this method are primitive because if a common factor d divides both (n-1) and √(2n-1), then d divides their difference? Check: n and n-1 are coprime, so gcd(n-1, n)=1. Since √(2n-1) < n, the triple is primitive. So this method yields only primitive triples.
Q3. Are there primitive triples that cannot be obtained through this method? If yes, give examples.
Answer:
Yes. This method only generates primitive triples where the hypotenuse is one more than one of the legs (since (n-1)² + (√(2n-1))² = n²). Many primitive triples do not have this property. Example: (8, 15, 17) - here 17 is not one more than 8 or 15. (20, 21, 29) - 29 is not one more than 20 or 21. (9, 40, 41) has 41 = 40+1, so this could be obtained (n=41, odd square 81=9²). But (5,12,13) works (13=12+1). Triples like (8,15,17) cannot be obtained by this method.
Q1. Find the diagonal of a square with sidelength 5 cm.
Answer:
Diagonal of a square = side × √2 = 5√2 cm ≈ 7.07 cm.
Q2. Find the missing sidelengths in the following right triangles:
Answer:
This question refers to a diagram with triangles. From the text: There are triangles labeled: 7, 9, 4, 10, 40, 41, 27, 45, 200, 10, 150, 10. Assuming these are side lengths:
- Triangle with sides 7 and 9: hypotenuse = √(49+81) = √130 ≈ 11.4
- Triangle with sides 4 and 10: hypotenuse = √(16+100) = √116 = 2√29 ≈ 10.77
- Triangle with sides 40 and 41: third side = √(1681-1600) = √81 = 9
- Triangle with sides 27 and 45: third side = √(2025-729) = √1296 = 36
- Triangle with sides 200 and 10: third side = √(40000-100) = √39900 ≈ 199.75
- Triangle with sides 150 and 10: third side = √(22500-100) = √22400 ≈ 149.67
Q3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Answer:
In a rhombus, diagonals bisect each other at right angles. Half of diagonals: 12 and 35. Side = √(12² + 35²) = √(144 + 1225) = √1369 = 37 units.
Q4. Is the hypotenuse the longest side of a right triangle? Justify your answer.
Answer:
Yes, the hypotenuse is the longest side. By Baudhāyana's theorem, a² + b² = c². Since a² and b² are positive, c² > a² and c² > b², so c > a and c > b.
Q5. True or False? — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.
Answer:
True. Any Baudhāyana triple (a, b, c) can be divided by the greatest common divisor (GCD) of a, b, c to get a primitive triple. Conversely, multiplying a primitive triple by any positive integer gives a non-primitive triple.
Q6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
Answer:
A rectangle with sides a, b and diagonal d satisfies a² + b² = d². Examples:
1) 3, 4, 5
2) 6, 8, 10
3) 5, 12, 13
4) 8, 15, 17
5) 7, 24, 25
Q7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
Answer:
Area difference = 7² - 5² = 49 - 25 = 24 sq units. So construct a square with area 24 sq units, meaning side = √24 = 2√6 ≈ 4.9 units. Using geometry: Draw a right triangle with hypotenuse 7 and one leg 5; the other leg is √(7²-5²) = √24. Construct a square on that leg.
Q8. (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?
Answer:
(i) (a) Area 2: Square with side √2. Take a square with vertices at (0,0), (1,1), (0,2), (-1,1) on a grid. (b) Area 3: side √3. This is not possible on integer grid because √3 cannot be expressed as √(a²+b²) with integers a,b? Actually 1²+1²=2, 2²+1²=5. No two perfect squares sum to 3. So area 3 is impossible. (c) Area 4: side 2. Square with vertices (0,0), (2,0), (2,2), (0,2). (d) Area 5: side √5 = √(2²+1²). Square with vertices (0,0), (2,1), (1,3), (-1,2).
(ii) Possible integer areas are sums of two squares: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50, ... (numbers that can be expressed as a²+b²).
Q9. 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.]
Answer:
In equilateral triangle of side 6, altitude bisects base. Half base = 3. Height h = √(6² - 3²) = √(36-9) = √27 = 3√3. Area = (1/2) × base × height = (1/2) × 6 × 3√3 = 9√3 sq units ≈ 15.59 sq units.
Try This
Q1. 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?
Answer:
Open the box labeled GREEN. Since no label is correct, this box does not contain green balls. It must contain either red or blue balls. If you find a red ball, then the GREEN-labeled box actually contains red balls. The box labeled RED cannot contain red (since no label is correct), so it must contain blue or green. But green cannot be in GREEN-labeled (since we found red there), so RED-labeled must contain green. Then BLUE-labeled contains blue. Similarly, if you find a blue ball in the GREEN-labeled box, then RED-labeled must contain green and BLUE-labeled contains red.
Math Talk
Q1. List down all the Baudhāyana triples with numbers less than or equal to 20.
Answer:
Baudhāyana triples (a,b,c) with a,b,c ≤ 20: (3,4,5), (5,12,13), (6,8,10), (8,15,17), (9,12,15), (12,16,20). Also (15,8,17) counts. (20,21,29) is out since 29>20. So primitive ones: (3,4,5), (5,12,13), (8,15,17). Non-primitive scaled versions: (6,8,10), (9,12,15), (12,16,20).
Q2. Is there an unending sequence of Baudhāyana triples?
Answer:
Yes, since (3k, 4k, 5k) gives infinitely many triples for k=1,2,3,...
Q3. Is (30, 40, 50) a Baudhāyana triple?
Answer:
Yes, since 30²+40² = 900+1600 = 2500 = 50². It is 10×(3,4,5).
Q4. Is (300, 400, 500) a Baudhāyana triple?
Answer:
Yes, since 300²+400² = 90000+160000 = 250000 = 500². It is 100×(3,4,5).
Q5. Do you see any pattern among them?
Answer:
Yes, (6,8,10), (9,12,15), (12,16,20) are all multiples of (3,4,5) by 2, 3, and 4 respectively.
Q6. Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive Baudhāyana triples with numbers less than or equal to 20?
Answer:
Yes, (5,12,13) is primitive because gcd(5,12,13)=1. Other primitive triples ≤20: (3,4,5), (8,15,17). Also (15,8,17). (20,21,29) has 29>20 so not included.
Q7. Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?
Answer:
For (3,4,5): (6,8,10), (9,12,15), (12,16,20), (15,20,25), (18,24,30). For (5,12,13): (10,24,26), (15,36,39), (20,48,52), (25,60,65), (30,72,78). For (8,15,17): (16,30,34), (24,45,51), (32,60,68), (40,75,85), (48,90,102). None of these scaled versions are primitive because they have a common factor greater than 1.
Q8. If (a, b, c) is non-primitive, and the integers have f — greater than 1 — as a common factor, then is (a/f, b/f, c/f) a Baudhāyana triple? Check this statement for (9, 12, 15). Justify this statement.
Answer:
Yes. For (9,12,15), f=3. (3,4,5) is a Baudhāyana triple since 3²+4²=9+16=25=5². In general, if (a,b,c) has common factor f, then a=fa', b=fb', c=fc' with integers a',b',c'. Since a²+b²=c², we have f²a'²+f²b'²=f²c'² → a'²+b'²=c'², so (a',b',c') is a Baudhāyana triple.