Figure it Out — Applications of the Theorem — The Baudhāyana–Pythagoras Theorem

1. 12 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 52

   Find the diagonal of a square with sidelength 5 cm.

   Hint. A diagonal splits the square into two right triangles whose legs are the sides.

   Set it up. The diagonal of a square cuts it into two right-angled triangles. In each one the two legs are sides of the square, both 5 cm, and the diagonal is the hypotenuse.

   Apply the theorem. d² = 5² + 5² = 25 + 25 = 50 d = √50 = 5√2 ≈ 7.07 cm

   Bounds. 7.0² = 49 < 50 and 7.1² = 50.41 > 50, so the diagonal lies between 7.0 cm and 7.1 cm.

   Sense check. The diagonal must be longer than a side (5) but shorter than two sides walked end to end (10), and 7.07 sits comfortably between. In general the diagonal of any square is its side multiplied by √2 ≈ 1.414, which is the shortcut worth carrying forward.

   ✦ The diagonal is √50 = 5√2 ≈ 7.07 cm (between 7.0 cm and 7.1 cm).

2. 26 marksGanita Prakash Cl-8 Part 2, Figure it Out, pages 52-53

   Find the missing sidelengths in the six right triangles shown: (i) legs 7 and 9 (ii) legs 4 and 10 (iii) leg 40 with hypotenuse 41 (iv) leg 10 with hypotenuse √200 (v) legs 10 and √150 (vi) leg 27 with hypotenuse 45.

   Hint. For each figure, first find where the small square marking the right angle sits — the side facing it is the hypotenuse.

   Read each figure before calculating. The little square marks the right angle, and the side facing it is the hypotenuse. If both labelled sides touch the right angle they are legs and you add; if one labelled side faces it, that one is the hypotenuse and you subtract.

   (i) Legs 7 and 9 (right angle between them) c² = 49 + 81 = 130, so c = √130 ≈ 11.40. 11.4² = 129.96 and 11.5² = 132.25, therefore 11.4 < c < 11.5.

   (ii) Legs 4 and 10 (right angle between them) c² = 16 + 100 = 116, so c = √116 = 2√29 ≈ 10.77. 10.7² = 114.49 and 10.8² = 116.64, therefore 10.7 < c < 10.8.

   (iii) Leg 40, hypotenuse 41 b² = 41² − 40² = 1681 − 1600 = 81, so b = 9 exactly. A quicker route: 41² − 40² = (41 + 40)(41 − 40) = 81 × 1 = 81. This is the triple (9, 40, 41) from the odd-square method.

   (iv) Leg 10, hypotenuse √200 b² = (√200)² − 10² = 200 − 100 = 100, so b = 10 exactly. Both legs equal — the triangle is the isosceles right triangle of §2.3, and √200 = 10√2 is just the diagonal of a square of side 10.

   (v) Legs 10 and √150 (right angle between them) c² = 100 + 150 = 250, so c = √250 = 5√10 ≈ 15.81. 15.8² = 249.64 and 15.9² = 252.81, therefore 15.8 < c < 15.9.

   (vi) Leg 27, hypotenuse 45 b² = 45² − 27² = 2025 − 729 = 1296, so b = 36 exactly. Spot the common factor and it is instant: (27, 36, 45) = 9 × (3, 4, 5).

   A note on the surd-labelled sides. In (iv) and (v) the given length is written as a square root, which makes the arithmetic easier, not harder — squaring √200 simply gives 200. Do not convert to decimals first; you would lose exactness for no gain.

   ✦ (i) √130 ≈ 11.40 (ii) √116 = 2√29 ≈ 10.77 (iii) 9 (iv) 10 (v) √250 = 5√10 ≈ 15.81 (vi) 36

3. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

   Hint. The diagonals of a rhombus bisect each other at right angles — so half of each diagonal forms a right triangle with a side.

   Use the property that makes this a right-triangle problem. In a rhombus the two diagonals bisect each other at right angles. So the point where they cross splits each diagonal in half and creates four congruent right-angled triangles, and each side of the rhombus is the hypotenuse of one of them.

   Halve the diagonals. half of 24 = 12 half of 70 = 35

   Apply the theorem. s² = 12² + 35² = 144 + 1225 = 1369 Since 37 × 37 = 1369, we get s = 37.

   Sense check. All four sides of a rhombus are equal, so the answer is a single number for all of them. And 37 is longer than 35, as any hypotenuse must be — while being much shorter than half the perimeter of the enclosing rectangle, which is right for a slim rhombus like this one.

   Worth noticing. (12, 35, 37) is a primitive Baudhayana triple, and it is one that the odd-square method of §2.5 cannot produce, since 37 − 35 = 2 rather than 1.

   ✦ Each side of the rhombus is 37 units.

4. 42 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   Is the hypotenuse the longest side of a right triangle? Justify your answer.

   Hint. Compare c² with a² and with b² separately, using the fact that a and b are positive.

   Yes, always. Here is the justification, straight from the theorem.

   Let the legs be a and b and the hypotenuse c, so that a² + b² = c². A side of a triangle has positive length, so b² > 0. Therefore c² = a² + b² > a², which gives c > a. In the same way a² > 0, so c² = a² + b² > b², which gives c > b. Since c is greater than both a and b, the hypotenuse is the longest of the three sides.

   A second reason, from angles. The largest angle of a triangle faces the longest side. The angles of a triangle add to 180°, and one of them here is already 90°, so the other two add to 90° and each must be less than 90°. The right angle is therefore strictly the largest, so the side facing it — the hypotenuse — is strictly the longest.

   Why it is worth stating. This gives a free check on every answer in this chapter: if a calculation ever returns a "hypotenuse" shorter than a leg, a leg and the hypotenuse have been swapped in the formula.

   ✦ Yes. Since c² = a² + b² with a and b both positive, c² is greater than a² and greater than b², so c > a and c > b — the hypotenuse is always the longest side.

5. 53 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   True or False — Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple. Justify your answer.

   Hint. Take the largest number that divides all three, and see what is left after dividing by it.

   True. And here is why no triple can escape.

   Take any Baudhayana triple (a, b, c) and let f be the largest whole number that divides all three.

   Case 1: f = 1. Then the three numbers share no factor above 1, so (a, b, c) is primitive by definition. The statement holds.

   Case 2: f > 1. Dividing a² + b² = c² by f² gives (a/f)² + (b/f)² = (c/f)², so (a/f, b/f, c/f) is itself a Baudhayana triple. Moreover it must be primitive: if some d > 1 divided all three of a/f, b/f and c/f, then df would divide a, b and c, contradicting the choice of f as the largest such divisor. Finally, (a, b, c) = f × (a/f, b/f, c/f), so the original triple is a scaled version of that primitive triple. The statement holds here too.

   Both cases are covered, so every triple is either primitive or a scaled version of one.

   Illustration. (12, 16, 20) has largest common factor 4; dividing gives (3, 4, 5), which is primitive, and 12, 16, 20 = 4 × (3, 4, 5). Similarly (30, 72, 78) has largest common factor 6 and reduces to the primitive (5, 12, 13).

   ✦ True. Dividing any triple by the largest number common to all three leaves a primitive triple, and the original is that primitive triple scaled by the factor removed.

6. 62 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

   Hint. A rectangle's diagonal makes a right triangle with its two sides — so every Baudhayana triple is such a rectangle.

   The link to make. A diagonal cuts a rectangle into two right-angled triangles whose legs are the length and the breadth, with the diagonal as hypotenuse. So a rectangle has whole-number sides and a whole-number diagonal exactly when (length, breadth, diagonal) is a Baudhayana triple. Every triple in the chapter therefore hands over a rectangle ready-made.

   Five examples:

   Length	Breadth	Diagonal	Check
   4	3	5	16 + 9 = 25 ✓
   8	6	10	64 + 36 = 100 ✓
   12	5	13	144 + 25 = 169 ✓
   15	8	17	225 + 64 = 289 ✓
   24	7	25	576 + 49 = 625 ✓

   As many as you like. Since scaling a triple gives another triple, every one of these can be doubled, trebled and so on — 8 by 6 with diagonal 10 is just 4 by 3 doubled. So there are infinitely many such rectangles, and the five above are only the smallest handful.

   ✦ Five such rectangles are 4 × 3 (diagonal 5), 8 × 6 (diagonal 10), 12 × 5 (diagonal 13), 15 × 8 (diagonal 17) and 24 × 7 (diagonal 25).

7. 74 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

   Hint. Baudhayana's combining method run backwards: put the larger length as the hypotenuse instead of as a leg.

   Work out the target first. The two areas are 7² = 49 and 5² = 25, and the difference is 49 − 25 = 24. So the square to be built has area 24 and side √24 = 2√6 ≈ 4.90 (since 4.8² = 23.04 and 4.9² = 24.01, the side lies between 4.8 and 4.9). The subtraction has to go this way round, because an area cannot be negative.

   Turn subtraction into the theorem. Rearranging a² + b² = c² gives b² = c² − a². So if we can build a right triangle whose hypotenuse is 7 and one of whose legs is 5, then the other leg squared is exactly 49 − 25 = 24 — which is the side we want. Baudhayana's rule for combining squares becomes a rule for subtracting them simply by putting the longer length on the hypotenuse.

   The construction.

   1. Draw a segment AB of length 5 units.
   2. At A, draw a line perpendicular to AB (an "east-west / north-south" pair, in Baudhayana's language).
   3. With centre B and radius 7 units, draw an arc cutting that perpendicular at C.
   4. Then AC is the required length, since angle A is a right angle and so AC² = BC² − AB² = 49 − 25 = 24.
   5. Build a square on AC. Its area is 24 square units.

   Check. Area of the new square (24) + area of the 5-square (25) = 49 = area of the 7-square ✓ — the new square is precisely what must be added to the smaller square to make the larger one.

   ✦ Build a right triangle with hypotenuse 7 and one leg 5; the remaining leg has length √24 = 2√6 ≈ 4.90, and the square on it has area 49 − 25 = 24 square units.

8. 85 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 53

   (i) Using the dots of a grid as vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, (d) 5 sq. units? (ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

   Hint. A tilted square's side is the hypotenuse of a right triangle whose legs run along the grid lines.

   The key idea. Take any square whose corners are grid dots. Going from one corner to the next means moving p steps across and q steps up, for some whole numbers p and q. That step is the hypotenuse of a right triangle with legs p and q, so by the theorem side² = p² + q², and therefore area = p² + q². So a grid square can have area A exactly when A can be written as the sum of two squares of whole numbers.

   (i) Testing the four values. (a) Area 2 — take p = 1, q = 1: 1 + 1 = 2 ✓. Draw the tilted square with corners (1, 0), (2, 1), (1, 2), (0, 1). Possible. (b) Area 3 — we need p² + q² = 3. The only squares not exceeding 3 are 0 and 1, and 0 + 0, 0 + 1, 1 + 1 give 0, 1, 2 — never 3. Impossible. (c) Area 4 — take p = 2, q = 0: 4 + 0 = 4 ✓. This is the ordinary 2 × 2 square sitting square-on to the grid. Possible. (d) Area 5 — take p = 1, q = 2: 1 + 4 = 5 ✓. Draw the tilted square with corners (0, 0), (2, 1), (1, 3), (−1, 2). Possible.

   So three of the four can be drawn, and area 3 cannot — a square of area 3 would need side √3, and √3 is not the hypotenuse of any right triangle with whole-number legs.

   (ii) Which areas are possible on an unlimited grid? Exactly the numbers of the form p² + q². Working through small values of p and q: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, … and the whole numbers up to 30 that are not achievable are 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28, 30. Notice that every perfect square is there (take q = 0) and every double of a perfect square is there (take p = q), but plenty of numbers in between are missing.

   A remark for the curious. There is a complete rule, found by Fermat — the same mathematician who appears in §2.6. A whole number is a sum of two squares exactly when every prime factor of the form 4k + 3 (that is, 3, 7, 11, 19, 23, …) appears an even number of times. That is why 3, 7, 21 and 6 fail while 9 = 3 × 3 and 18 = 2 × 9 succeed. The rule is well beyond Class 8, but the grid experiment is what it grew out of.

   ✦ (i) Areas 2, 4 and 5 can be drawn; area 3 cannot. (ii) Exactly those whole numbers expressible as p² + q² for whole numbers p and q — 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, … — while 3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28 and 30 are impossible.

9. 94 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 54

   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.]

   Hint. Drop the altitude, prove the two halves are congruent, and then the theorem gives the height.

   Step 1 — show the altitude bisects the base. Let the triangle be ABC with AB = BC = CA = 6, and drop the altitude AD from A onto BC. This creates two triangles ABD and ACD. In them: · AB = AC (the triangle is equilateral) · AD is common · angle ADB = angle ADC = 90° (AD is an altitude) By RHS congruence, triangle ABD is congruent to triangle ACD. Corresponding sides are therefore equal, so BD = DC. Since BD + DC = 6, each is 3. The altitude bisects the base.

   Step 2 — find the height. Triangle ABD is right-angled at D with hypotenuse AB = 6 and base BD = 3, so AD² = 6² − 3² = 36 − 9 = 27 AD = √27 = 3√3 ≈ 5.196. (Bounds: 5.1² = 26.01 and 5.2² = 27.04, so 5.1 < AD < 5.2.)

   Step 3 — find the area. Area = ½ × base × height = ½ × 6 × 3√3 = 9√3 ≈ 15.59 square units. (Bounds: 15.5² = 240.25 and 15.6² = 243.36, while (9√3)² = 81 × 3 = 243, so 15.5 < area < 15.6.)

   Sense check. The triangle sits inside a 6 × 5.2 rectangle, of area about 31, and fills a little over half of it — 15.59 is about right. Note also that the height 5.196 is less than the side 6, as it must be, since the altitude is a leg while the side is the hypotenuse.

   The general formula this gives. Repeating the working with side a gives height a√3/2 and area a²√3/4 — a result used constantly in later classes, and it comes entirely from the Baudhayana-Pythagoras theorem.

   ✦ The height is √27 = 3√3 ≈ 5.196 units and the area is 9√3 ≈ 15.59 square units (between 15.5 and 15.6).

Solutions written by the tuition.in editorial team and checked against the NCERT Class 8 Mathematics textbook Ganita Prakash Part 2, Reprint 2026-27 (hegp202.pdf), where this is Chapter 2 (pages 33-54) — the ninth chapter of the Class 8 course. Like the rest of Part 2, this PDF carries NO printed answer key, so every answer here was derived from first principles and then independently recomputed in Python before being written — including all the √2-style one-decimal bounds, the six figure triangles, the rhombus side, the odd-square triple generator, and the complete list of Baudhāyana triples with all numbers at most 20. TWO POINTS WHERE THE BOOK'S OWN TEXT NEEDS CARE ARE FLAGGED IN PLACE: (1) on page 48 the book lists four triples with numbers at most 20 and says the list 'contains' them — the complete list has six, since (5, 12, 13) and (8, 15, 17) also qualify and are not multiples of (3, 4, 5); (2) the six right triangles in Figure it Out Q2 on pages 52-53 are labelled only in the printed figure, so each one's right-angle position was read directly off the rendered PDF page before solving, and the reading is stated in the solution so a student can check it against the book.. Questions are referenced from the NCERT textbook for identification.