CLASS 8 · GANITA PRAKASH PART II · CHAPTER 2

The Baudhayana-Pythagoras Theorem

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39 worked answers

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39 worked answers · short steps first, explanations when you need them

Figure it Out — Isosceles triangles

Textbook pages 39–40

Question 1

Arrange the four equal right triangles cut from two identical squares into one square.

Steps only

Place the four right angles together at the centre.

new side = a√2; new area = (a√2)² = 2a²

Four equal triangles around O form the new square · Original Learnivelle diagram. Equal scale on both axes.

O(0, 0) · A(1, 0) · B(0, 1) · C(-1, 0) · D(0, -1)

Answer

A square whose side is the diagonal of an original square.

Question 2(1)

The equal legs are 3 units. Find the hypotenuse and bounds to one decimal place.

Steps only

c² = 3² + 3² = 18

c = 3√2

4.2² < 18 < 4.3²

4.2 < c < 4.3

Answer

3√2 units; 4.2 < c < 4.3.

Question 2(2)

The equal legs are 4 units. Find the hypotenuse and bounds to one decimal place.

Steps only

c² = 4² + 4² = 32

c = 4√2

5.6² < 32 < 5.7²

5.6 < c < 5.7

Answer

4√2 units; 5.6 < c < 5.7.

Question 2(3)

The equal legs are 6 units. Find the hypotenuse and bounds to one decimal place.

Steps only

c² = 6² + 6² = 72

c = 6√2

8.4² < 72 < 8.5²

8.4 < c < 8.5

Answer

6√2 units; 8.4 < c < 8.5.

Question 2(4)

The equal legs are 8 units. Find the hypotenuse and bounds to one decimal place.

Steps only

c² = 8² + 8² = 128

c = 8√2

11.3² < 128 < 11.4²

11.3 < c < 11.4

Answer

8√2 units; 11.3 < c < 11.4.

Question 2(5)

The equal legs are 9 units. Find the hypotenuse and bounds to one decimal place.

Steps only

c² = 9² + 9² = 162

c = 9√2

12.7² < 162 < 12.8²

12.7 < c < 12.8

Answer

9√2 units; 12.7 < c < 12.8.

Question 3

An isosceles right triangle has hypotenuse 10 units. Find its other sides.

Steps only

a² + a² = 10²

2a² = 100; a² = 50

a = √50 = 5√2

Answer

Each other side is 5√2 units.

Figure it Out — Using the theorem

Textbook pages 47

Question 1

Find the hypotenuse of a right triangle with legs 5 cm and 12 cm.

Steps only

c² = 5² + 12² = 169

c = 13 cm

Right triangle with legs 5 cm and 12 cm · Original Learnivelle diagram. Equal scale on both axes.

O(0, 0) · A(12, 0) · B(0, 5)

Answer

13 cm.

Question 2

Find the other leg of a right triangle with one leg 8 cm and hypotenuse 17 cm.

Steps only

b² = 17² − 8² = 225

b = 15 cm

Right triangle with legs 8 cm and 15 cm · Original Learnivelle diagram. Equal scale on both axes.

O(0, 0) · A(15, 0) · B(0, 8)

Answer

15 cm.

Question 3(i)

Construct a square with three times the area of a square of side a.

Steps only

Construct the original diagonal a√2.

Make a right triangle with legs a and a√2.

c² = a² + (a√2)² = 3a²

Construct the new square on hypotenuse c.

Answer

The new square has side a√3 and area 3a².

Question 3(ii)

Construct a square with five times the area of a square of side a.

Steps only

Make a right triangle with legs a and 2a.

c² = a² + (2a)² = 5a²

Construct a square on the hypotenuse.

Answer

The new square has side a√5 and area 5a².

Question 4(1)

Find the missing side when a = 5, b = 7; c is the hypotenuse.

Steps only

c² = 5² + 7² = 74

c = √74

Answer

c = √74 units.

Question 4(2)

Find the missing side when a = 8, b = 12; c is the hypotenuse.

Steps only

c² = 8² + 12² = 208

c = 4√13

Answer

c = 4√13 units.

Question 4(3)

Find the missing side when a = 9, c = 15; c is the hypotenuse.

Steps only

b² = 15² − 9² = 144

b = 12

Answer

b = 12 units.

Question 4(4)

Find the missing side when a = 7, b = 12; c is the hypotenuse.

Steps only

c² = 7² + 12² = 193

c = √193

Answer

c = √193 units.

Question 4(5)

Find the missing side when a = 1.5, b = 3.5; c is the hypotenuse.

Steps only

c² = 1.5² + 3.5² = 14.5

c = √58/2

Answer

c = √58/2 units.

Figure it Out — Integer triples

Textbook pages 50

Question 1

Give five more integer right-triangle triples using a leg one less than the hypotenuse.

Steps only

a = n; b = (n²−1)/2; c = (n²+1)/2, n odd

(11,60,61), (13,84,85), (15,112,113), (17,144,145), (19,180,181)

Answer

Five examples: (11,60,61), (13,84,85), (15,112,113), (17,144,145), (19,180,181).

Question 2

Can the preceding method produce a non-primitive triple?

Steps only

The two larger sides differ by 1.

gcd(b,c) = 1

No common factor greater than 1 can divide all three sides.

Answer

No; these triples are always primitive.

Question 3

Give primitive triples that are missed by the preceding method.

Steps only

8² + 15² = 17²; 17−15 = 2

20² + 21² = 29²; 29−21 = 8

Answer

Examples: (8,15,17) and (20,21,29).

Figure it Out — Further applications

Textbook pages 52–54

Question 1

Find the diagonal of a square of side 5 cm.

Steps only

d² = 5² + 5² = 50

d = 5√2 cm

Answer

5√2 cm.

Question 2(1)

Find the missing side in the triangle with legs 7 and 9.

Steps only

c² = 7²+9² = 130

missing length = √130

Answer

√130 units.

Question 2(2)

Find the missing side in the triangle with legs 4 and 10.

Steps only

c² = 4²+10² = 116

missing length = 2√29

Answer

2√29 units.

Question 2(3)

Find the missing side in the triangle with leg 40 and hypotenuse 41.

Steps only

b² = 41²−40² = 81

missing length = 9

Answer

9 units.

Question 2(4)

Find the missing side in the triangle with leg 10 and hypotenuse √200.

Steps only

b² = 200−10² = 100

missing length = 10

Answer

10 units.

Question 2(5)

Find the missing side in the triangle with legs 10 and √150.

Steps only

c² = 10²+(√150)² = 250

missing length = 5√10

Answer

5√10 units.

Question 2(6)

Find the missing side in the triangle with leg 27 and hypotenuse 45.

Steps only

b² = 45²−27² = 1296

missing length = 36

Answer

36 units.

Question 3

A rhombus has diagonals 24 and 70 units. Find its side.

Steps only

half-diagonals = 12, 35

s² = 12²+35² = 1369

s = 37

Answer

37 units.

Question 4

Explain why the hypotenuse is the longest side of a right triangle.

Steps only

c² = a²+b² > a² and c² > b²

c > a and c > b

Answer

Yes; the hypotenuse is longer than either leg.

Question 5

Is every integer right-triangle triple primitive or a multiple of a primitive triple?

Steps only

g = gcd(a,b,c)

(a/g)²+(b/g)² = (c/g)²

gcd(a/g,b/g,c/g) = 1

Answer

True.

Question 6

Give five rectangles with integer side lengths and integer diagonals.

Steps only

(width, height, diagonal):

(3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29)

Answer

Examples: 3×4, 5×12, 8×15, 7×24 and 20×21 rectangles.

Question 7

Construct a square whose area is the positive difference between squares of sides 7 and 5 units.

Steps only

required area = 7²−5² = 24

b²+5² = 7²; b = √24 = 2√6

Answer

Construct the new square on a side of length 2√6 units.

Question 8(i)(a)

Can a square on integer grid points have area 2?

Steps only

Use the four labelled vertices in the diagram.

side² = 2; area = 2

An original grid square with area 2 · Original Learnivelle diagram. Equal scale on both axes.

A(0, 0) · B(1, 1) · C(0, 2) · D(-1, 1)

Answer

Yes; an example is shown on the grid.

Question 8(i)(c)

Can a square on integer grid points have area 4?

Steps only

Use the four labelled vertices in the diagram.

side² = 4; area = 4

An original grid square with area 4 · Original Learnivelle diagram. Equal scale on both axes.

A(0, 0) · B(2, 0) · C(2, 2) · D(0, 2)

Answer

Yes; an example is shown on the grid.

Question 8(i)(d)

Can a square on integer grid points have area 5?

Steps only

Use the four labelled vertices in the diagram.

side² = 5; area = 5

An original grid square with area 5 · Original Learnivelle diagram. Equal scale on both axes.

A(0, 0) · B(2, 1) · C(1, 3) · D(-1, 2)

Answer

Yes; an example is shown on the grid.

Question 8(i)(b)

Can a square on integer grid points have area 3 square units?

Steps only

area = p²+q², with p,q integers

p²,q² ∈ {0,1} when p²+q² = 3

No two such squares sum to 3.

Answer

No.

Question 8(ii)

Which integer areas are possible for squares whose vertices are grid points?

Steps only

area = p²+q², p,q ∈ ℤ, (p,q) ≠ (0,0)

examples: 1,2,4,5,8,9,10,13,16,17,…

Answer

Exactly the positive integers expressible as a sum of two integer squares.

Question 9

Find the area of an equilateral triangle of side 6 units.

Steps only

half-base = 3

h² = 6²−3² = 27; h = 3√3

area = (1/2)×6×3√3 = 9√3

Answer

9√3 square units.

Explore the ideas

Textbook pages 36, 48

Question Triples up to 20

List the positive right-triangle triples with every side at most 20.

Steps only

List each triangle once, with the smaller leg first.

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

Answer

The six triples are (3,4,5), (6,8,10), (5,12,13), (9,12,15), (8,15,17) and (12,16,20).

Question Square halving

How can you construct a square with half the area of a given square?

Steps only

Join the midpoints of all four sides.

new side² = (a/2)²+(a/2)² = a²/2

Answer

Join the side midpoints; the inner square has area a²/2.