Ganita Prakash | Part 2 | Class 8 | Ch 2 The Baudhayana-Pythagoras Theorem

Ganita Prakash | Part 2 | Class 8 | Ch 2  The Baudhayana-Pythagoras Theorem

Solutions

2.1 Doubling A Square, 2.2 Halving A Square & 2.3 Hypotenuse of an Isosceles Right Triangle

Figure It Out (Pages 39-40)

Question 1.
Earlier, we saw a method to create a square with double the area of a given square piece of 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?
Solution:
Given: Two identical squares
These two are cut diagonally, forming two equal triangles, as shown in the figure.

Thus, we have four identical triangles from two identical squares.
Area of two triangles = Area of the given square
Area of four triangles = double the area of the given square.

Question 2.
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
Solution:
If a is the length of two equal sides of an isosceles right triangle, then
hypotenuse = a2+a2−−−−−−√=2a2−−−√=a2–√

(i) a = 3
Hypotenuse = a√2 = 3√2 = 3×3×2−−−−−−−√ = √18
√18 lies between √16 and √25
√16 < √18 < √25
⇒ 4 < 3√2 < 5
In one decimal point
4.1² = 16.81, 4.2² = 17.64, 4.3² = 18.49
So, 4.2 < 3√2 < 4.3

(ii) a = 4
Hypotenuse = a√2 = 4√2 = √32
√25 < √32 < √36
⇒ 5 < 4√2 < 6
In one decimal point
5.1² = 26.01, 5.2² = 27.04, 5.3² = 28.09, 5.4² = 29.16, 5.5² = 30.25, 5.6² = 31.36, 5.7² = 32.49
So, 5.6 < 4√2 < 5.7

(iii) a = 6
Hypotenuse = a√2 = 6√2 = 6×6×2−−−−−−−√ = √72
√64 < √72 < √81
⇒ 8 < 6√2 < 9
In one decimal point
8.1² = 65.61, 8.2² = 67.24, 8.3² = 68.89, 8.4² = 70.56, 8.5² = 72.25
So, 8.4 < 6√2 < 8.5.

(iv) a = 8
Hypotenuse = a√2 = 8√2 = √128
√121 < √128 < √144
⇒ 11 < 8√2 < 12
In one decimal point
11.1² = 123.21, 11.2² = 125.44, 11.3² = 127.69, 11.4² = 129.96
So, 11.3 < 8√2 < 11.4.

(v) a = 9
Hypotenuse = a√2 = 9√2 = √162
√144 < √162 < √169
⇒ 12 < 9√2 < 13
In one decimal point
12.1² = 146.41, 12.2² = 148.84, 12.3² = 151.29, 12.4² = 153.76, 12.5² = 156.25, 12.6² = 158.76, 12.7² = 161.29, 12.8² = 163.84
So, 12.7 < 9√2 < 12.8

Question 3.
The hypotenuse of an isosceles right triangle is 10. What are its other two side lengths?
[Hint: Find the area of the square composed of two such right triangles.]
Solution:

If a is the length of two equal sides of an isosceles right triangle, then
a² + a² = 102
⇒ 2a² = 100
⇒ a² = 50
⇒ a = √50
⇒ a = 5√2

2.4 Combining Two Different Squares

Figure It Out (Page 47)

Question 1.
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 Baudhayana’s Theorem.
Solution:
Given AB = 5 cm
BC = 12 cm
AC = 13 cm (by measurement)

Now, using Baudhayana’s Theorem
AC² = AB² + BC²
= 5² + 12²
= 25 + 144
= 169
= 13²
∴ AC = 13 cm
Therefore, the hypotenuse of the right-angled triangle is 13 cm, according to Baudhayana’s Theorem.

Question 2.
If a right-angled triangle has a short side of length 8 cm and a 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 Baudhayana’s Theorem.
Solution:
Here, AB = 8 cm
AC = 17 cm
BC = 15 cm (by measurement)

Now, using Baudhayana’s Theorem
AB² + BC² = AC²
8² + BC² = 17²
BC² = 289 – 64
= 225
= 15²
∴ BC = 15 cm
Therefore, the other side of the right-angled triangle is 15 cm, which satisfies Baudhayana’s Theorem.

Question 3.
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? (Baudhayana’s Sulba-Sutra, Verse 1.10)
Solution:
(a) ABCD is a square with side a.
AC = a√2
ACEF is a rectangle with sides a√2 and a.
Now AE = a√3

AEGH is a square with a side of a√3
Then Ar AEGH = 3a²
Ar ABCD = a²
∴ Area of AEGH = 3 × Area of ABCD

(b) ABCD and CFED are squares with side ‘a’.
BE is a diagonal of the rectangle ABFE.
In rectangle ABFE
EF = a and BF = BC + CF = a + a = 2a

Using Baudhayana’s Theorem
BE² = EF² + BF²
= a² + (2a)²
= a² + 4a²
= 5a²
BE = √5 a
BEFG is a square with side BE.
Area BEFG = BE²
= (√5a)²
= 5a²
Then the area of BEFG = 5 × the area of ABCD

Question 4.
Let a, b, and c denote the lengths 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
Solution:
Here, c² = a² + b²

(i) Now, c² = 5² + 7²
= 25 + 49
= 74
⇒ c = √74

(ii) Now, c² = 8² + 12²
= 64 + 144
= 208
⇒ c = √208
= 2×2×2×2×13−−−−−−−−−−−−−−√
= 4√13

(iii) Here, 15² = 9² + b²
b² = 15² – 9²
= 225 – 81
= 144
⇒ b = √144
= 2×2×2×2×3×3−−−−−−−−−−−−−−−−−√
= 12

(iv) c² = 7² + 12²
= 49 + 144
= 193
⇒ c = √193

(v) c² = 1.5² + 3.5²
= 2.25 + 12.25
= 14.5
⇒ c = √14.5

2.5 Right-Triangles Having Integer Sidelengths

Figure It Out (Page 50)

Question 1.
Find 5 more Baudhayana triples using this idea.
Solution:
(1 + 3 + 5 + … + 47) + 49 = 25²
24² + 7² = 25² (24, 7, 25)
(1 + 3 + 5 + … + 79) + 81 = 41²
40² + 9² = 41² (40, 9, 41)
(1 + 3 + 5 + … + 119) + 121 = 61²
60² + 11² = 61² (60, 11, 61)
(1 + 3 + 5 + … + 167) + 169 = 85²
842 + 132 = 852 (84, 13, 85)
(1 + 3 + 5 + …+ 223) + 225 = 113²
112² + 15² = 113² (112, 15, 113)

Question 2.
Does this method yield non-primitive Baudhayana triples?
[Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]
Solution:
(24, 7, 25)
HCF of 24, 7, 25 is 1.
(40, 9, 41)
HCF of 40, 9, 41 is 1, etc.
The triples generated by the above method are primitive in nature.

Question 3.
Are there primitive triples that cannot be obtained through this method? If yes, give examples.
Solution:
In each of the above case we have taken the sum of the first ‘n’ odd numbers where ‘n’ is a perfect square.
We observe that the smallest number in the triple is always odd.
Consider triples such as (8, 15, 17), (16, 63, 65), etc.
Such triples cannot be generated by this method.

2.6 A Long-Standing Open Problem, 2.7 Further Applications of the Baudhayana-Pythagoras Theorem

Figure It Out (Pages 52-54)

Question 1.
Find the diagonal of a square with sidelength 5 cm.
Solution:

BD² = 5² + 5²
= 25 + 25
= 50
BD = √50
= 5×5×2−−−−−−−√
= 5√2
= 5 × 1.414
= 7.07 (~ 7.1)
Hence, the length of the diagonal is 5√2 (7.1 cm) approx.

Question 2.
Find the missing sidelengths in the following right triangles:

Solution:

(a) a2 = 72 + 92
= 49 + 81
= 130
⇒ a = √130

(b) b² = 4² + 10²
= 16 + 100
= 116
⇒ b = √116
= 2×2×29−−−−−−−−√
= 2√29

(c) 40² + c² = 41²
⇒ 1600 + c2 = 1681
⇒ c² = 1681 – 1600
⇒ c² = 81
⇒ c = √81
⇒ c = 9

(d) 10² + d² = (√200)²
⇒ 100 + d² = 200
⇒ d² = 200 – 100
⇒ d² = 100
⇒ d = √100
⇒ d = 10

(e) e² = 10² + (√150)²
⇒ e² = 100 + 150
⇒ e² = 250
⇒ e = √250
⇒ e = 5×5×5×2−−−−−−−−−−−√
⇒ e = 5√10

(f) 27² + f² = 45²
⇒ 729 + f² = 2025
⇒ f² = 2025 – 729
⇒ f² = 1296
⇒ f = √1296
= 2×2×2×2×3×3×3×3−−−−−−−−−−−−−−−−−−−−−−−√
= 2 × 2 × 3 × 3
= 36

Question 3.
Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Solution:

OA = 12 × 24 = 12
OD = 12 × 70 = 35
In ΔAOD,
a² = 12² + 35²
= 144 + 1225
= 1369
a = √1369 = 37
∴ The side of the rhombus is 37 units.

Question 4.
Is the hypotenuse the longest side of a right triangle? Justify your answer.
Solution:

c² = a² + b²
∴ c² > a² and c² > b²
or c > a and c > b
Hence, ‘c’ is the longest side of the right triangle.

Question 5.
True or False — Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
Solution:
Let (a, b, c) be a Baudhayana triple.
Then HCF (a, b, c) = 1 or HCF (a, b, c) ≠ 1
(3, 4, 5): HCF is 1
(10, 24, 26): HCF = 2
If HCF is 1, then the triple is primitive.
If HCF is a number other than 1, then the triple is scaled.

Question 6.
Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
Solution:

Question 7.
Construct a square whose area is equal to the difference of the areas of squares of side lengths 5 units and 7 units.
Solution:
Area of square = 7² – 5²
= 49 – 25
= 24 sq. units

1. Construct a square ABCD with a side of 2 cm.
Then DB = 2√2 units
2. Draw BE ⊥ DB at B such that BE = 4 units
3. Join DE.
DE² = (2√2)² + 4²
= 8 + 16
= 24
⇒ DE = √24
4. Draw a square with side DE (DEFG).
Area of DEFG = (√24)² = 24 square units.

Question 8.

(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. units?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
Solution:
(i) (a) 2 = 1² + 1²

Mark dots A, B, C, and D as shown.
Join AB, BC, CD, and DA.
Then ABCD is a square and area ABCD = 2 sq. units
(b) Square with area 3 units is not possible as 3 ≠ a² + a² for any integer ‘a’.
(c) 4 = 2 × 2

Mark dots A, B, C, D as shown.
Join AB, BC, CD, DA.
Then ABCD is a square.
and ar ABCD = 2 × 2 = 4 sq. units
(d) (i)

Mark dots A, B, C, and D as shown.
Join A, B, C, and D
AB² = 2² + 1² = 5
AB = √5 units
Hence, ABCD is a square with an area of 5 sq units.

(ii) Let the given value of area be x, where ‘x’ is an integer.
Then, x = a² + b², where ‘a’ and ‘b’ are integers or x is a perfect square, we can create squares with vertices as dots of the grid.

Question 9.
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.]
Solution:
Let ΔABC be an equilateral triangle.
AB = BC = CA = 6 cm

Let AD be perpendicular to BC.
Then ∠1 = ∠2 (each = 90°)
AB = AC (each = 6 cm)
AD = AD (common)
ΔADB ≅ ΔADC (RHS)
BD = DC (CPCT)
∴ BD = DC = 12 × 6 cm = 3 cm
In ΔADC,
h² + 3² = 6² (Baudhayana’s triple)
⇒ h² = 36 – 9 = 27
⇒ h = 3×3×3−−−−−−−√
⇒ h = 3√3 cm
Ar ABC = 12 × BC × AD
= 12 × 6 × 3√3 sq. units
= 9√3 sq. units