Class 9 Answer Key

Answer Key & Step-by-Step Solutions

Exercise Set 1.1

1. (i) The door is 8 units (feet) away from the left wall (y-axis).
Since the door is on the line OA (x-axis), it is 0 units from the x-axis.
(ii) The coordinates of D_1 are (8, 0).
(iii) The width of the door = 3.5 ft.
Yes, this is a very comfortable width. Standard room doors are usually about 3 feet wide. A 3.5 ft door (42 inches) is wider than standard requirements, making it very easy for a wheelchair user to enter.
(iv) The width of the bathroom door is 2.5 ft.
The bathroom door is narrower than the room door.

Exercise Set 1.2

1. (i) The fourth foot is at (8, 7).
(ii) Yes! Looking at the floor plan, placing it between x=8 to 11 and y=7 to 9 tucks it nicely into the top right area of the room, keeping the central walking space clear.
(iii) Length = 11 - 8 = 3 ft. Width = 9 - 7 = 2 ft.
Because we are strictly looking at a 2D floor plan (xy-plane), we cannot determine the height of the table.

2. The door will not hit the wardrobe.
If the door is made wider (e.g., 3.5 ft), its swing would reach x = 3.5, which means it would hit the wardrobe.
Suggestion: Slide the wardrobe further to the right or have the door open outwards into the bathroom.

3.(i) Coordinates of the four corners O, F, R, and P:
The coordinates are O(0, 0), F(0, 10), R(-6, 10), and P(-6, 0).
(ii) It is a square/rectangle.
The coordinates are S(-6, 5), H(-2, 5), W(-2, 10), and R(-6, 10).
(iii) Washbasin (3 \text{ ft} \times 2 \text{ ft} in the bottom corner): (-6, 0), (-3, 0), (-3, 2), (-6, 2).

Toilet (2 \text{ ft} \times 3 \text{ ft} above the basin): (-6, 2), (-4, 2), (-4, 5), (-6, 5).

4. Other rooms in the house (Dining Room):
(i) The room spans 18 ft from P to A.
If P is (-6, 0) and A is (12, 0), the length is 12 - (-6) = 18 ft.
If it has a width of 15 ft extending downwards, the coordinates of the corners are (-6, 0), (12, 0), (12, -15), and (-6, -15).

(ii) Table placed perfectly in the center, its corners (feet) will be (0.5, -6), (5.5, -6), (5.5, -9), and (0.5, -9).

End-of-Chapter Exercises
1. The coordinates are (0, 0).
2. Point H will have coordinates (-5, y).
H can lie in Quadrant II or Quadrant III.
3. (i) Sides perpendicular to each other: Line segments AM and MP.
(ii) Parallel to axes: AM is parallel to the x-axis, and MP is parallel to the y-axis.
(iii) Mirror images: Points M(-5, -2) and P(-5, 2) are mirror images of each other across the x-axis.

4. Plot Z(5, -6) and construct right-angled triangle IZN:
Let’s choose I(5, 0) on the x-axis and N(0, -6) on the y-axis.

Lengths of sides:

• IZ = 6 units
• ZN = \sqrt{(5-0)^2 + (-6 - (-6))^2} = 5 units
• Hypotenuse IN = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} units.

5. If we didn’t have negative numbers, our coordinate system would be limited to only the first quadrant.
We would not be able to locate all points in a 2-D plane, making concepts like a four-quadrant Cartesian plane impossible.

6. Yes.
You can check this mathematically without plotting by comparing the distance MG and the sum of distances MA and AG

7. Are R(-5, -1), B(-2, -5), and C(4, -12) collinear?
No R, B and C are not collinear.

8. Plotting vertices using the origin (0,0):

• (i) Right-angled isosceles triangle: (0, 0), (a, 0), and (0, a).

• (ii) Isosceles triangle with vertices in Q3 and Q4: (0, 0), (-a, -b), and (a, -b).

9. Midpoint Table Verification:
Formula Check: Midpoint = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

10. Coordinates of B are (-17, 6).

11. Points of Trisection:
P(8, 4)
Q(12, 1)

12. Circle K:
(i) Radius R = \sqrt{65}.

(ii)OD = \sqrt{61}.
D lies inside the circle.

OE = 9.
E lies outside the circle.

13. A(1, 7), B(-1, -1), and C(11, 3).

14. City Map intersections:

• (a) There is exactly 1 street intersection referred to as (4, 3).
• (b) There is exactly 1 street intersection referred to as (3, 4).

15. (i) No part lies outside.
(ii) Distance between centers A and B: AB = 170 \text{ pixels}
The sum of their radii is 80 + 100 = 180 pixels.

Since the distance between centers (170) is less than the sum of the radii (180), the two circles intersect.

16. Yes, ABCD is a square.
Area = \text{10 square units}.

Answer Key & Step-by-Step Solutions (Exercises 2.1 to 2.3)

EXERCISE SET 2.1

1. (i) The degree is 2.
(ii) The degree is 3.
(iii) The degree is 0.
(iv) The degree is 1.

2. (Answers may vary. Here are standard examples)

• Degree 1 (Linear): 5x + 2
• Degree 2 (Quadratic): 3y^2 - 4y + 7
• Degree 3 (Cubic): 2z^3 + z^2 - 8

3. The term with x^2 is 6x^2. Its coefficient is 6.
The term with x^3 is -3x^3. Its coefficient is -3.

4. The coefficient of z is 0.

5. The constant term is -10.

EXERCISE SET 2.2

1. The value of the linear polynomial 5x - 3 is:
(i) -3
(ii) -8
(iii) 7

2. Find the value of the quadratic polynomial 7s^2 - 4s + 6 is:
(i) 6
(ii) 81
(iii) 102

3. Salil’s present age is 15 years, and his mother’s present age is 45 years.
4. The two integers are 42 and 105.
5. Ruby has 8 coins of ₹5, and 24 coins of ₹2.
6. The shorter piece is 60 feet and the longer piece is 240 feet.
7. The dimensions of the rectangle are Length = 9 cm and Width = 3 cm.

EXERCISE SET 2.3

1. Amount at the end of every month from the 2nd month onwards:

• End of Month 1: 500 + 150 = \text{₹650}
• End of Month 2: 650 + 150 = \text{₹800}
• End of Month 3: 800 + 150 = \text{₹950}

Linear Expression: \text{Amount in } n^{\text{th}} \text{ month} = 500 + 150n

2. Members remaining after 1, 2, 3 hours:

• After 1 hour: 120 - 9 = 111 members
• After 2 hours: 111 - 9 = 102 members
• After 3 hours: 102 - 9 = 93 members

\text{Members at the end of } n^{\text{th}} \text{ hour} = 120 - 9n

3. Area
(i) Area = 156 \text{ cm}^2
(ii) Area = 130 \text{ cm}^2
(iii) Area = 104 \text{ cm}^2

The linear pattern representing the area is: \text{Area} = 13b

4. Volume
(i) Volume = 385 \text{ cm}^3
(ii) Volume = 693 \text{ cm}^3
(iii) Volume = 1001 \text{ cm}^3

The linear pattern representing the volume is: \text{Volume} = 77h

5. Pages left after 15 days = 200
Because the pages remaining decrease every day, this is a linear decay pattern.
\text{Pages left after } n \text{ days} = 500 - 20n

Chapter 2: Introduction to Linear Polynomials

Answer Key & Step-by-Step Solutions

EXERCISE SET 2.4

1. (i) The height after 7 months: = 5.25 \text{ feet}
(ii) Table of values for t varying from 0 to 10 months:

(iii) Expression h(t) = 1.75 + 0.5t
This represents linear growth because the height increases by a constant amount (0.5 feet) over equal intervals (every month).

2. (i) The value of the phone after 3 years: = \text{₹7600}
(ii) Table of values for t varying from 0 to 8 years:

(iii) Expression: v(t) = 10000 - 800t
This represents linear decay because the value decreases by a fixed, constant amount (₹800) over equal intervals (every year).

3. (i) The population after 6 years: 1050 \text{ people}
(ii) Table of values for t varying from 0 to 10 years:

(iii) Expression: P(t) = 750 + 50t
This represents linear growth because the population increases by a constant amount (50 people) each year.

4. (i) Equation: b(x) = 600 - 15x
This is linear decay because the balance is reduced by a constant amount (₹15) every single day.
(ii) The balance runs out after 40 \text{ days}
(iii) Table of values for x varying from 1 to 10 days:

EXERCISE SET 2.5

1. a = 25, b = 150. (The linear relation is y = 25x + 150).

2. a = 60, b = 200. (The linear relation is y = 60x + 200).

3. a = \frac{5}{9}, b = -\frac{160}{9}. (The relation is C = \frac{5}{9}F - \frac{160}{9}).

EXERCISE SET 2.6

(Note for Blog: Have students draw these on graph paper to see the visual reflections).

1. Draw the graphs and reflect on the role of ‘a’ (slope) and ‘b’ (y-intercept):

• (i)

Reflection: The value of a (slope) is 3 for all of them. Since the slopes are identical, these three lines are perfectly parallel. The value of b changes where they cross the y-axis ([/latex]-1, 0, \text{ and } 1[/latex]).

• (ii) y = 4x, y = 2x, y = x

Reflection: The value of b is 0 for all, so they all pass exactly through the origin (0,0). The changing value of a (4, 2, 1) changes the steepness of the line. The larger the a, the steeper the graph.

• (iii) y = 5x, y = -5x

Reflection: The number for the slope is the same, but the signs are opposite. y = 5x slopes upwards (linear growth), while y = -5x slopes downwards (linear decay) at the exact same steepness.

(iv) y = 3x - 1, y = 3x, y = 3x + 1

(v) y = -2x - 3, y = -2x, y = 2x + 3

(Students will plot the remaining lines noting that a negative ‘a’ creates a downward sloping line, and changing ‘b’ shifts those parallel lines up and down the y-axis).

END-OF-CHAPTER 2 EXERCISES

1. x^3 - 7x^2 + 5x + 2
(Any cubic equation works as long as the x^2 term is -7x^2.)

2. (i) 9
(ii) 4a^3 - a^2 + 6

3. The number is -\frac{1}{2}.

4. The numbers are 7 and 35.

5. (i) After 6 months, amount = \text{₹2300}
(ii) After 2 years (24 months), amount = \text{₹6800}
Linear pattern: Amount = 800 + 250n
(where n is months)

6. The numbers are 85 or 58.

7. Identify slopes, y-intercepts, and see if parallel:
(i) y = -3x + 4: Slope = -3, cuts y-axis at (0, 4).

(ii) 2y = 4x + 7 \implies y = 2x + 3.5: Slope = 2, cuts y-axis at (0, 3.5).

(iii) 5y = 6x - 10 \implies y = 1.2x - 2: Slope = 1.2, cuts y-axis at (0, -2).

(iv) 3y = 6x - 11 \implies y = 2x - \frac{11}{3}: Slope = 2, cuts y-axis at (0, -\frac{11}{3}).

Answer: Yes, lines (ii) and (iv) are parallel because they both have the exact same slope of 2.

8. (i) 104^\circ\text{F}
(ii) 343\text{ K}
9. Work w and distance d with constant force = 3.

Formula: w = F \times d \implies \boldsymbol{w = 3d}
If d = 2 units, w = 3(2) = \boldsymbol{6 \text{ units}}.
(Students will plot this straight line passing through (0,0) and (2,6)).

10. (i) p(x) = 3x + 2
(ii) Cuts y-axis at \boldsymbol{(0, 2)}
Cuts x-axis at \boldsymbol{(-\frac{2}{3}, 0)}

11. p(x) = 2x + 5 and q(x) = 4x - 1

12. Growing pattern of hexagons.
(i) Next two stages:

Stage 4 will have 16 + 5 = 21 matchsticks.

Stage 5 will have 21 + 5 = 26 matchsticks.

(ii)

(iii) Rule for n^{\text{th}} stage: 5n + 1
(iv) Matchsticks for 15th stage: 5(15) + 1 = 75 + 1 = \boldsymbol{76}
(v) No. Since n must be a whole number (you can’t have a fraction of a stage), 200 matchsticks will not form a complete stage.

13. p(x) = 2x - 1
q(x) = 2x - 9.

Coordinates where they meet the x-axis:
For p(x): \boldsymbol{(\frac{1}{2}, 0)}
For q(x): \boldsymbol{(\frac{9}{2}, 0)}

14. The linear functions all represent straight lines passing through the exact same point: (-1, 0) on the x-axis.

EXERCISE SET 3.1

1. The merchant will leave with 90 copper ingots.

2. The next three prime numbers following 19 are 23, 29, and 31.

3. No, Natural Numbers are not closed under subtraction.
Example: If we take two natural numbers, say 5 and 8, and subtract them: 5 - 8 = -3.
Since -3 is an integer but not a natural number, the closure property fails for subtraction.

4. Using the thumb to point to each joint allows you to count to 4 \times 3 = 12 on a single hand.
Relation: This simple anatomical fact is exactly why ancient cultures developed base-12 (dozenal) counting systems. Measuring in dozens and grosses is a direct result of counting finger joints!

EXERCISE SET 3.2

1. The midnight temperature is -11^\circ\text{C}.

2. Equation: Debt is negative, profit is positive, and loss is negative. -850 + 1200 + (-450)
The final standing is -100, which means the trader has a net debt of ₹100.

3. (i) The product of a debt and a fortune is a debt. Answer: -60
(ii) The product of two debts is a fortune. Answer: 56
(iii) Subtracting a debt is equivalent to adding a fortune. Answer: 14
(iv) Dividing a debt by a fortune yields a debt. Answer: -5

4. Imagine you have ₹10 in your pocket, but you also owe your friend ₹5. That ₹5 you owe is a debt, represented as -5. If your friend suddenly forgives your debt and takes it away (subtracts the -5), your financial standing improves! Removing a ₹5 debt makes you ₹5 richer. Therefore, taking away a debt of 5 is exactly the same as gaining a fortune of 5: 10 - (-5) = 15.

EXERCISE SET 3.3

1. (i) \frac{2}{3} and \frac{4}{6}
2 \times 6 = 12 and 3 \times 4 = 12.
Since 12 = 12, they are equal.

(ii) \frac{5}{4} and \frac{10}{8}
5 \times 8 = 40 and 4 \times 10 = 40.
Since 40 = 40, they are equal.

(iii) -\frac{3}{5} and -\frac{6}{10}
-3 \times 10 = -30 and 5 \times (-6) = -30.
Since -30 = -30, they are equal.

(iv) \frac{9}{3} and 3
9 \times 1 = 9 and 3 \times (3) = 9.
Since 9 = 9, they are equal.

2. (i) \frac{7}{10}

(ii) \frac{29}{24}

(iii) -\frac{5}{14}

3. (i) \frac{7}{12}

(ii) \frac{5}{8}

(iii) -\frac{1}{9}

4. (i) \frac{1}{5}

(ii) \frac{35}{88}

(iii) -\frac{10}{49}

5. (i) \frac{20}{9}

(ii) \frac{56}{55}

(iii) -\frac{8}{5}

6. Show that: \left(\frac{1}{2} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{1}{2} \times \frac{8}{3} + \frac{3}{4} \times \frac{8}{3}

LHS (Left Hand Side):
First solve the bracket.
\left(\frac{2}{4} + \frac{3}{4}\right) \times \frac{8}{3} = \frac{5}{4} \times \frac{8}{3} = \frac{40}{12} = \frac{10}{3}

RHS (Right Hand Side):
Solve multiplications first.
\left( \frac{8}{6} \right) + \left( \frac{24}{12} \right) = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}

Conclusion: LHS = RHS, Hence Proved

7. \frac{1}{12}

8. x can be ANY rational number.
Every rational number will satisfy this equation because it perfectly demonstrates the distributive property.

EXERCISE SET 3.4

1. Represent the rational numbers \frac{2}{3}, -\frac{5}{4}, and 1\frac{1}{2} on a single number line.

• Step 1: Convert mixed fractions to improper fractions for easier plotting. 1\frac{1}{2} = \frac{3}{2}.
• Step 2: Find a common denominator to make marking the number line easier. The LCM of 3, 4, and 2 is 12.
  • \frac{2}{3} = \frac{8}{12}
  • -\frac{5}{4} = -\frac{15}{12}
  • \frac{3}{2} = \frac{18}{12}
• Step 3: Draw a number line with 0 in the middle. Divide each 1-unit interval into 12 equal parts. Mark the points at 8 units right, 15 units left, and 18 units right of 0, respectively.

2. -\frac{9}{32}, -\frac{10}{32}, -\frac{11}{32} (Answers may vary).

3. \frac{1}{6}.

4. He can make exactly 7 kurtas.

5. 3.14151, 3.14152, 3.14155 (Answers may vary).

6. Yes! In addition to the “mean” or “average” method (\frac{a+b}{2}), you can convert the rational numbers to decimals and pick a terminating decimal between them (as we did in Q5). You can also make their denominators very large (equivalent fractions) and pick integer numerators in between (as we did in Q2).

EXERCISE SET 3.5

1. \frac{7}{20}: terminating.

\frac{4}{15}: non-terminating repeating.

\frac{13}{250}: terminating.

2. \frac{1}{13} = 0.076923076923... = 0.\overline{076923}. The repeating block is 076923.

\frac{2}{13} = 0.\overline{153846}.

\frac{3}{13} = 0.\overline{230769}.

Notice that unlike \frac{1}{7}, the digits for \frac{1}{13} and \frac{2}{13} are NOT part of the same cyclic circle.
13 produces two separate cyclic groups of 6 digits!

3. (i) \sqrt{81}: Rational

(ii) \sqrt{12}: Irrational

(iii) 0.33333…: Rational

(iv) 0.123451234512345: Rational

(v) 1.01001000100001…: Irrational

(vi) 23.560185612239874790120: Rational

4. Let x = 0.9999... … (1)
Multiply by 10 (since 1 digit repeats): 10x = 9.9999... … (2)

Subtract the first equation from the second:
10x - x = 9.9999... - 0.9999...
9x = 9
x = 1

Conclusion: Therefore, 0.\overline{9} is mathematically identical to 1.

5. Other numbers whose reciprocals produce cyclic numbers include 17, 19, 23, and 29.

END-OF-CHAPTER 3 EXERCISES

1. (i) \frac{3}{50}: Terminating. 3 \div 50 = \mathbf{0.06}.

(ii) \frac{2}{9}: Non-terminating repeating. 2 \div 9 = 0.222... = \mathbf{0.\overline{2}}.

2. Prove that \sqrt{5} is an irrational number.

Step 1 (Assumption):
Assume \sqrt{5} is rational.
So, \sqrt{5} = \frac{p}{q}, where p and q are co-prime integers and q \neq 0.

Step 2: Square both sides:
5 = \frac{p^2}{q^2} \implies p^2 = 5q^2.

Step 3: This means p^2 is a multiple of 5.
Therefore, p must also be a multiple of 5.
Let p = 5k.

Step 4: Substitute p back in: p^2 = 5q^2
(5k)^2 = 5q^2
25k^2 = 5q^2
5k^2 = q^2.
q^2 = 5k^2.

Step 5: This means q^2 is a multiple of 5, so q must also be a multiple of 5.

Conclusion: If both p and q are multiples of 5, they share a common factor of 5. This contradicts our assumption that they are co-prime.
This contradiction arises because of our wrong assumption and
Therefore, \sqrt{5} must be irrational.

3. (i) 12.6 = \frac{63}{5}

(ii) 0.0120 = \frac{3}{250}

(iii) 3.0\overline{52} = \frac{1511}{495}

(iv) 1.235 = \frac{247}{200}

(v) 0.\overline{23} = \frac{23}{99}

5. \frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}. Answers may be different

6. \frac{13}{30}, \frac{14}{30}, \frac{15}{30}, \frac{16}{30}, \frac{17}{30}. Answers may be different

7. \frac{6}{30}, \frac{7}{30}, \frac{8}{30}, \frac{9}{30}, \frac{10}{30}. Answers may be different

8. 2

9. The product ab must be negative.

10. Yes, it is absolutely necessary.

11. Terminating.
It will terminate after 3 decimal places.

12. Because the denominator can be expressed as 10^3, the decimal expansion will have exactly 3 decimal places.

13. \frac{22}{36}, \frac{23}{36}, \frac{24}{36}, \frac{25}{36}, \frac{26}{36}
Explanation of the rule: If you need to fit exactly n integers between two endpoints (k_1 and k_2), the “gap” or difference between those endpoints must be at least n + 1. For example, to fit 5 integers between two numbers, the difference must be at least 6. The condition k_2 - k_1 > n (or \ge n + 1) ensures there is enough mathematical “space” to pull out n unique integer numerators!

16. \sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{9}, \sqrt{10}, \sqrt{11}, \sqrt{12}, \text{and } \sqrt{13}

Chapter 4: Exploring Algebraic Identities

Step-by-Step Solutions (Exercises 4.1 to 4.3)

EXERCISE SET 4.1

1. (i) = 49x^2 + 56xy + 16y^2

(ii) = \frac{49}{25}x^2 + \frac{21}{5}xy + \frac{9}{4}y^2

(iii) = 6.25p^2 + 7.5pq + 2.25q^2

(iv) = \frac{9}{16}s^2 + 12st + 64t^2

(v) = x^2 + \frac{x}{y} + \frac{1}{4y^2}

(vi) = \frac{1}{x^2} + \frac{2}{xy} + \frac{1}{y^2}

2. (i) = 4096
(ii) = 11025
(iii) = 42025

EXERCISE SET 4.2

1. (i) = (3x + 4y)^2
(ii) = (2s + 5t)^2
(iii) = (7x + 2y)^2
(iv) = \left(8p + \frac{2}{3}q\right)^2
(v) = 3\left(a + \frac{2}{3}b\right)^2
(vi) = \frac{1}{5}(3s + 5v)^2

2. (i) = 6241
(ii) = 37249

EXERCISE SET 4.3

1. (i) 13689
(ii) 6084
(iii) 39204
(iv) 45796
(v) 1218816
(vi) 1254400

2. (i) (4y - 3)^2

(ii) \left(\frac{3}{2}s + 2t\right)^2

(iii) \left(\frac{m}{3} + \frac{k}{2} + 3n\right)^2

(iv) \left(\frac{p}{4} - \frac{4}{p}\right)^2

(v) (3a - 2b + c)^2

3. (i) p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr
(ii) 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz

4. No, this is not an identity.

EXERCISE SET 4.4

1. (i) (s - 8)(s - 3)
(ii) (3x - 7)
(iii) (2x - 3)(5x + 2)
(iv) (3x + 2)(2x + 1)

2. (i) 1681
(ii) 729
(iii) 391
(iv) 18225
(v) 9409
(vi) 522
(vii) 1462
(viii) 42025

EXERCISE SET 4.5

1. (i) \frac{3(p-3q)(p+2q)}{(p+5q)(p-2q)}

(ii) \frac{(n - m)^3}{5(n - m)^2} = \frac{n - m}{5}

(iii) \frac{w^2 + v^2 + x^2 + wv + vx - wx}{w - v + x}

(iv) \frac{5z - 2y}{5z + 2y}

(v) 1

(vi) \frac{(p + 2)(p^2 + 4)}{p - 2}

END-OF-CHAPTER EXERCISES

1. (i) 9x^2 - 24x + 16

(ii) 4s^2 - 49

(iii) p^4 - \frac{1}{4}

(iv) 4n^2 - 49

(v) s^3 - 8t^3

(vi) \frac{1}{4r^2} - 4 + 16r^2

(vii) 9m^2 + 16k^2 + l^2 - 24mk + 6ml - 8kl

(viii) x^3 - x^2y + \frac{1}{3}xy^2 - \frac{1}{27}y^3

(ix) \frac{343}{8}k^3 - \frac{49}{2}k^2m + \frac{14}{3}km^2 - \frac{8}{27}m^3

2. (i) 357
(ii) 9984
(iii) 384
(iv) 3176523
(v) 7880599
(vi) 2048383
(vii) -1225043
(viii) -26730899

3. (i) (2y + \frac{1}{4y})^2

(ii) (3m - \frac{1}{5n})(3m + \frac{1}{5n})

(iii) (3b - \frac{1}{4b})(9b^2 + \frac{3}{4} + \frac{1}{16b^2})

(iv) (x + \frac{1}{2})(x + \frac{1}{3})

(v) (3u - \frac{1}{5})^3

(vi) (4y + \frac{1}{5}z)(16y^2 - \frac{4}{5}yz + \frac{1}{25}z^2)

(vii) (p + 3q + r)(p^2 + 9q^2 + r^2 - 3pq - 3qr - rp)

(viii) \frac{1}{3}(3x - 2y + z)(9x^2 + 4y^2 + z^2 + 6xy + 2yz - 3xz)

(ix) (3m - 2)^2

(x) (2x + 3y + 6z)^2

(xi) (3u - \frac{1}{6})^3

4. (i) \frac{2x + 1}{2x - 1}

(ii) \frac{3(a^2 + 2ab + 4b^2)}{a + 2b}

(iii) \frac{s^2 - 5st + 25t^2}{s - 7t}

5. (i) Length = 5a - 3b, Breadth = 5a - 3b

(ii) Length = 6s + 7t, Breadth = 6s - 7t

6. (i) Dimensions: 6, (a - 2b), and (a + 2b).

(ii) Dimensions: 3p, (s - 1), and (s - 4).

7. 4s^2 + 160s square metres.

8. The number is 3 or \frac{1}{3}.

9. Length = x + 3.

12. Factorise: n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1).
These are three consecutive natural numbers.
Reason: In any three consecutive numbers, at least one is even (divisible by 2) and exactly one is a multiple of 3.
Therefore, their product is always divisible by 2 \times 3 = 6.

13. (i) The value is 0.
(ii) The value is 0.

EXERCISE SET 5.1

1. Draw \Delta ABC with AB=5 cm, \angle A=70^{\circ} and \angle B=60^{\circ}. Draw the circumcircle of \Delta ABC. Is the centre inside or outside the triangle?

• Step 1: Calculate the third angle. Using the angle sum property of a triangle, \angle C = 180^{\circ} - (70^{\circ} + 60^{\circ}) = 50^{\circ}.
• Step 2: Because all three angles (70^{\circ}, 60^{\circ}, and 50^{\circ}) are less than 90^{\circ}, \Delta ABC is an acute-angled triangle.
• Answer: For an acute-angled triangle, the circumcentre lies inside the triangle.

2. Draw \Delta ABC with AB=5 cm, \angle A=100^{\circ}, AC=4 cm. Draw the circumcircle of \Delta ABC. Is the centre inside or outside the triangle?

• Step 1: Look at the given angles. We are given that \angle A = 100^{\circ}.
• Step 2: Since one angle is greater than 90^{\circ}, this is an obtuse-angled triangle.
• Answer: For an obtuse-angled triangle, the circumcentre lies outside the triangle.

3. Draw \Delta ABC, with AB=6 cm, BC=7 cm and CA=7 cm. Draw the circumcircle of \Delta ABC. Let the circumcentre be O. Measure OA, OB, OC.

• Explanation: The circumcentre (O) is defined as the centre of the circle that passes through all three vertices of the triangle: A, B, and C.
• Answer: Because A, B, and C all lie on the boundary of the same circle, the distances from the centre O to each of these points represent the radii of the circumcircle. Therefore, when you measure them, you will find that OA = OB = OC.

4. \frac{AB}{2}.

EXERCISE SET 5.3

3. The distance between the midpoints of the chords is 7 cm.

EXERCISE SET 5.5

1. 2\sqrt{13} cm.

3. No, we cannot conclude that CD = 2AB.
Reason: The relationship between the length of a chord and its distance from the centre is not linear; it is governed by squares and square roots in the Pythagorean theorem.
If d_{AB} = 2d_{CD}, the lengths will be AB = 2\sqrt{r^2 - 4d_{CD}^2} and CD = 2\sqrt{r^2 - d_{CD}^2}.
These are clearly not multiples of 2 of each other!

EXERCISE SET 5.6

1. AB = 12 cm.

2. (i) No. Angles subtended by an arc at any point on the remaining part of the circle (the same segment) are always equal.
(ii) Yes. If the angles are equal and not right angles, they must lie in the same major or minor segment, and thus on the same side of the chord AB.
(iii) Yes. By Theorem 10, if a line segment subtends equal angles at two points on the same side, all four points are concyclic.

3. 80^\circ.

END-OF-CHAPTER EXERCISES

1. 24 cm.
2. 35^\circ .
3. 5 cm.
4. 24 cm.

6. Measure: 90^\circ.
Reasoning: The angle subtended by a diameter at any point on the circle is 90^\circ (angle in a semicircle).
7. \angle C = 105^\circ.
\angle D = 70^\circ.
8. x = 38.
P = 86^\circ
R = 94^\circ.

9. r = 10 cm.
10. Area = 60 square units.
11. Check the angles of the triangles formed by the diagonals. If any three vertices form an obtuse-angled triangle, the circumcentre will fall outside the triangle (and potentially outside the quadrilateral). For an acute-angled triangle, it lies inside.

16. Thus, the shape formed is a smaller concentric circle.
18. The radius of the circle is 13 cm.
19. The distance of each side from the centre is \frac{\sqrt{3}}{2}r.

20. Explanation: The angle subtended by a diameter at any point on the circle is 90^{\circ}. Because MN is the diameter, it subtends angles at the remaining vertices O and P on the circumference.
Answer: Therefore, both \angle MON = 90^{\circ} and \angle MPN = 90^{\circ}.

22. Explanation: Let the distance from the centre to a chord be d, and the radius be r. The length of any chord is 2\sqrt{r^2 - d^2}.
Conclusion: To maximize this length, we must subtract the smallest possible value for d^2. Since distance squared cannot be negative, its minimum value is d = 0.
When d = 0, the chord passes through the centre and its length becomes 2\sqrt{r^2} = 2r, which is the diameter. Any chord where d > 0 will strictly be less than 2r.

Chapter 6: Measuring Space: Perimeter and Area

Step-by-Step Solutions

EXERCISE SET 6.1

1. The radius is 7 cm.

2. (i) 44. cm
(ii) \frac{440}{7} \approx 62.9 cm.
(iii) \frac{528}{7} \approx 75.4 cm.

3. (i) \frac{22}{6} \approx 3.67 cm.
(ii) \frac{66}{5} = \mathbf{13.2} m.

4. \frac{139}{3} \mathbf{46.33} cm.

6. (i) Distance for one revolution = Circumference = 176 cm
(ii) Revolutions = \frac{1,000,000}{176} \approx \mathbf{5681.8}[/latex] revolutions.

8. The ratio of their radii is 5:4.

EXERCISE SET 6.2

1. Area = 40 sq cm.
2. Area = 720 sq cm.
3. Area = 8\sqrt{30} sq cm.
4. Area = 1500\sqrt{3} sq m.
5. Shorter Diagonal = 8\sqrt{2} cm.

EXERCISE SET 6.3

1. Find the area of a sector of a circle with radius 7 cm if the angle of the sector is 60^{\circ}.

• Formula: Area = \pi r^2 \times \frac{\theta^{\circ}}{360^{\circ}}.
• Area = \frac{22}{7} \times 7^2 \times \frac{60}{360} = \frac{22}{7} \times 49 \times \frac{1}{6} = \frac{154}{6} = \mathbf{25.67} sq cm.

2. Find the area of a quadrant of a circle whose circumference is 44 cm.

• Step 1: Find radius. 2 \times \frac{22}{7} \times r = 44 \implies r = 7 cm.
• Step 2: Area of a quadrant is \frac{1}{4}\pi r^2.
• Area = \frac{1}{4} \times \frac{22}{7} \times 49 = \frac{154}{4} = \mathbf{38.5} sq cm.

3. The length of the minute hand of a clock is 7 cm. Find the area swept by the minute hand in 10 minutes.

• Step 1: In 60 minutes, the hand sweeps 360^{\circ}. In 10 minutes, it sweeps \frac{10}{60} \times 360^{\circ} = 60^{\circ}.
• Step 2: Use the sector area formula with r = 7 and \theta = 60^{\circ}.
• Area = \frac{22}{7} \times 7^2 \times \frac{60}{360} = \frac{154}{6} \approx \mathbf{25.67} sq cm.

4. A chord of a circle of radius 10 cm subtends 90^{\circ} at the centre. Find the area of the corresponding (i) minor sector and (ii) major sector. (Use \pi \approx 3.14).

• (i) Minor Sector Area: 3.14 \times 10^2 \times \frac{90}{360} = 314 \times \frac{1}{4} = \mathbf{78.5} sq cm.
• (ii) Major Sector Area: Total Area – Minor Area = (3.14 \times 100) - 78.5 = 314 - 78.5 = \mathbf{235.5} sq cm.

5. A chord of a circle of radius 15 cm subtends an angle of 60^{\circ} at the centre. Find the areas of the corresponding minor and major segments. (Use \pi \approx 3.14 and \sqrt{3} \approx 1.73).

• Step 1: Area of the minor sector = 3.14 \times 15^2 \times \frac{60}{360} = 3.14 \times 225 \times \frac{1}{6} = 117.75 sq cm.
• Step 2: Area of the triangle formed by radii and chord. Since the central angle is 60^{\circ} and radii are equal, it’s an equilateral triangle. Area = \frac{\sqrt{3}}{4} r^2 = \frac{1.73}{4} \times 225 \approx 97.31 sq cm.
• (i) Minor Segment: Sector Area – Triangle Area = 117.75 - 97.31 = \mathbf{20.44} sq cm.
• (ii) Major Segment: Total Circle Area – Minor Segment = (3.14 \times 225) - 20.44 = 706.5 - 20.44 = \mathbf{686.06} sq cm.

8. An equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to \frac{3\sqrt{3}}{4\pi}.

• Step 1: An equilateral triangle inscribed in a circle of radius r has a side length a = \sqrt{3}r.
• Step 2: The area of this equilateral triangle is \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (\sqrt{3}r)^2 = \frac{3\sqrt{3}}{4} r^2.
• Step 3: The area of the circle is \pi r^2.
• Step 4: Ratio = \frac{\frac{3\sqrt{3}}{4} r^2}{\pi r^2} = \mathbf{\frac{3\sqrt{3}}{4\pi}}.

END-OF-CHAPTER EXERCISES

2. An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.

• Step 1: Find the base: 40 - (15 + 15) = 10 cm.
• Step 2: Semi-perimeter s = 20 cm.
• Step 3: Heron’s Formula: \text{Area} = \sqrt{20(20-15)(20-15)(20-10)} = \sqrt{20 \times 5 \times 5 \times 10} = \sqrt{5000} = \mathbf{50\sqrt{2}} sq cm.

3. An isosceles triangle has a base of 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

• Step 1: Use \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}. So, 60 = \frac{1}{2} \times 10 \times h \implies h = 12 cm.
• Step 2: Use the Baudhāyana-Pythagoras theorem on the right triangle formed by the height, half the base (5 cm), and the equal side a.
• a = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = \mathbf{13} cm.

4. The area of a right-angled triangle is 54 sq cm. One of its legs has a length of 12 cm. Find its perimeter.

• Step 1: Let the other leg be b. Area = \frac{1}{2} \times 12 \times b = 54 \implies 6b = 54 \implies b = 9 cm.
• Step 2: Find the hypotenuse c using Baudhāyana-Pythagoras: c = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15 cm.
• Step 3: Perimeter = 12 + 9 + 15 = \mathbf{36} cm.

6. The sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.

• Method 1 (Heron’s Formula): s = \frac{7 + 24 + 25}{2} = 28. Area = \sqrt{28(28-7)(28-24)(28-25)} = \sqrt{28 \times 21 \times 4 \times 3} = \sqrt{7056} = \mathbf{84} sq cm.
• Method 2 (Right Triangle Check): Notice that 7^2 + 24^2 = 49 + 576 = 625 = 25^2. It is a right-angled triangle! Area = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times 24 = \mathbf{84} sq cm.

10. Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?

• Yes. If two rectangles have the same perimeter, the sum of their length and width is identical (L_1 + W_1 = L_2 + W_2). If their areas are the same, the product of their length and width is identical (L_1 \times W_1 = L_2 \times W_2). Algebraically, two numbers that have the same sum and the same product must be the exact same two numbers. Therefore, the dimensions are identical, and they are completely congruent.

14. Show that the area of a kite is half the product of its diagonals.

• Explanation: Let the diagonals of the kite be d_1 and d_2. A kite’s diagonals intersect at right angles (90^{\circ}). One diagonal (d_1) perfectly bisects the kite into two identical triangles.
• The base of each triangle is d_1, and the height of each triangle is exactly half of the other diagonal (\frac{d_2}{2}).
• Area of one triangle = \frac{1}{2} \times d_1 \times \frac{d_2}{2}.
• Since there are two identical triangles, the total area = 2 \times \left( \frac{1}{2} \times d_1 \times \frac{d_2}{2} \right) = \mathbf{\frac{1}{2} \times d_1 \times d_2}.