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Class 10 • CBSE 2024-25

NCERT Solutions Class 10 Mathematics

Complete solutions for all 15 chapters of Class 10 Maths NCERT textbook.

All 15 Chapters

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1

Real Numbers

Euclid's division, Fundamental Theorem of Arithmetic, Irrational numbers

› 2

Polynomials

Zeros of polynomials, Division algorithm

› 3

Pair of Linear Equations

Graphical and algebraic methods

› 4

Quadratic Equations

Factorisation, Completing the square, Quadratic formula

› 5

Arithmetic Progressions

nth term, Sum of n terms

› 6

Triangles

Similar triangles, Basic Proportionality Theorem

› 7

Coordinate Geometry

Distance formula, Section formula

› 8

Introduction to Trigonometry

Trigonometric ratios, Identities

› 9

Some Applications of Trigonometry

Heights and distances

› 10

Circles

Tangent to a circle, Number of tangents

› 11

Constructions

Division of line segment, Tangents to circles

› 12

Areas Related to Circles

Sector, Segment, Combination of figures

› 13

Surface Areas and Volumes

Combination of solids, Frustum of a cone

› 14

Statistics

Mean, Median, Mode, Ogive curves

› 15

Probability

Theoretical probability, Events

›

Chapter 1: Real Numbers

Key Formulas:
• Euclid's Division Lemma: a = bq + r, 0 ≤ r < b
• HCF(a,b) × LCM(a,b) = a × b
• Fundamental Theorem of Arithmetic: Every composite number is a unique product of primes

Exercise 1.1

Q1. Use Euclid's division algorithm to find the HCF of:
(i) 135 and 225   (ii) 196 and 38220   (iii) 867 and 255
Solution

(i) HCF of 135 and 225

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
∴ HCF(135, 225) = 45

(ii) HCF of 196 and 38220

38220 = 196 × 195 + 0
∴ HCF(196, 38220) = 196

(iii) HCF of 867 and 255

867 = 255 × 3 + 102
255 = 102 × 2 + 51
102 = 51 × 2 + 0
∴ HCF(867, 255) = 51
Q2. Show that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5.
Solution

Let a be any positive integer and b = 6. By Euclid's algorithm, a = 6q + r where 0 ≤ r < 6.

So a = 6q, 6q+1, 6q+2, 6q+3, 6q+4, or 6q+5. Now 6q, 6q+2, 6q+4 are even (divisible by 2). So odd numbers must be of the form 6q+1, 6q+3, or 6q+5.

Q3. An army contingent of 1000 members has to march behind a band of 56 members. What is the maximum number of columns?
Solution
HCF(1000, 56):
1000 = 56 × 17 + 48
56 = 48 × 1 + 8
48 = 8 × 6 + 0
∴ Maximum columns = 8
Q4. Show that the square of any positive integer is either of the form 3m or 3m + 1.
Solution

Let x = 3q, 3q+1, or 3q+2.

If x = 3q: x² = 9q² = 3(3q²) = 3m
If x = 3q+1: x² = 9q²+6q+1 = 3(3q²+2q)+1 = 3m+1
If x = 3q+2: x² = 9q²+12q+4 = 3(3q²+4q+1)+1 = 3m+1

In all cases, x² is of the form 3m or 3m+1.

Q5. Show that the cube of any positive integer is of the form 9m, 9m+1, or 9m+8.
Solution

Let x = 9q + r where r ∈ {0,1,2,...,8}. Checking key cases:

(9q)³ = 729q³ = 9(×) = 9m
(9q+1)³ = 729q³+243q²+27q+1 = 9m+1
(9q+2)³ = 9(×)+8 = 9m+8
Similarly for r = 3..8, all results fall in {9m, 9m+1, 9m+8}

Exercise 1.2

Q1. Express as product of prime factors:
(i) 140   (ii) 156   (iii) 3825   (iv) 5005   (v) 7429
Solution
(i) 140 = 2² × 5 × 7
(ii) 156 = 2² × 3 × 13
(iii) 3825 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
Q2. Find LCM and HCF and verify LCM × HCF = product:
(i) 12, 15   (ii) 21, 28   (iii) 8, 12   (iv) 36, 48
Solution
(i) 12=2²×3, 15=3×5 → HCF=3, LCM=60. Check: 3×60=180=12×15 ✓
(ii) 21=3×7, 28=2²×7 → HCF=7, LCM=84. Check: 7×84=588 ✓
(iii) 8=2³, 12=2²×3 → HCF=4, LCM=24. Check: 4×24=96 ✓
(iv) 36=2²×3², 48=2⁴×3 → HCF=12, LCM=144. Check: 12×144=1728 ✓
Q3. Find LCM and HCF by prime factorisation:
(i) 12, 15, 21   (ii) 17, 23, 29   (iii) 8, 9, 25
Solution
(i) HCF=3, LCM=2²×3×5×7=420
(ii) All prime → HCF=1, LCM=17×23×29=11339
(iii) HCF=1, LCM=2³×3²×5²=1800
Q4. Given HCF(306, 657) = 9, find LCM(306, 657).
Solution
LCM = (306 × 657) / 9 = 306 × 73 = 22338
Q5. Check whether 6³ can end with digit 0.
Solution

6³ = 2³ × 3³. Since 5 is not a prime factor, 6³ cannot be divisible by 5 and thus cannot end with 0.

Q6. Explain why 7×11×13+13 and 7!+5 are composite numbers.
Solution
7×11×13+13 = 13(77+1) = 13×78 = composite
7!+5 = 5(7!/5+1) = 5×1009 = composite

Exercise 1.3

Q1. Prove that √5 is irrational.
Solution

Assume √5 = a/b (co-prime integers). Then a² = 5b², so 5|a. Let a=5c. Then 25c²=5b², so b²=5c² and 5|b. This contradicts that a and b are co-prime. ∴ √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.
Solution

Assume 3+2√5 = a/b. Then √5 = (a-3b)/2b = rational. But √5 is irrational. Contradiction. ∴ 3+2√5 is irrational.

Q3. Prove these are irrational: (i) 1/√2   (ii) 7√5   (iii) 6 + √2
Solution
(i) 1/√2 = a/b → √2 = b/a = rational. Contradiction. √2 is irrational.
(ii) 7√5 = a/b → √5 = a/7b = rational. Contradiction. √5 is irrational.
(iii) 6+√2 = a/b → √2 = (a-6b)/b = rational. Contradiction. √2 is irrational.

Exercise 1.4

Q1. Determine if 987/10500 is terminating or non-terminating repeating.
Solution
987/10500 = 333/3500
3500 = 2² × 5³ × 7
Since 7 is a prime factor other than 2 and 5, it is non-terminating repeating.
Q3. Decide rational or irrational: (i) 43.123456789   (ii) 0.120120012000120000...
Solution

(i) Terminating decimal → rational, denominator has factors of 2 and 5 only.
(ii) Neither terminating nor repeating → irrational.

Chapter 2: Polynomials

Key Formulas:
• Quadratic: ax²+bx+c, Sum of zeros = -b/a, Product = c/a
• Cubic: ax³+bx²+cx+d, Sum = -b/a, Sum of products = c/a, Product = -d/a
• Division Algorithm: Dividend = Divisor × Quotient + Remainder

Exercise 2.1

Q1. Find the number of zeroes from the given graphs of y = p(x). (i) Graph doesn't meet x-axis (ii) Meets at 1 point (iii) 3 points (iv) 2 points (v) 4 points (vi) 3 points
Solution
Number of zeroes = number of x-axis intersections:
(i) 0 zeroes   (ii) 1 zero   (iii) 3 zeroes
(iv) 2 zeroes   (v) 4 zeroes   (vi) 3 zeroes

Exercise 2.2

Q1. Find zeroes and verify relationships:
(i) x²-2x-8   (ii) 4s²-4s+1   (iii) 6x²-7x-3   (iv) t²-15   (v) 3x²+x-4
Solution

(i) x²-2x-8 = 0

(x-4)(x+2)=0 → x=4, x=-2
Sum = 4+(-2) = 2 = -(-2)/1 ✓
Product = 4×(-2) = -8 = -8/1 ✓

(ii) 4s²-4s+1 = 0

(2s-1)²=0 → s=1/2, 1/2
Sum = 1/2+1/2 = 1 = -(-4)/4 ✓
Product = 1/4 = 1/4 ✓

(iii) 6x²-7x-3 = 0

6x²-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)=0
x = -1/3, 3/2
Sum = 7/6 = -(-7)/6 ✓
Product = -1/2 = -3/6 ✓

(iv) t²-15 = 0

t = ±√15
Sum = 0 ✓ Product = -15 ✓

(v) 3x²+x-4 = 0

3x²+4x-3x-4 = x(3x+4)-1(3x+4) = (x-1)(3x+4)=0
x = 1, -4/3
Sum = -1/3 ✓ Product = -4/3 ✓
Q2. Find quadratic polynomials given sum and product of zeroes:
(i) 1/4, 1   (ii) √2, 1/3   (iii) 0, √5   (iv) 1, 1   (v) -1/4, 1/4   (vi) 4, 1
Solution
Required polynomial: x² - (sum)x + product
(i) x²-x/4+1 or 4x²-x+4
(ii) x²-√2x+1/3 or 3x²-3√2x+1
(iii) x²+√5
(iv) x²-x+1
(v) x²+x/4+1/4 or 4x²+x+1
(vi) x²-4x+1

Exercise 2.3

Q1. Divide and find quotient and remainder:
(i) x³-3x²+5x-3 ÷ x²-2
(ii) x⁴-3x²+4x+5 ÷ x²-x+1
Solution
(i) x³-3x²+5x-3 = (x-3)(x²-2) + 7x-9
Quotient: x-3, Remainder: 7x-9

(ii) Quotient: x²+x-3, Remainder: x+8
Q2. Check whether first polynomial is a factor:
(i) t²-3 divides 2t⁴+3t³-2t²-9t-12?
(ii) x²+3x+1 divides 3x⁴+5x³-7x²+2x+2?
Solution
(i) Quotient=2t²+3t+4, Remainder=0 → Yes, it is a factor
(ii) Quotient=3x²-4x+2, Remainder=0 → Yes, it is a factor
Q3. Find other zeroes of 3x⁴+6x³-2x²-10x-5 if two zeroes are √(5/3) and -√(5/3).
Solution
Factor from given zeroes: x²-5/3, multiply by 3: 3x²-5
Dividing: (3x⁴+6x³-2x²-10x-5) = (3x²-5)(x²+2x+1)
x²+2x+1 = (x+1)² → Other zeroes: x = -1, -1
Q4. Dividing x³-3x²+x+2 by g(x) gives quotient x-2 and remainder -2x+4. Find g(x).
Solution
g(x)(x-2) = x³-3x²+x+2-(-2x+4) = x³-3x²+3x-2
g(x) = (x³-3x²+3x-2)/(x-2) = x²-x+1

Exercise 2.4 (Optional)

Q1. Verify zeroes of 2x³+x²-5x+2 are 1/2, 1, -2 and verify coefficient relationships.
Solution
p(1/2)=2(1/8)+1/4-5/2+2 = 1/4+1/4-5/2+2 = 0 ✓
p(1)=2+1-5+2 = 0 ✓
p(-2)=-16+4+10+2 = 0 ✓
α+β+γ = 1/2+1+(-2) = -1/2 = -b/a ✓
αβ+βγ+γα = 1/2-2-1 = -5/2 = c/a ✓
αβγ = 1/2×1×(-2) = -1 = -d/a ✓
Q3. If zeroes of x³-3x²+x+1 are a-b, a, a+b, find a and b.
Solution
Sum: (a-b)+a+(a+b) = 3a = 3 → a=1
Product: a(a²-b²) = -1 → 1-b²=-1 → b²=2
∴ a=1, b=±√2

Chapter 3: Pair of Linear Equations in Two Variables

Key Conditions for a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0:
• Unique solution: a₁/a₂ ≠ b₁/b₂
• Infinite solutions: a₁/a₂ = b₁/b₂ = c₁/c₂
• No solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂
• Cross multiplication: x/(b₁c₂-b₂c₁) = y/(c₁a₂-c₂a₁) = 1/(a₁b₂-a₂b₁)

Exercise 3.1

Q1. Aftab tells his daughter: 7 years ago I was 7 times as old as you. 3 years from now I shall be 3 times as old. Represent algebraically.
Solution
Let father's age = x, daughter's age = y.
7 years ago: x-7 = 7(y-7) → x-7y = -42
3 years from now: x+3 = 3(y+3) → x-3y = 6
∴ Equations: x-7y+42=0 and x-3y-6=0
Q2. 7 bats and 6 balls cost Rs 3800. 3 bats and 5 balls cost Rs 1750. Find cost of each.
Solution
7x+6y = 3800 ...(i)
3x+5y = 1750 ...(ii)
From (ii): x = (1750-5y)/3
Substituting: 7(1750-5y)/3+6y = 3800
12250-35y+18y = 11400
-17y = -850 → y = 50
x = (1750-250)/3 = 500
∴ Bat = Rs 500, Ball = Rs 50
Q5. Half the perimeter of a rectangular garden is 36m, length is 4m more than width. Find dimensions.
Solution
x = y+4 and (x+y)/2 = 36 → x+y = 72
(y+4)+y = 72 → 2y = 68 → y = 34
x = 38
∴ Length = 38m, Width = 34m

Exercise 3.2

Q1. 10 students in math quiz. Girls are 4 more than boys. Find boys and girls.
Solution
x+y=10, y=x+4
x+(x+4)=10 → 2x=6 → x=3, y=7
∴ Boys = 3, Girls = 7
Q2. 5 pencils and 7 pens cost Rs 50. 7 pencils and 5 pens cost Rs 46. Find cost of each.
Solution
5x+7y=50, 7x+5y=46
Multiply (i) by 5 and (ii) by 7:
25x+35y=250, 49x+35y=322
Subtracting: -24x=-72 → x=3
5(3)+7y=50 → 7y=35 → y=5
∴ Pencil = Rs 3, Pen = Rs 5
Q3. Determine consistent/inconsistent: (i) x+y=5; 2x+2y=10   (ii) x-y=8; 3x-3y=16
Solution
(i) 1/2=1/2=5/10 → all equal → Consistent, infinitely many solutions
(ii) 1/3=(-1)/(-3)=1/3 ≠ 8/16=1/2 → Inconsistent (no solution)
Q5. Half perimeter of rectangle is 36m, length 4m more than width. Find dimensions.
Solution

Same as Exercise 3.1 Q5: Length=38m, Width=34m

Exercise 3.3

Q1. Solve by substitution: (i) x+y=14; x-y=4   (ii) s-t=3; s/3+t/2=6
Solution

(i)

x=14-y
(14-y)-y=4 → 14-2y=4 → y=5
x=14-5=9
∴ x=9, y=5

(ii)

2s+3t=36 (multiplying by 6)
s=t+3
2(t+3)+3t=36 → 5t+6=36 → t=6
s=6+3=9
∴ s=9, t=6
Q2. Solve 2x+3y=11 and 2x-4y=-2, find m if y=mx+3.
Solution
Subtracting: (2x+3y)-(2x-4y)=11-(-2)
7y=13 → y=13/7
2x=11-3(13/7)=38/7 → x=19/7
For y=mx+3: 13/7=m(19/7)+3
13/7-21/7=19m/7
-8/7=19m/7 → m=-8/19
Q4. Solve by elimination: (i) x+y=5; 2x-3y=4   (ii) 3x+4y=10; 2x-2y=2
Solution

(i)

Multiply (i) by 3: 3x+3y=15
Add to (ii): 5x=19 → x=19/5
y=5-19/5=6/5
∴ x=19/5, y=6/5

(ii)

Multiply (ii) by 2: 4x-4y=4
Add to (i): 7x=14 → x=2
2(2)-2y=2 → y=1
∴ x=2, y=1
Q5. A library charges fixed + per day. Sheela paid Rs 27 for 7 days, Sunita paid Rs 21 for 5 days. Find charges.
Solution
Fixed charge = x, extra per day = y
x+4y = 27 (3 days fixed + 4 extra)
x+2y = 21 (3 days fixed + 2 extra)
Subtracting: 2y=6 → y=3
x=21-6=15
∴ Fixed = Rs 15, Extra = Rs 3/day

Exercise 3.4

Q1. Solve by cross multiplication: (i) x-3y-3=0; 3x-9y-2=0   (ii) 2x+y-5=0; 3x+2y-8=0
Solution
(i) 1/3=(-3)/(-9)=1/3 ≠ -3/(-2)=3/2 → No solution

(ii) x/(-8+10) = y/(-15+16) = 1/(4-3)
x/2 = y/1 = 1/1
∴ x=2, y=1

Exercise 3.5

Q2. Find a and b for infinitely many solutions: 2x+3y=7; (a-b)x+(a+b)y=3a+b-2
Solution
2/(a-b) = 3/(a+b) = 7/(3a+b-2)
From first two: 2(a+b)=3(a-b)
2a+2b=3a-3b → a=5b ...(i)
From first and third: 2(3a+b-2)=7(a-b)
6a+2b-4=7a-7b → a=9b-4 ...(ii)
From (i) and (ii): 5b=9b-4 → 4b=4 → b=1
a=5(1)=5
∴ a=5, b=1
Q3. For what k does 3x+y=1; (2k-1)x+(k-1)y=2k+1 have no solution?
Solution
3/(2k-1) = 1/(k-1) ≠ 1/(2k+1)
3(k-1) = 2k-1
3k-3 = 2k-1 → k=2
Check: 3/(3)=1/(1)=1 ≠ 1/(5) ✓
∴ k=2

Exercise 3.6

Q1. Solve: 1/x+1/y=10; 1/x-1/y=2
Solution
Let p=1/x, q=1/y
p+q=10, p-q=2
Adding: 2p=12 → p=6, q=4
∴ x=1/6, y=1/4
Q2. Solve: 6/x+3/y=6; 4/x+9/y=5
Solution
Let p=1/x, q=1/y
6p+3q=6 → 2p+q=2 ...(i)
4p+9q=5 ...(ii)
From (i): q=2-2p
4p+9(2-2p)=5
4p+18-18p=5
-14p=-13 → p=13/14 → x=14/13
q=2-26/14=2/14=1/7 → y=7
∴ x=14/13, y=7

Exercise 3.7 (Optional)

Q4. A boat goes 30km upstream and 44km downstream in 10h. It goes 40km upstream and 55km downstream in 13h. Find speed of boat and stream.
Solution
Let boat speed=x, stream speed=y
30/(x-y)+44/(x+y)=10 ...(i)
40/(x-y)+55/(x+y)=13 ...(ii)
Let p=1/(x-y), q=1/(x+y)
30p+44q=10, 40p+55q=13
Multiply first by 4, second by 3:
120p+176q=40, 120p+165q=39
11q=1 → q=1/11 → x+y=11
30p=10-44/11=6 → p=1/5 → x-y=5
x=(11+5)/2=8, y=(11-5)/2=3
∴ Boat=8 km/h, Stream=3 km/h

Chapter 4: Quadratic Equations

Key Formulas:
• Standard form: ax²+bx+c=0
• Quadratic formula: x=(-b±√(b²-4ac))/2a
• Discriminant D=b²-4ac: D>0 two distinct roots, D=0 equal roots, D<0 no real roots
• Sum of roots = -b/a, Product of roots = c/a

Exercise 4.1

Q1. Check if quadratic: (i) (x+1)²=2(x-3)   (ii) x²-2x=(-2)(3-x)   (iii) (x-2)(x+1)=(x-1)(x+3)
Solution
(i) x²+2x+1=2x-6 → x²+7=0 → Quadratic (Yes)
(ii) x²-2x=-6+2x → x²-4x+6=0 → Quadratic (Yes)
(iii) x²-x-2=x²+2x-3 → -3x+1=0 → Not quadratic (linear)

Exercise 4.2

Q1. Find roots by factorisation:
(i) x²-3x-10=0   (ii) 2x²+x-6=0   (iii) √2x²+7x+5√2=0   (iv) 2x²-x+1/8=0
Solution

(i) x²-3x-10=0

x²-5x+2x-10=0
(x-5)(x+2)=0
∴ x=5 or x=-2

(ii) 2x²+x-6=0

2x²+4x-3x-6=0
2x(x+2)-3(x+2)=0
(2x-3)(x+2)=0
∴ x=3/2 or x=-2

(iii) √2x²+7x+5√2=0

√2x²+2x+5x+5√2=0
√2x(x+√2)+5(x+√2)=0
(√2x+5)(x+√2)=0
∴ x=-5/√2 or x=-√2

(iv) 2x²-x+1/8=0

Multiply by 8: 16x²-8x+1=0
(4x-1)²=0
∴ x=1/4 (repeated root)
Q4. Sum of squares of two consecutive positive integers is 365. Find them.
Solution
x²+(x+1)²=365
x²+x²+2x+1=365
2x²+2x-364=0
x²+x-182=0
(x+14)(x-13)=0
x=13 (positive)
∴ Integers are 13 and 14
Q5. Altitude of right triangle is 7cm less than base, hypotenuse 13cm. Find sides.
Solution
Base=x, altitude=x-7
x²+(x-7)²=13²
x²+x²-14x+49=169
2x²-14x-120=0
x²-7x-60=0
(x-12)(x+5)=0 → x=12
Base=12cm, Altitude=5cm
∴ Sides are 12cm and 5cm
Q6. Cost per article is 3 more than twice number of articles. Total cost Rs 90. Find number and cost.
Solution
x(2x+3)=90
2x²+3x-90=0
2x²+15x-12x-90=0
x(2x+15)-6(2x+15)=0
(x-6)(2x+15)=0 → x=6
Cost=2(6)+3=15
∴ Articles=6, Cost=Rs 15 each

Exercise 4.3

Q1. Solve by completing the square: (i) 2x²+x-4=0   (ii) 4x²+4√3x+3=0   (iii) 2x²+x+4=0
Solution

(i)

x²+x/2=2
x²+x/2+1/16=2+1/16
(x+1/4)²=33/16
x=(-1±√33)/4

(ii)

x²+√3x+3/4=0
(x+√3/2)²=0
∴ x=-√3/2 (equal roots)

(iii)

D=1-4(2)(4)=1-32=-31<0
∴ No real roots
Q5. Find nature of roots: (i) 2x²-3x+5=0   (ii) 3x²-4√3x+4=0   (iii) 2x²-6x+3=0
Solution
(i) D=9-40=-31<0 → No real roots
(ii) D=48-48=0 → Equal roots: x=4√3/6=2/√3
(iii) D=36-24=12>0 → Two distinct: x=(6±2√3)/4=(3±√3)/2

Exercise 4.4

Q2. Find k for equal roots: (i) 2x²+kx+3=0   (ii) kx(x-2)+6=0
Solution
(i) D=k²-24=0 → k²=24 → k=±2√6
(ii) kx²-2kx+6=0
D=4k²-24k=0 → 4k(k-6)=0
k≠0 so k=6
Q3. Can a rectangular grove have length twice breadth and area 800m²?
Solution
Let breadth=x, length=2x
2x×x=800 → 2x²=800 → x²=400 → x=20
∴ Yes. Length=40m, Breadth=20m
Q4. Sum of ages of two friends is 20. Four years ago product was 48. Is this possible?
Solution
(x-4)(20-x-4)=48
(x-4)(16-x)=48
-x²+20x-64=48
x²-20x+112=0
D=400-448=-48<0
∴ Not possible (no real roots)
Q5. Can a park have perimeter 80m and area 400m²?
Solution
l+b=40, l×b=400
b(40-b)=400
b²-40b+400=0
(b-20)²=0 → b=20, l=20
∴ Yes, it is a square of side 20m

Chapter 5: Arithmetic Progressions

Key Formulas:
• nth term: aₙ = a + (n-1)d
• Sum of n terms: Sₙ = n/2 [2a + (n-1)d] = n/2 (a + l)
• Common difference: d = aₙ - aₘ⁺⁴

Exercise 5.1

Q2. Write first 4 terms of AP: (i) a=10,d=10   (ii) a=-2,d=0   (iii) a=4,d=-3   (iv) a=-1,d=1/2   (v) a=-1.25,d=-0.25
Solution
(i) 10, 20, 30, 40
(ii) -2, -2, -2, -2
(iii) 4, 1, -2, -5
(iv) -1, -1/2, 0, 1/2
(v) -1.25, -1.50, -1.75, -2.00
Q3. Write first term and common difference: (i) 3, 1, -1, -3...   (ii) -5, -1, 3, 7...   (iii) 1/3, 5/3, 9/3...   (iv) 0.6, 1.7, 2.8...
Solution
(i) a=3, d=-2
(ii) a=-5, d=4
(iii) a=1/3, d=4/3
(iv) a=0.6, d=1.1
Q4. Which are APs? (i) 2,4,8,16...   (ii) 2,5/2,3,7/2...   (iii) -1.2,-3.2,-5.2,-7.2...   (iv) -10,-6,-2,2...
Solution
(i) Not AP (d not constant: 2,4,8)
(ii) AP, d=1/2, next 3 terms: 4, 9/2, 5
(iii) AP, d=-2, next 3 terms: -9.2, -11.2, -13.2
(iv) AP, d=4, next 3 terms: 6, 10, 14

Exercise 5.2

Q2. 30th term of AP: 10, 7, 4... is: (A) 97 (B) 77 (C) -77 (D) -87
Solution
a=10, d=-3
aₙ⁴₄=10+29(-3)=10-87=-77
∴ (C) -77
Q4. Which term of AP: 3, 8, 13, 18... is 78?
Solution
78=3+(n-1)5
75=5(n-1) → n-1=15 → n=16
∴ 16th term
Q5. Find number of terms: 7, 13, 19, ..., 205
Solution
205=7+(n-1)6
198=6(n-1) → n-1=33 → n=34
∴ 34 terms
Q8. Find the 20th term if 5th term is 21 and d=6.
Solution
a+4(6)=21 → a=21-24=-3
aₙ⁴⁵=-3+19(6)=-3+114=111
Q14. How many three-digit numbers are divisible by 7?
Solution
AP: 105, 112, ..., 994
994=105+(n-1)7
889=7(n-1) → n-1=127 → n=128
∴ 128 numbers
Q16. Sum of first 22 terms with d=7 and aₙ⁴⁵=149.
Solution
a+21(7)=149 → a=2
Sₙ⁴⁵=22/2[2(2)+21(7)]=11×151=1661
Q18. Rs 700 given as 7 prizes, each Rs 20 less than preceding. Find each prize.
Solution
Sₙ=7/2[2a+6(-20)]=700
7/2(2a-120)=700
7(a-60)=700 → a=160
∴ Prizes: Rs 160, 140, 120, 100, 80, 60, 40

Exercise 5.3

Q1. Find sum: 2, 7, 12, ..., to 10 terms
Solution
a=2, d=5, n=10
Sₙ=10/2[2(2)+9(5)]=5[4+45]=5×49=245
Q4. How many terms of AP: 9, 17, 25... give sum 636?
Solution
636=n/2[18+(n-1)8]
1272=n(10n+8)
8n²+10n-1272=0
4n²+5n-636=0
n=(-5+√(25+10176))/8=(-5+101)/8=12
∴ n=12
Q5. First term 5, last term 45, sum 400. Find n and d.
Solution
400=n/2(5+45)=25n → n=16
45=5+15d → d=40/15=8/3
Q7. Sum of first 40 positive integers divisible by 6.
Solution
AP: 6, 12, 18, ..., 240
Sₙ=40/2[2(6)+39(6)]=20[12+234]=20×246=4920
Q9. 600 units in 3rd year, 700 in 7th year. Find (i) 1st year (ii) 10th year (iii) total in 7 years.
Solution
a+2d=600, a+6d=700
4d=100 → d=25, a=550
(i) 550 units
(ii) 550+9(25)=775 units
(iii) Sₙ=7/2(550+700)=7/2×1250=4375 units
Q10. Sum of first 24 terms of AP: aₙ=3+4n.
Solution
aₙ=3+4n
a=3+4=7, d=4
Sₙ=24/2[2(7)+23(4)]=12[14+92]=12×106=1272

Exercise 5.4 (Optional)

Q5. Students plant trees: class k plants k per section, 3 sections per class, classes 1-12. Total trees?
Solution
Total = 3×(1+2+3+...+12)
= 3×12/2×(1+12)
= 3×6×13
= 234 trees

Chapter 6: Triangles

Key Theorems:
• BPT (Thales): Line parallel to one side divides other two proportionally
• AA Similarity: Two angles equal → similar triangles
• SSS Similarity: All corresponding sides proportional → similar
• SAS Similarity: Two sides proportional + included angle equal → similar
• Area ratio: Area(ΔABC)/Area(ΔPQR) = (AB/PQ)²

Exercise 6.1

Q1. Fill in blanks: (i) All circles are ___. (ii) All squares are ___. (iii) All ___ triangles are similar.
Solution
(i) similar (same shape)
(ii) similar (same shape)
(iii) equilateral triangles are similar
Q2. Give two examples: (i) similar figures (ii) non-similar figures
Solution
(i) Similar: Two equilateral triangles, two circles
(ii) Non-similar: A square and a rectangle, a triangle and a parallelogram
Q3. Is ΔABC (6,8,10) similar to ΔPQR (3,4,5)?
Solution
AB/PQ = 6/3 = 2
AC/PR = 8/5 = 1.6
BC/QR = 10/4 = 2.5
Ratios not equal → Not similar

Exercise 6.2

Q1. In ΔABC, DE || BC. AD=3.6cm, DB=5.4cm, AE=2.4cm. Find EC.
Solution
By BPT: AD/DB = AE/EC
3.6/5.4 = 2.4/EC
EC = 2.4 × 5.4/3.6 = 3.6 cm
Q2. DE || BC, AD/DB=3/5, AC=4.8cm. Find AE.
Solution
AD/DB = AE/EC = 3/5
AE=3k, EC=5k
3k+5k=4.8 → 8k=4.8 → k=0.6
AE=3(0.6)=1.8 cm
Q5. DE || BC. AD=x, DB=x-2, AE=x+2, EC=x-1. Find EC.
Solution
x/(x-2) = (x+2)/(x-1)
x(x-1) = (x+2)(x-2)
x²-x = x²-4
-x=-4 → x=4
EC = 4-1 = 3
Q7. AD/AB=AE/AC=3/8. Prove DE || BC.
Solution
AD/AB = AE/AC = 3/8
By converse of Basic Proportionality Theorem,
DE || BC.

Exercise 6.3

Q1. ΔABC: ∠A=60°, ∠B=80°. ΔPQR: ∠Q=60°, ∠R=40°. Are they similar?
Solution
ΔABC: ∠C = 180-60-80 = 40°
∠A=∠Q=60°, ∠C=∠R=40°
By AA similarity: ΔABC ~ ΔPQR
Q2. ΔDEF: DE=3cm, DF=6cm, EF=5cm. ΔPQR: PQ=5cm, PR=10cm, QR=7.5cm. Similar?
Solution
DE/PQ = 3/5 = 0.6
DF/PR = 6/10 = 0.6
EF/QR = 5/7.5 = 2/3 ≠ 0.6
∴ Not similar
Q6. In ΔABC, ∠B=90°, BD ⊥ AC. Prove ΔDBA ~ ΔDBC.
Solution
In ΔABC with ∠B=90°:
Let ∠A=α, then ∠C=90°-α
In ΔABD: ∠ADB=90°, ∠ABD=90°-α
In ΔBDC: ∠BDC=90°, ∠DBC=α
In ΔDBA and ΔDBC:
∠ADB=∠BDC=90°
∠ABD=90°-α=∠C
By AA: ΔDBA ~ ΔDBC

Exercise 6.4

Q1. ΔABC ~ ΔDEF, AB=2.5cm, DE=5cm, perimeter of ΔABC=20cm. Find perimeter of ΔDEF.
Solution
AB/DE = Perimeter(ABC)/Perimeter(DEF)
2.5/5 = 20/P
P = 20 × 5/2.5 = 40 cm
Q3. Similar triangles have sides ratio 3:5. Area of first is 81cm². Find area of second.
Solution
Area ratio = (3/5)² = 9/25
81/Area = 9/25
Area = 81 × 25/9 = 225 cm²
Q4. ΔABC ~ ΔDEF, BC=3cm, EF=4cm. Area(ΔABC)=12cm². Find Area(ΔDEF).
Solution
BC/EF = 3/4
Area(ABC)/Area(DEF) = (3/4)² = 9/16
12/Area = 9/16
Area = 12 × 16/9 = 64/3 cm²

Exercise 6.5

Q1. Which are right-angled? (i) 7,24,25   (ii) 3,8,6   (iii) 50,80,100   (iv) 13,12,5
Solution
(i) 7²+24²=49+576=625=25² ✓ Right-angled
(ii) 3²+6²=45≠64 ✗ Not right-angled
(iii) 50²+80²=8900≠10000 ✗ Not right-angled
(iv) 12²+5²=169=13² ✓ Right-angled
Q3. In ΔABC, AD ⊥ BC. Prove: AB²+CD²=BD²+AC².
Solution
In ΔABD: AB² = AD²+BD² ...(i)
In ΔACD: AC² = AD²+CD² ...(ii)
(i)-(ii): AB²-AC² = BD²-CD²
∴ AB²+CD² = BD²+AC²
Q7. AD ⊥ BC, BD=3CD. Prove 2AB²=2AC²+BC².
Solution
Let CD=x, BD=3x, BC=4x
AB²=AD²+9x² ...(i)
AC²=AD²+x² ...(ii)
(i)-(ii): AB²-AC²=8x²
AB²=AC²+8x²
2AB²=2AC²+16x²
BC²=16x²
∴ 2AB²=2AC²+BC²

Chapter 7: Coordinate Geometry

Key Formulas:
• Distance: d = √[(x₂-x₁)² + (y₂-y₁)²]
• Section formula: P = ((mx₂+nx₁)/(m+n), (my₂+ny₁)/(m+n))
• Midpoint: M = ((x₁+x₂)/2, (y₁+y₂)/2)
• Area of triangle = ½|x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Exercise 7.1

Q1. Find distance: (i) (2,3) and (4,1)   (ii) (-5,7) and (-1,3)   (iii) (a,b) and (-a,-b)
Solution
(i) √[(4-2)²+(1-3)²] = √(4+4) = 2√2
(ii) √[(-1+5)²+(3-7)²] = √(16+16) = 4√2
(iii) √[(-a-a)²+(-b-b)²] = √(4a²+4b²) = 2√(a²+b²)
Q2. Find distance of (3,-4) from origin.
Solution
√(3²+(-4)²) = √(9+16) = √25 = 5 units
Q3. Find point on x-axis equidistant from (5,9) and (-4,6).
Solution
Let point be (x,0)
√[(x-5)²+81] = √[(x+4)²+36]
(x-5)²+81 = (x+4)²+36
x²-10x+106 = x²+8x+52
-18x = -54 → x=3
∴ Point = (3, 0)
Q5. Find x if distance between (x,-1) and (1,3) is 5 units.
Solution
√[(1-x)²+(3+1)²] = 5
(1-x)²+16 = 25
(1-x)² = 9
1-x = ±3
x = -2 or x = 4
∴ x = -2 or x = 4
Q7. Find coordinates dividing (-1,7) and (4,-3) in ratio 2:3.
Solution
P = ((2×4+3×(-1))/(2+3), (2×(-3)+3×7)/5)
= ((8-3)/5, (-6+21)/5)
= (5/5, 15/5) = (1, 3)
Q9. Find points of trisection of line joining (4,-1) and (-2,-3).
Solution
First point (1:2): ((1×(-2)+2×4)/3, (1×(-3)+2×(-1))/3)
= (6/3, -5/3) = (2, -5/3)
Second point (2:1): ((2×(-2)+1×4)/3, (2×(-3)+1×(-1))/3)
= (0/3, -7/3) = (0, -7/3)

Exercise 7.2

Q1. Find coordinates dividing (-1,7) and (4,-3) internally in ratio 2:3.
Solution
P = ((2×4+3×(-1))/5, (2×(-3)+3×7)/5)
= (5/5, 15/5) = (1, 3)
Q5. Find centroid of ΔABC: A(-1,3), B(1,-1), C(5,1).
Solution
G = ((-1+1+5)/3, (3-1+1)/3)
= (5/3, 3/3) = (5/3, 1)

Exercise 7.3

Q1. Find area: (i) A(7,2), B(9,10), C(3,6)   (ii) A(-2,-3), B(3,2), C(-1,-8)
Solution
(i) Area = ½|7(10-6)+9(6-2)+3(2-10)|
= ½|28+36-24| = ½|40| = 20 sq. units

(ii) Area = ½|-2(2+8)+3(-8+3)+(-1)(-3-2)|
= ½|-20-15+5| = ½|-30| = 15 sq. units
Q4. Find k if A(-2,3), B(k,1), C(-1,4) are collinear.
Solution
Area = 0 for collinear points
½|-2(1-4)+k(4-3)+(-1)(3-1)| = 0
½|6+k-2| = 0
k+4 = 0
k = -4

Chapter 8: Introduction to Trigonometry

Key Formulas:
• sin θ = Perpendicular/Hypotenuse, cos θ = Base/Hypotenuse, tan θ = P/B
• cosec θ = 1/sinθ, sec θ = 1/cosθ, cot θ = 1/tanθ
• sin²θ + cos²θ = 1
• 1 + tan²θ = sec²θ
• 1 + cot²θ = cosec²θ

Exercise 8.1

Q1. In ΔABC, right-angled at B, AB=24cm, BC=7cm. Find sin A, cos A, sin C, cos C.
Solution
AC = √(24²+7²) = √(576+49) = √625 = 25cm
sin A = BC/AC = 7/25
cos A = AB/AC = 24/25
sin C = AB/AC = 24/25
cos C = BC/AC = 7/25
Q3. If sin A = 3/4, find cos A and tan A.
Solution
sin²A+cos²A=1
9/16+cos²A=1
cos²A=7/16
cos A=√7/4
tan A=sin A/cos A = 3/√7 = 3√7/7
Q4. Given 15 cot A = 8, find sin A and sec A.
Solution
cot A = 8/15 = Adjacent/Opposite
Hypotenuse = √(8²+15²) = √(64+225) = √289 = 17
sin A = 15/17
sec A = 17/8
Q5. Given sec θ = 13/12, find all trigonometric ratios.
Solution
sec θ=13/12 → cos θ=12/13
sin²θ=1-144/169=25/169 → sin θ=5/13
tan θ=5/12
cosec θ=13/5
cot θ=12/5
Q9. ΔPQR, right-angled at Q, PR+QR=25cm, PQ=5cm. Find sin P, cos P, tan P.
Solution
PQ=5, let QR=x, PR=25-x
5²+x²=(25-x)²
25+x²=625-50x+x²
50x=600 → x=12
QR=12, PR=13
sin P=12/13, cos P=5/13, tan P=12/5

Exercise 8.2

Q1. Evaluate: sin 60° cos 30° + sin 30° cos 60°
Solution
= (√3/2)(√3/2) + (1/2)(1/2)
= 3/4 + 1/4 = 1
Q2. Evaluate: 2 tan² 45° + cos² 30° - sin² 60°
Solution
= 2(1) + 3/4 - 3/4 = 2
Q3. Evaluate: (cos 0°+sin 45°)/(sin 0°+cos 45°)
Solution
= (1+1/√2)/(0+1/√2)
= (√2+1)/√2 × √2/√2 = √2+1
Q4. Evaluate: (5cos² 60°+4sec² 30°-tan² 45°)/(sin² 30°+cos² 30°)
Solution
= (5/4+16/3-1)/1
= (15+64-12)/12
= 67/12
Q6. Evaluate: (sin 30°/tan 45°)+(cos 60°/tan 30°)
Solution
= (1/2)/1 + (1/2)/(1/√3)
= 1/2 + √3/2
= (1+√3)/2

Exercise 8.3

Q1. Evaluate: (i) sin² 30°+cos² 30°   (ii) sin² 60²+cos² 60°
Solution
(i) (1/2)²+(√3/2)² = 1/4+3/4 = 1
(ii) (√3/2)²+(1/2)² = 3/4+1/4 = 1
Q4. Evaluate: cos² 0°+cos² 30°+cos² 45°+cos² 60°+cos² 90°
Solution
= 1+3/4+1/2+1/4+0
= 1+0.75+0.5+0.25
= 5/2

Exercise 8.4

Q1. Express sin 67°+cos 75° in terms of angles between 0° and 45°.
Solution
sin 67°=sin(90°-23°)=cos 23°
cos 75°=cos(90°-15°)=sin 15°
∴ sin 67°+cos 75° = cos 23°+sin 15°
Q3. If tan 2A=cot(A-18°), find A.
Solution
tan 2A=cot(A-18°)=tan(90°-(A-18°))
tan 2A=tan(108°-A)
2A=108°-A
3A=108°
A=36°
Q4. If tan A=cot B, prove A+B=90°.
Solution
tan A=cot B=tan(90°-B)
A=90°-B
∴ A+B=90°
Q5. If sec 4A=cosec(A-20°), find A.
Solution
sec 4A=cosec(A-20°)
cos(4A)=sin(A-20°)
sin(90°-4A)=sin(A-20°)
90°-4A=A-20°
110°=5A
A=22°
Q6. In ΔABC, show sin((B+C)/2)=cos(A/2).
Solution
A+B+C=180°
B+C=180°-A
(B+C)/2=90°-A/2
sin((B+C)/2)=sin(90°-A/2)=cos(A/2)

Chapter 9: Some Applications of Trigonometry

Key Concepts:
• Angle of elevation: angle above horizontal when looking up
• Angle of depression: angle below horizontal when looking down
• Line of sight: line from observer to object
• Use sin, cos, tan ratios in right triangles

Exercise 9.1

Q1. A circus artist climbs a 20m rope from top of a vertical pole to ground. Rope makes 30° angle with ground. Find height of pole.
Solution
sin 30° = h/20
1/2 = h/20
h = 20/2 = 10 m
Q2. A tower stands vertically. From 30m away, angle of elevation of top is 30°. Find height of tower.
Solution
tan 30° = h/30
1/√3 = h/30
h = 30/√3 = 30√3/3 = 10√3 m
Q3. Kite is at 60m height. String makes 60° with ground. Find length of string.
Solution
sin 60° = 60/l
√3/2 = 60/l
l = 120/√3 = 120√3/3 = 40√3 m
Q4. From 12m away from building, angle of elevation of top is 30°. Find height.
Solution
tan 30° = h/12
h = 12/√3 = 4√3 m
Q5. From top of 7m building, angle of elevation of cable tower top is 60°, depression of foot is 45°. Find tower height.
Solution
Let tower height=H, distance=d
tan 45°=7/d → d=7m
tan 60°=(H-7)/7
√3=(H-7)/7
H-7=7√3
H=7(1+√3)=7(1+√3) m
Q6. 1.5m tall boy, angle of elevation to top of 30m building goes from 30° to 60° as he walks towards building. Find distance walked.
Solution
Height above boy = 30-1.5 = 28.5m
At 60°: tan 60°=28.5/d₁ → d₁=28.5/√3
At 30°: tan 30°=28.5/d₂ → d₂=28.5√3
Distance walked = 28.5√3 - 28.5√3/3
= 28.5√3 × 2/3 = 19√3 m
Q7. From ground, angles of elevation of bottom and top of tower on 20m building are 45° and 60°. Find tower height.
Solution
tan 45°=20/d → d=20m
tan 60°=(20+h)/20
√3=(20+h)/20
h=20√3-20=20(√3-1) m
Q10. Two poles of equal height on either side of 80m road. Angles of elevation 60° and 30°. Find height and distances.
Solution
Let height=h, distance from 60° pole=x
tan 60°=h/x → h=x√3
tan 30°=h/(80-x)
x√3=(80-x)/√3
3x=80-x → 4x=80 → x=20
h=20√3 m
∴ Height=20√3 m, distances=20m and 60m

Chapter 10: Circles

Key Theorems:
• Tangent is perpendicular to radius at point of contact
• Lengths of tangents from an external point are equal
• Two tangents can be drawn from an external point
• Angle between tangents from P = 180° - (angle subtended at centre)

Exercise 10.1

Q1. How many tangents can be drawn through a point inside a circle?
Solution
Zero. Tangents can only be drawn from a point on or outside the circle.
Q2. How many tangents through a point on the circle?
Solution
Exactly one. The tangent at any point on a circle is unique.
Q3. How many tangents through a point outside the circle?
Solution
Exactly two. Two tangents can be drawn from an external point.

Exercise 10.2

Q1. From Q, tangent length=24cm, distance from centre=25cm. Find radius.
Solution
r²+24²=25²
r²+576=625
r²=49
r=7 cm
Q2. TP and TQ are tangents, ∠PTQ=60°. Find ∠OPQ.
Solution
TP=TQ → ΔTPQ is isosceles
∠TPQ=∠TQP=(180-60)/2=60°
OP ⊥ TP → ∠OPT=90°
∠OPQ=90°-60°=30°
Q3. PA and PB are tangents inclined at 80°. Find ∠POA.
Solution
In quadrilateral OAPB:
∠OAP=∠OBP=90°
∠AOB=360-90-90-80=100°
∠POA=100/2=50°
Q4. Prove tangent segments from external point are equal.
Solution
Let PA and PB be tangents from P.
In ΔOAP and ΔOBP:
OA=OB (radii)
OP=OP (common)
∠OAP=∠OBP=90°
By RHS congruence, ΔOAP ≅ ΔOBP
∴ PA=PB
Q6. Prove ∠PTQ=2∠OPQ.
Solution
Let ∠OPQ=x
In ΔOPQ: OP=OQ → ∠OQP=x
In ΔTPQ: ∠TPQ=∠TQP=90°-x
∠PTQ=180-2(90-x)=2x
∴ ∠PTQ=2∠OPQ
Q7. Tangents at ends of a diameter are parallel. Prove.
Solution
Let AB be diameter. Tangents at A and B are perpendicular to AB. Since both are perpendicular to the same line AB, they are parallel.
∴ Tangents at ends of diameter are parallel.
Q8. Angle between tangents is supplementary to angle subtended at centre.
Solution
In quadrilateral OAPB:
∠OAP=∠OBP=90°
∠AOB+∠APB+90+90=360
∠AOB+∠APB=180
∴ Supplementary.
Q11. Tangent PQ at P, radius 5cm, OQ=12cm. Find PQ.
Solution
OP=5cm, OQ=12cm
PQ²=OQ²-OP²=144-25=119
PQ=√119 cm

Chapter 11: Constructions

Key Constructions:
• Division of a line segment in given ratio
• Construction of similar triangles (scale factor)
• Construction of tangents to a circle from external point

Exercise 11.1

Q1. Divide a line segment of length 7.6cm in the ratio 5:8. Measure the two parts.
Solution
Steps of Construction:
1. Draw AB = 7.6cm
2. Draw ray AX making acute angle with AB
3. Mark 5+8=13 equal points on AX: A₁, A₂,...,A₁₃
4. Join A₁₃ to B
5. Draw line through A₅ parallel to A₁₃B meeting AB at C

AC = 5/13 × 7.6 ≈ 2.92 cm
CB = 8/13 × 7.6 ≈ 4.68 cm
Q2. Construct ΔABC with sides 4cm, 5cm, 6cm and then a similar triangle with ratio 2/3.
Solution
Steps:
1. Draw ΔABC: AB=4cm, BC=6cm, AC=5cm
2. Draw ray BX making acute angle with BC
3. Mark 3 equal points on BX (denominator=3)
4. Join 3rd point (B₃) to C
5. Draw line through 2nd point parallel to B₃C
6. This meets BC at C₂. Draw C₂A₂ || CA

New triangle sides: 8/3 cm, 10/3 cm, 4 cm
Q3. Construct ΔABC with sides 5cm, 6cm, 7cm and similar triangle with sides 7/5 of corresponding sides.
Solution
Steps:
1. Draw ΔABC: AB=5cm, BC=7cm, AC=6cm
2. Draw ray BX, mark 5 equal points
3. Join B₅ to C
4. Draw line through B₇ (extends beyond) parallel to B₅C
5. This meets BC extended at C₄
6. Draw C₄A₄ || CA to meet BA extended

New triangle sides: 7cm, 49/5 cm, 42/5 cm

Exercise 11.2

Q1. Draw circle radius 6cm. From point 10cm away, construct tangents. Measure lengths.
Solution
Steps:
1. Draw circle with centre O, radius 6cm
2. Mark point P such that OP=10cm
3. Bisect OP at M (midpoint)
4. Draw circle with centre M, radius MO
5. This circle meets given circle at A and B
6. Join PA and PB

PA=PB=√(10²-6²)=√64=8cm
Q2. Construct tangents to circle radius 4cm from point on concentric circle radius 6cm.
Solution
Steps:
1. Draw two concentric circles with centre O, radii 4cm and 6cm
2. Mark point P on outer circle
3. Bisect OP at M
4. Draw circle centre M radius MO
5. Meets inner circle at A and B
6. Join PA and PB

PA=PB=√(6²-4²)=√20=2√5 cm
Q3. Draw circle radius 3cm. Take point P 7cm from centre. Construct pair of tangents.
Solution
Steps:
1. Draw circle centre O, radius 3cm
2. OP = 7cm
3. Bisect OP at M
4. Draw circle centre M, radius 3.5cm
5. Meets given circle at A and B
6. PA and PB are required tangents

Length = √(7²-3²) = √40 = 2√10 cm

Chapter 12: Areas Related to Circles

Key Formulas:
• Area of circle = πr²
• Perimeter (circumference) = 2πr
• Area of sector = θ/360 × πr² (or ½r²θ in radians)
• Length of arc = θ/360 × 2πr
• Area of segment = Area of sector - Area of corresponding triangle

Exercise 12.1

Q1. The radii of two circles are 19cm and 9cm. Find radius of circle having circumference equal to sum of circumferences of two circles.
Solution
2π(19)+2π(9)=2πr
19+9=r
r=28 cm
Q2. Radii are 8cm and 6cm. Find radius of circle with area equal to sum of areas.
Solution
π(8²)+π(6²)=πr²
64+36=r²
r²=100
r=10 cm
Q3. Diameter of goldsmith's circular sheet is 3cm. How many sheets of diameter 3mm are made from it?
Solution
π(30²) = n × π(0.3²)
900 = n × 0.09
n = 900/0.09 = 10000 sheets
Q4. Find π from: √(5/7)(r+21)=r-5
Solution
√(5/7)(r+21)=r-5
5(r+21)/7=(r-5)²
5r+105=7(r²-10r+25)
5r+105=7r²-70r+175
7r²-75r+70=0
r=(75±√(5625-1960))/14
r=(75±√3665)/14
Since r must be positive and valid:
r≈ 4.5 cm (taking positive root that satisfies equation)

Exercise 12.2

Q1. Find area of sector with radius 14cm and angle 90°.
Solution
Area = θ/360 × πr²
= 90/360 × π × 14²
= 1/4 × 196π
= 49π cm²
Q2. Find area of sector with radius 21cm and angle 120°.
Solution
Area = 120/360 × π × 21²
= 1/3 × 441π
= 147π cm²
Q3. Find area of quadrant of circle with circumference 22cm.
Solution
2πr=22 → r=22/2π=7/2 cm
Area of quadrant = 1/4 × π × (7/2)²
= π/4 × 49/4
= 49π/16 cm²
Q4. Find area of sector with radius 15cm and angle 60°.
Solution
Area = 60/360 × π × 15²
= 1/6 × 225π
= 75π/2 cm²

Exercise 12.3

Q1. Find area of shaded region where PQ=24cm, PR=7cm, O is centre of circle.
Solution
In ΔPQR: ∠R=90° (angle in semicircle)
RQ = √(24²-7²) = √(576-49) = √527...
Actually PQ is diameter=24, so radius=12cm
PR=7cm, RQ=√(24²-7²)=√527
Area of ΔPQR = 1/2 × 7 × √527...

Let me recalculate: PQ=24 (diameter), so radius=12
Area of semicircle = π × 12²/2 = 72π
Area of ΔPQR = 1/2 × PR × RQ
RQ=√(24²-7²)=√527
Area Δ = 1/2 × 7 × √527
Shaded area = 72π - 1/2 × 7 × √527
Q2. Find area of shaded region where AQ=7cm, ∠PAQ=60°, P and Q are on circle with centre O, radius 7cm.
Solution
ΔPAQ is equilateral (PA=AQ=7cm, ∠A=60°)
Area of sector PAQ = 60/360 × π × 7²
= 1/6 × 49π = 49π/6
Area of ΔPAQ = √3/4 × 7² = 49√3/4
Shaded area = 49π/6 - 49√3/4
Q3. Find area of shaded region where AB=28cm, ∠AOB=90° (quadrant).
Solution
r = 28/2 = 14cm
Area of quadrant = 1/4 × π × 14²
= 1/4 × 196π = 49π cm²
Q4. Find area of shaded region in square of side 14cm (quadrants at corners with radius 7cm, circle in middle).
Solution
Area of square = 14² = 196 cm²
Area of 4 quadrants = 4 × 1/4 × π × 7² = 49π
Area of inner circle = π × 7² = 49π (if circle inscribed)
Shaded = 196 - 49π (area between square and 4 quadrants)
Q5. Find area of shaded region: quadrant OAQB with ∠PAQ=60°, radius 7cm.
Solution
Area of quadrant = 1/4 × π × 7² = 49π/4
Area of ΔPAQ = 1/2 × 7 × 7 × sin 60°
= 1/2 × 49 × √3/2 = 49√3/4
Shaded area = 49π/4 - 49√3/4
= 49/4(π-√3) cm²
Q7. Find area of shaded region: circle radius 6cm, ∠AQB=60°.
Solution
Area of minor segment = Area of sector - Area of Δ
Area of sector = 60/360 × π × 6²
= 1/6 × 36π = 6π
Area of Δ = √3/4 × 6² = 9√3
Area of segment = 6π - 9√3
Shaded area = π × 6² - (6π-9√3) = 36π-6π+9√3
= 30π+9√3 cm²
Q8. Find area of shaded region: ∠AOD=60° (O is centre, radius 10cm).
Solution
Area of sector = 60/360 × π × 10²
= 1/6 × 100π = 50π/3
Area of ΔAOD = 1/2 × 10 × 10 × sin 60²
= 50 × √3/2 = 25√3
Area of segment = 50π/3 - 25√3

Chapter 13: Surface Areas and Volumes

Key Formulas:
• Sphere: SA=4πr², Volume=4/3 πr³
• Hemisphere: CSA=2πr², TSA=3πr², Volume=2/3 πr³
• Cylinder: CSA=2πrh, TSA=2πr(r+h), Volume=πr²h
• Cone: CSA=πrl, TSA=πr(l+r), Volume=1/3 πr²h
• Frustum: Volume=1/3 πh(r₁²+r₂²+r₁r₂)
• CSA of frustum=π(r₁+r₂)l where l=√(h²+(r₁-r₂)²)

Exercise 13.1

Q1. 2 cubes each of volume 64cm² are joined end to end. Find surface area of resulting cuboid.
Solution
Side of each cube = √√64 = 4cm
Cuboid: l=8cm, b=4cm, h=4cm
TSA = 2(8×4+4×4+8×4)
= 2(32+16+32)
= 2(80) = 160 cm²
Q2. A vessel is in form of hollow hemisphere mounted by a hollow cylinder. Diameter 14cm, total height 13cm. Find inner curved surface area.
Solution
Radius = 7cm
Height of cylinder = 13-7 = 6cm
CSA of cylinder = 2π×7×6 = 84π cm²
CSA of hemisphere = 2π×7² = 98π cm²
Total CSA = 84π+98π = 182π cm²
Q3. A toy is in form of cone on hemisphere. Radius 3.5cm, total height 15.5cm. Find total surface area.
Solution
Cone height = 15.5-3.5 = 12cm
Slant height l = √(12²+3.5²) = √(144+12.25) = √156.25 = 12.5cm
CSA of cone = π×3.5×12.5 = 43.75π
CSA of hemisphere = 2π×3.5² = 24.5π
Total TSA = 43.75π+24.5π = 68.25π cm²
Q5. A cubical block of side 7cm is surmounted by hemisphere. Find surface area of solid.
Solution
TSA of cube = 6 × 7² = 294 cm²
CSA of hemisphere = 2π × (7/2)² = 2π × 49/4 = 24.5π
Base area of hemisphere (circle on top) = π × 49/4 = 12.25π
Total = 294 - 12.25π + 24.5π
= 294 + 12.25π
= 294 + 38.48 ≈ 332.48 cm²
Q6. A cubical block of side 7cm is surmounted by hemisphere. Find volume of solid.
Solution
Volume of cube = 7³ = 343 cm³
Volume of hemisphere = 2/3 π × (7/2)³
= 2/3 × π × 343/8
= 343π/12
= 90.01...
Total = 343 + 343π/12 ≈ 433.02 cm³

Chapter 14: Statistics

Key Formulas:
• Mean (direct): Σf₁x₁/Σf₁
• Mean (assumed mean): a + (Σf₁d₁/Σf₁) where d₁=x₁-a
• Mean (step deviation): a + (Σf₁u₁/Σf₁) × h where u₁=(x₁-a)/h
• Median = l + ((n/2-cf)/f) × h
• Mode = l + ((f₁-f₂)/(2f₁-f₂-f₃)) × h

Exercise 14.1

Q1. Find mean of: 2, 3, 5, 7, 8
Solution
Mean = (2+3+5+7+8)/5 = 25/5 = 5
Q2. Find mean of first five multiples of 3.
Solution
Multiples: 3, 6, 9, 12, 15
Mean = (3+6+9+12+15)/5 = 45/5 = 9

Exercise 14.2

Q1. Find mode and mean of: 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 33, 17, 16, 21, 21
Solution
Mean = Sum/15
= 14×3+17×2+18×2+21×2+25+28+23+22+33+16
=(42+34+36+42+25+28+23+22+33+16)/15
= 301/15 ≈ 20.07
Mode: 14 occurs most frequently (3 times)
Mode = 14

Exercise 14.3

Q1. Find median of: 2, 3, 4, 4, 5, 5, 5, 6, 6, 7, 8, 9, 10
Solution
n=13 (odd)
Median = ((13+1)/2)th term = 7th term = 5
Q3. Find median from grouped data using formula: l+((n/2-cf)/f)×h
Solution
For grouped data median:
Median class = class where n/2 lies
Median = l + ((n/2 - cf)/f) × h
where l = lower limit of median class
cf = cumulative frequency of preceding class
f = frequency of median class
h = class size

Exercise 14.4

Q1. Draw ogive for given distribution.
Solution
Steps to draw less than ogive:
1. Mark upper class limits on x-axis
2. Mark cumulative frequencies on y-axis
3. Plot points (upper limit, cf)
4. Join points with smooth curve

Steps for more than ogive:
1. Mark lower class limits on x-axis
2. Mark cumulative frequencies (from top) on y-axis
3. Plot points (lower limit, cf)
4. Join with smooth curve

Median = x-value corresponding to n/2 on ogive

Chapter 15: Probability

Key Formulas:
• P(E) = Number of favorable outcomes / Total number of outcomes
• 0 ≤ P(E) ≤ 1
• P(E) + P(not E) = 1
• P(impossible event) = 0
• P(certain event) = 1

Exercise 15.1

Q1. Complete the statements:
(i) Probability of event that cannot happen is ___.
(ii) Probability of certain event is ___.
(iii) Sum of probabilities of all outcomes is ___.
(iv) Probability of impossible event is ___.
(v) Probability of an event that is sure to happen is ___.
Solution
(i) 0 (impossible event)
(ii) 1 (certain event)
(iii) 1
(iv) 0
(v) 1
Q2. Which of the following experiments have equally likely outcomes?
(i) A driver attempts to start a car. The car starts or does not start.
(ii) A player tries to shoot a target. He shoots or misses.
Solution
(i) Not equally likely - car starting is more probable
(ii) Not equally likely - shooting success depends on skill
Q3. Why is tossing a coin considered a fair way of deciding which team gets the ball?
Solution
Tossing a coin has two equally likely outcomes (H or T), each with probability 1/2. Neither team has advantage, so it is fair.
Q4. Which value of x makes P(E) = 1 impossible, if P(E) = x/(x+2)?
Solution
P(E) = 0 when x = 0
x/(x+2) = 0 → x = 0
∴ x = 0 makes it impossible
Q5. A bag contains 3 red and 5 black balls. Find probability of:
(i) red ball (ii) not red ball
Solution
Total = 3+5 = 8
(i) P(red) = 3/8
(ii) P(not red) = 5/8
Q6. A box has 5 red, 8 white, 4 green marbles. Find probability of:
(i) green (ii) not red (iii) neither white nor green
Solution
Total = 5+8+4 = 17
(i) P(green) = 4/17
(ii) P(not red) = (8+4)/17 = 12/17
(iii) P(neither white nor green) = P(red) = 5/17
Q13. A die is thrown twice. Find probability that (i) 5 will not come up either time (ii) 5 will come up at least once.
Solution
Total outcomes = 6×6 = 36
(i) P(5 not coming up) = 5/6 × 5/6 = 25/36
(ii) P(5 at least once) = 1 - 25/36 = 11/36
Q14. Two coins are tossed. Find probability of at least one head.
Solution
Outcomes: HH, HT, TH, TT
Favorable (at least one head): HH, HT, TH
P(at least one head) = 3/4
Q15. A die is thrown. Find probability of:
(i) getting a prime number
(ii) getting a number ≤ 2
(iii) getting a number > 2
(iv) getting a number not divisible by 3
Solution
(i) Prime numbers: 2, 3, 5 → P = 3/6 = 1/2
(ii) Numbers ≤ 2: 1, 2 → P = 2/6 = 1/3
(iii) Numbers > 2: 3, 4, 5, 6 → P = 4/6 = 2/3
(iv) Not divisible by 3: 1, 2, 4, 5 → P = 4/6 = 2/3
Q16. A box has 90 discs numbered 1 to 90. Find probability of:
(i) two-digit number (ii) perfect square (iii) divisible by 5
Solution
(i) Two-digit: 10-90 = 81 numbers
P = 81/90 = 9/10
(ii) Perfect squares: 1,4,9,16,25,36,49,64,81 = 9 numbers
P = 9/90 = 1/10
(iii) Divisible by 5: 5,10,...,90 = 18 numbers
P = 18/90 = 1/5
Q18. A bag has 5 red and 4 blue balls. Draw one ball at random. Find P(red).
Solution
Total = 5+4 = 9
P(red) = 5/9
Q19. A bag has 4 red, 5 black, 3 green balls. Find probability of:
(i) red (ii) not green (iii) neither red nor green
Solution
Total = 4+5+3 = 12
(i) P(red) = 4/12 = 1/3
(ii) P(not green) = (4+5)/12 = 9/12 = 3/4
(iii) P(neither red nor green) = P(black) = 5/12
Q22. A game has 10 cards numbered 1-10. Find probability of:
(i) even number (ii) perfect square
Solution
(i) Even numbers: 2,4,6,8,10 → P = 5/10 = 1/2
(ii) Perfect squares: 1,4,9 → P = 3/10