MATHEMATICS (BASIC): MINIMUM LEVEL LEARNING MATERIAL
Class X (2025-26)
Prepared by: GYANPOINTS
CHAPTER 6: TRIANGLES
Important Theorems
Basic Proportionality Theorem (Thales Theorem):
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
[Diagram: Triangle ABC with line l parallel to BC intersecting AB at D and AC at E]
\[ \frac{AD}{DB} = \frac{AE}{EC} \]
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Important Questions
- 1. If a line intersects sides AB and AC of a \(\Delta ABC\) at D and E respectively and is parallel to BC, prove that \(\frac{AD}{AB} = \frac{AE}{AC}\).
- 2. ABCD is a trapezium with \(AB \parallel DC\). E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that \(\frac{AE}{ED} = \frac{BF}{FC}\).
- 3. In \(\Delta PQR\), \(\frac{PS}{SQ} = \frac{PT}{TR}\) and \(\angle PST = \angle PRQ\). Prove that PQR is an isosceles triangle.
- 4. In the figure (not shown), if \(LM \parallel CB\) and \(LN \parallel CD\), prove that \(\frac{AM}{AB} = \frac{AN}{AD}\).
- 5. In the figure, \(DE \parallel AC\) and \(DF \parallel AE\). Prove that \(\frac{BF}{FE} = \frac{BE}{EC}\).
- 6. In the figure, \(DE \parallel OQ\) and \(DF \parallel OR\). Show that \(EF \parallel QR\).
- 7. A, B and C are points on OP, OQ and OR respectively such that \(AB \parallel PQ\) and \(AC \parallel PR\). Show that \(BC \parallel QR\).
- 8. ABCD is a trapezium in which \(AB \parallel DC\) and its diagonals intersect each other at the point O. Show that \(\frac{AO}{BO} = \frac{CO}{DO}\).
- 9. The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{AO}{BO} = \frac{CO}{DO}\). Show that ABCD is a trapezium.
Important 1 Mark Questions (MCQs & Short Answer)
1. In \(\Delta ABC\), D and E are points on sides AB and AC respectively such that \(DE \parallel BC\) and \(AD : DB = 3 : 1\). If \(EA = 6.6\) cm then find AC.
3. The perimeter of two similar triangles ABC and LMN are 60 cm and 48 cm respectively. If \(LM = 8\) cm, then what is the length of AB?
7. A vertical stick 12 m long casts a shadow 8 m long on the ground. At the same time a tower casts the shadow 40 m long on the ground. Determine the height of the tower.
11. The areas of two similar triangles are in the ratio 4 : 9. The corresponding sides of these triangles are in the ratio:
CHAPTER 10: CIRCLES
Theorems
- The tangent to a circle is perpendicular to the radius through the point of contact.
- The lengths of tangents drawn from an external point to a circle are equal.
[Diagram: Circle with centre O, external point P, tangents PQ and PR]
To Prove: \(PQ = PR\)
Proof: In \(\Delta OQP\) and \(\Delta ORP\):
\(OQ = OR\) (radii)
\(OP = OP\) (common)
\(\angle Q = \angle R = 90^\circ\) (Tangent \(\perp\) Radius)
Hence \(\Delta OQP \cong \Delta ORP\) (RHS Criterion)
\(\therefore PQ = PR\) (CPCT).
Important Questions
- 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. Find the radius of the circle.
- 2. If TP and TQ are the two tangents to a circle with centre O so that \(\angle POQ = 110^\circ\), then find \(\angle PTQ\).
- 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of \(80^\circ\), then find \(\angle POA\).
- 4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
- 8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that \(AB + CD = AD + BC\).
- 10. Prove that the parallelogram circumscribing a circle is a rhombus.
- 15. Two tangents TP and TQ are drawn to a circle with centre O from an external point T. Prove that \(\angle PTQ = 2 \angle OPQ\).
MCQs (1 Mark)
1. Find the length of tangent drawn to a circle with radius 7 cm from a point 25 cm away from the centre.
11. TP and TQ are the two tangents to a circle with center O so that angle \(\angle POQ = 130^\circ\). Find \(\angle PTQ\).
CHAPTER 14: STATISTICS
Solved Examples
Question: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate: 45–55, 55–65, 65–75, 75–85, 85–95
Number of cities: 3, 10, 11, 8, 3
Literacy rate: 45–55, 55–65, 65–75, 75–85, 85–95
Number of cities: 3, 10, 11, 8, 3
Solution (Step Deviation Method):
Assume Mean \(A = 70\), \(h = 10\).
Calculating \(u_i = \frac{x_i – A}{h}\):
\(x_i\): 50, 60, 70, 80, 90
\(f_i\): 3, 10, 11, 8, 3
\(u_i\): -2, -1, 0, 1, 2
\(f_i u_i\): -6, -10, 0, 8, 6. Total \(\sum f_i u_i = -2\). Total \(N = 35\).
Mean \(\bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 70 + \left( \frac{-2}{35} \right) \times 10\)
\(= 70 – 0.57 = 69.43\) %
Assume Mean \(A = 70\), \(h = 10\).
Calculating \(u_i = \frac{x_i – A}{h}\):
\(x_i\): 50, 60, 70, 80, 90
\(f_i\): 3, 10, 11, 8, 3
\(u_i\): -2, -1, 0, 1, 2
\(f_i u_i\): -6, -10, 0, 8, 6. Total \(\sum f_i u_i = -2\). Total \(N = 35\).
Mean \(\bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h = 70 + \left( \frac{-2}{35} \right) \times 10\)
\(= 70 – 0.57 = 69.43\) %
Questions for Practice
- 1. Find the mean of the data: Class 10–25 (Freq 2), 25–40 (3), 40–55 (7), 55–70 (6), 70–85 (6), 85–100 (6).
- Mode Q1: Find the mode of agriculture holdings: 1–3 (20), 3–5 (45), 5–7 (80), 7–9 (55), 9–11 (40), 11–13 (12).
- Median Q1: Marks 30–35 (14), 35–40 (16), 40–45 (18), 45–50 (23), 50–55 (18), 55–60 (8), 60–65 (3). Determine median.
MCQs
1. For a frequency distribution, mean, median and mode are connected by the relation:
16. Construction of cumulative frequency table is useful in determining the:
CHAPTER 15: PROBABILITY
Solved Examples
Question: Two dice are thrown together. Find the probability that the sum of the numbers on the top of the dice is (i) 9 (ii) 10.
Total outcomes \(n(S) = 36\).
(i) Event A (Sum 9): \((3,6), (4,5), (5,4), (6,3)\). \(n(A) = 4\).
\(P(A) = \frac{4}{36} = \frac{1}{9}\).
(ii) Event B (Sum 10): \((4,6), (5,5), (6,4)\). \(n(B) = 3\).
\(P(B) = \frac{3}{36} = \frac{1}{12}\).
(i) Event A (Sum 9): \((3,6), (4,5), (5,4), (6,3)\). \(n(A) = 4\).
\(P(A) = \frac{4}{36} = \frac{1}{9}\).
(ii) Event B (Sum 10): \((4,6), (5,5), (6,4)\). \(n(B) = 3\).
\(P(B) = \frac{3}{36} = \frac{1}{12}\).
Practice Questions
- 1. Two dice are thrown together. Find the probability that the product of the numbers is (i) 6 (ii) 12 (iii) 7.
- 8. A box contains 5 red, 8 white and 4 green marbles. One is taken out. Probability it is (i) red? (ii) white? (iii) not green?
- 13. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
MCQs
3. A die is thrown once. What will be the probability of getting a prime number?
5. Cards marked 1 to 25. Probability of getting a number less than 11?
CHAPTER 1: REAL NUMBERS
Solved Examples
Question: Find the LCM and HCF of 510 and 92 and verify that \(LCM \times HCF = \text{product of the two numbers}\).
\(510 = 2 \times 3 \times 5 \times 17\)
\(92 = 2^2 \times 23\)
\(HCF = 2\).
\(LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460\).
Verification: \(510 \times 92 = 46920\). \(HCF \times LCM = 2 \times 23460 = 46920\).
\(92 = 2^2 \times 23\)
\(HCF = 2\).
\(LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460\).
Verification: \(510 \times 92 = 46920\). \(HCF \times LCM = 2 \times 23460 = 46920\).
Question: Prove that \(\sqrt{5}\) is an irrational number.
Let \(\sqrt{5} = \frac{p}{q}\) (p, q are co-primes).
\(5 = \frac{p^2}{q^2} \Rightarrow p^2 = 5q^2\). \(p^2\) is divisible by 5, so \(p\) is divisible by 5.
Let \(p = 5m\). \( (5m)^2 = 5q^2 \Rightarrow 25m^2 = 5q^2 \Rightarrow q^2 = 5m^2\).
\(q\) is divisible by 5. Both p and q have common factor 5, contradicting co-prime assumption.
Hence \(\sqrt{5}\) is irrational.
\(5 = \frac{p^2}{q^2} \Rightarrow p^2 = 5q^2\). \(p^2\) is divisible by 5, so \(p\) is divisible by 5.
Let \(p = 5m\). \( (5m)^2 = 5q^2 \Rightarrow 25m^2 = 5q^2 \Rightarrow q^2 = 5m^2\).
\(q\) is divisible by 5. Both p and q have common factor 5, contradicting co-prime assumption.
Hence \(\sqrt{5}\) is irrational.
MCQs
1. If HCF and LCM of two numbers are 4 and 9696, then the product of the two numbers is:
CHAPTER 7: COORDINATE GEOMETRY
Distance Formula: \(AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)
Section Formula: \(P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)\)
Section Formula: \(P(x, y) = \left( \frac{m_1 x_2 + m_2 x_1}{m_1 + m_2}, \frac{m_1 y_2 + m_2 y_1}{m_1 + m_2} \right)\)
Solved Examples
Question: Show that points (1, 7), (4, 2), (-1, -1) and (-4, 4) are vertices of a square.
Calculate distances AB, BC, CD, DA and diagonals AC, BD.
\(AB = \sqrt{3^2 + (-5)^2} = \sqrt{34}\).
All sides \(\sqrt{34}\). Diagonals \(AC = \sqrt{68}, BD = \sqrt{68}\).
Since sides are equal and diagonals are equal, it is a square.
\(AB = \sqrt{3^2 + (-5)^2} = \sqrt{34}\).
All sides \(\sqrt{34}\). Diagonals \(AC = \sqrt{68}, BD = \sqrt{68}\).
Since sides are equal and diagonals are equal, it is a square.
Questions for Practice
- 6. Find the point on x-axis which is equidistant from (7, 6) and (–3, 4).
- 10. Find the values of y for which the distance between P(2, – 3) and Q(10, y) is 10 units.
- 17. Find the ratio in which the line \(2x + 3y – 5 = 0\) divides the line segment joining the points (8, –9) and (2, 1).
CHAPTER 2: POLYNOMIALS
Solved Examples
Question: Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.
\(p(x) = x^2 – (\alpha + \beta)x + \alpha\beta\)
\(p(x) = x^2 – (-3)x + 2 = x^2 + 3x + 2\).
\(p(x) = x^2 – (-3)x + 2 = x^2 + 3x + 2\).
Question: Find the zeroes of \(x^2 – 2x – 8\) and verify.
\(x^2 – 4x + 2x – 8 = 0 \Rightarrow (x-4)(x+2) = 0\). Zeroes: 4, -2.
Sum: \(4 + (-2) = 2\). \( -b/a = -(-2)/1 = 2 \). Verified.
Product: \(4(-2) = -8\). \( c/a = -8/1 = -8 \). Verified.
Sum: \(4 + (-2) = 2\). \( -b/a = -(-2)/1 = 2 \). Verified.
Product: \(4(-2) = -8\). \( c/a = -8/1 = -8 \). Verified.
CHAPTER 4: QUADRATIC EQUATIONS
Solved Examples
Question: Solve by factorization: \(x^2 + 2x – 8 = 0\).
\(x^2 + 4x – 2x – 8 = 0 \Rightarrow (x+4)(x-2) = 0 \Rightarrow x = -4, 2\).
Question: Find the discriminant of \(2x^2 – 4x + 3 = 0\).
\(D = b^2 – 4ac = (-4)^2 – 4(2)(3) = 16 – 24 = -8\).
\(D < 0\), so no real roots.
\(D < 0\), so no real roots.
CHAPTER 3: PAIR OF LINEAR EQUATIONS
Solved Examples
Question: Solve graphically: \(x + 3y = 6\) and \(2x – 3y = 12\).
Plot points for \(x+3y=6\): (0, 2), (3, 1), (6, 0).
Plot points for \(2x-3y=12\): (0, -4), (3, -2), (6, 0).
Intersection point is B(6, 0). Solution: \(x = 6, y = 0\).
Plot points for \(2x-3y=12\): (0, -4), (3, -2), (6, 0).
Intersection point is B(6, 0). Solution: \(x = 6, y = 0\).
Important Questions
- 6. Find k for no solution: \(3x – y – 5 = 0, 6x – 2y + k = 0\).
- 13. For what value of k does \(kx – 3y + 6 = 0, 4x – 6y + 15 = 0\) represent parallel lines?
CHAPTER 5: ARITHMETIC PROGRESSION
Solved Examples
Question: Which term of the AP: 3, 9, 15… is 99?
\(a = 3, d = 6\). \(a_n = 3 + (n-1)6 = 99\).
\(6(n-1) = 96 \Rightarrow n-1 = 16 \Rightarrow n = 17\).
\(6(n-1) = 96 \Rightarrow n-1 = 16 \Rightarrow n = 17\).
Question: Find the sum of the first 22 terms of the AP: 8, 3, -2…
\(a = 8, d = -5, n = 22\).
\(S_{22} = \frac{22}{2}[2(8) + (21)(-5)] = 11 = 11(-89) = -979\).
\(S_{22} = \frac{22}{2}[2(8) + (21)(-5)] = 11 = 11(-89) = -979\).
CHAPTER 8 & 9: TRIGONOMETRY
Solved Examples
Question: If \(\tan A = 4/3\), find all T-ratios.
\(BC = 4k, AB = 3k\). By Pythagoras, \(AC = 5k\).
\(\sin A = 4/5, \cos A = 3/5, \cot A = 3/4\), etc.
\(\sin A = 4/5, \cos A = 3/5, \cot A = 3/4\), etc.
Question (Identities): Prove \(\frac{\cos A}{1 – \tan A} + \frac{\sin A}{1 – \cot A} = \sin A + \cos A\).
[Proof steps provided in text involving converting tan/cot to sin/cos and simplifying].
Question (Heights): Angles of depression of top and bottom of 8m tall building from a multi-storey building are \(30^\circ\) and \(45^\circ\). Find height of multi-storey building.
Height \( = 4(3 + \sqrt{3})\) m.
CHAPTER 12: AREAS RELATED TO CIRCLES
Solved Examples
Question: Find the area of sector with radius 4 cm and angle \(30^\circ\).
Area \(= \frac{30}{360} \times 3.14 \times 4 \times 4 = 4.19 \text{ cm}^2\).
Question: In a square ABCD of side 56m, two circular flower beds are on two sides… Find total area.
Total area = Area of sectors + Area of triangles.
Calculation yields \(4032 \text{ m}^2\).
Calculation yields \(4032 \text{ m}^2\).
CHAPTER 13: SURFACE AREAS AND VOLUMES
Solved Examples
Question: Decorative block made of cube (edge 5cm) and hemisphere (diameter 4.2cm). Find TSA.
\(TSA = \text{TSA Cube} – \text{Base Area Hemisphere} + \text{CSA Hemisphere}\)
\(= 150 – \pi r^2 + 2\pi r^2 = 150 + \pi (2.1)^2 = 163.86 \text{ cm}^2\).
\(= 150 – \pi r^2 + 2\pi r^2 = 150 + \pi (2.1)^2 = 163.86 \text{ cm}^2\).
Question: Bird-bath cylinder (ht 1.45m, r 30cm) with hemispherical depression. Find TSA.
\(TSA = CSA_{cyl} + CSA_{hem} = 2\pi rh + 2\pi r^2 = 3.3 \text{ m}^2\).