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Quadratic Equations

🔹 Definition

A quadratic equation in one variable is an equation of the form:

$$ax^2 + bx + c = 0$$

where

• x is a variable

• $a, b, c$ are real numbers, and

• $a \ne 0$.

🔹 Standard Form

$$ax^2 + bx + c = 0$$

Here:

• Coefficient of $x^2 = a$
  

• Coefficient of $x = b$
  

• Constant term $= c$
  

🔹 Methods of Solving a Quadratic Equation

🧭 Method 1: Factorisation (Splitting the Middle Term)

✅ Stepwise Procedure

Step 1: Write the equation in standard form:

$$ax^2 + bx + c = 0$$

Step 2: Identify coefficients.

• x=variable

• Coefficient of $x^2 = a$
  

• Coefficient of $x = b$
  

• Constant term $= c$
  

Step 3: Multiply

$$(\text{Coefficient of } x^2) \times (\text{Constant term}) = a \times c$$

Step 4: Find two numbers $m$ and $n$ such that

$$m + n = b, \quad m \times n = a \times c$$

Step 5: Rewrite the middle term $bx$ as $mx + nx$.

Step 6: Group and factorise the terms.

Step 7: Equate each factor to zero and find the value of $x$.

🧩 Rule Box: Splitting the Middle Term

Find two numbers $m$ and $n$ such that

$$m + n = \text{Coefficient of } x$$

$$$$and also
$$m \times n = (\text{Coefficient of } x^2) \times (\text{Constant term})$$Then rewrite $bx$ as $mx + nx$,

Now group and factorise.

✳️ Example 1 (Easy)

Solve $x^2 + 5x + 6 = 0$

Step 1: Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 1, b = 5, c = 6$

$$Step 2: Multiply $a \times c = 1 \times 6 = 6$

Step 3: Find $m, n$ such that:
$m + n = 5, \quad m \times n = 6$
So, $m = 2, n = 3$

Step 4: Rewrite middle term:

$$x^2 + 2x + 3x + 6 = 0$$

Step 5: Group and factorise:

$$x(x + 2) + 3(x + 2) = 0$$
$$(x + 2)(x + 3) = 0$$Step 6:

$$x + 2 = 0 \Rightarrow x = -2 \quad \text{and} \quad x + 3 = 0 \Rightarrow x = -3$$

✅ Roots: $x = -2, -3$

✳️ Example 2 (Moderate)

Solve $2x^2 + 5x + 2 = 0$

Step 1:

Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 2, b = 5, c = 2$

Step 2: $a \times c = 4$

Step 3: Find $m, n$ such that:
$m+n=5,m\times n=4$

$m + n = 5, m \times n = 4 \Rightarrow m = 4, n = 1$

Step 4: Rewrite middle term:

$$2x^2 + 4x + x + 2 = 0$$

Step 5: Group and factorise:

$$2x(x + 2) + 1(x + 2) = 0$$
$$(x + 2)(2x + 1) = 0$$Step 6:

$$x = -2 \quad \text{or} \quad x = -\frac{1}{2}$$

✅ Roots: $x = -2, -\frac{1}{2}$

✳️ Example 3 (Hard)

Solve $3x^2 – 8x – 3 = 0$

Step 1:

Compare the Coefficients of the given qudratic equation with the standard quadratic equation

$$a{x}^2+bx+c=0$$$\mathrm{On\; comparision\; we\; get:}$

$a = 3, b = -8, c = -3$

Step 2: $a \times c = -9$

Step 3: Find $m, n$ such that:
$m + n = -8, m \times n = -9$
$m = -9, n = 1$

Step 4: Rewrite middle term:

$$3x^2 – 9x + x – 3 = 0$$

Step 5: Group and factorise:

$$3x(x – 3) + 1(x – 3) = 0$$

$$$$$$(x – 3)(3x + 1) = 0$$

Step 6:

$$x = 3, \quad x = -\frac{1}{3}$$

✅ Roots: $x = 3, -\frac{1}{3}$

Method 2: Completing the Square

🧭 Concept

A quadratic equation

$$ax^2 + bx + c = 0$$

can be solved by expressing it as a perfect square of a binomial.

✅ Step-by-Step Procedure

Step 1: Bring the equation to the form

$$ax^2 + bx = -c$$

Step 2: Divide each term by $a$ to make the coefficient of $x^2$ equal to 1.

$$x^2+\frac{b}{a}x=-\frac{c}{a}$$Step 3: Add $\left(\frac{b}{2a}\right)^2$to both sides to make a perfect square.

Step 4: Write the LHS as a square:

$${(x+\frac{b}{2a})}^2=\frac{b^2-4ac}{4a^2}$$Step 5: Take square root on both sides.

Step 6: Solve for $x$.

🧩 Rule Box: Completing the Square

To complete the square for $x^2 + px$,
add and subtract $\left(\frac{p}{2}\right)^2$.

$$x^2 + px + \left(\frac{p}{2}\right)^2 = \left(x + \frac{p}{2}\right)^2$$

✳️ Example 1 (Easy)

Solve $x^2 + 6x + 5 = 0$

Step 1: Move constant to RHS:

$$x^2 + 6x = -5$$

Step 2: Add $\left(\frac{6}{2}\right)^2 = 9$ to both sides:

$$x^2+6x+9=-5+9$$

Step 3: Write as perfect square:

$$(x+3{)}^2=4$$

Step 4: Take square roots:

$$x + 3 = \pm 2$$

Step 5:

$$x = -3 \pm 2$$

✅ Roots: $x = -1, -5$

✳️ Example 2 (Moderate)

Solve $2x^2 + 3x – 2 = 0$

Step 1: Divide by 2:$$x^2 + \frac{3}{2}x – 1 = 0$$

Step 2: Move constant:

$$x^2+\frac{3}{2}x=1$$

Step 3: Add $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$:

$$x^2 + \frac{3}{2}x + \frac{9}{16} = 1 + \frac{9}{16}$$

Step 4: Write as square:

$$\left(x + \frac{3}{4}\right)^2 = \frac{25}{16}$$

Step 5: Take square roots:

$$x + \frac{3}{4} = \pm \frac{5}{4}$$

Step 6:

$$x = \frac{-3 \pm 5}{4}$$

✅ Roots: $x = \tfrac{1}{2}, -2$.

✳️ Example 3 (Hard)

Solve $3x^2 – 10x + 7 = 0$

Step 1: Divide by 3:

$$x^2 – \frac{10}{3}x + \frac{7}{3} = 0$$

Step 2: Move constant:

$$x^2 – \frac{10}{3}x = -\frac{7}{3}$$

Step 3: Add $\left(\frac{-10}{6}\right)^2 = \frac{25}{9}$:

$$x^2 – \frac{10}{3}x + \frac{25}{9} = -\frac{7}{3} + \frac{25}{9}$$

Step 4: Simplify RHS:

$$-\frac{7}{3} + \frac{25}{9} = \frac{4}{9}$$$$\left(x – \frac{5}{3}\right)^2 = \frac{4}{9}$$

Step 5:

$$x – \frac{5}{3} = \pm \frac{2}{3}$$

Step 6:

$$x = \frac{5 \pm 2}{3}$$

✅ Roots: $x = \frac{7}{3}, 1$

🔹 Method 3: Quadratic Formula

🧭 Concept

For any quadratic equation

$$ax^2 + bx + c = 0,$$

the solution is given by the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

Here,

$$D = b^2 – 4ac$$

s called the Discriminant.

✅ Step-by-Step Procedure

Step 1: Identify $a, b, c$.
Step 2: Compute discriminant $D = b^2 – 4ac$.
Step 3: Substitute in the formula

$$x = \frac{-b \pm \sqrt{D}}{2a}$$

Step 4: Simplify to get the roots.

🧩 Rule Box: Quadratic Formula

$$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$Let $D = b^2 – 4ac$:

✳️ Example 1 (Easy)

Solve $x^2 – 5x + 6 = 0$

Step 1: $a = 1, b = -5, c = 6$

Step 2: $D = (-5)^2 – 4(1)(6) = 25 – 24 = 1$

Step 3:

$$x = \frac{-(-5) \pm \sqrt{1}}{2(1)} = \frac{5 \pm 1}{2}$$

Step 4:

$$x = 3 \text{ or } 2$$

✅ Roots: $x = 2, 3$.

✳️ Example 2 (Moderate)

Solve $2x^2 + 3x – 2 = 0$

Step 1: $a = 2, b = 3, c = -2$

Step 2: $D = 3^2 – 4(2)(-2) = 9 + 16 = 25$

Step 3:

$$x = \frac{-3 \pm \sqrt{25}}{4}$$

Step 4:

$$x = \frac{-3 \pm 5}{4}$$

Step 5:

$$x = \frac{1}{2}, -2$$

Roots: $x = \tfrac{1}{2}, -2$

✳️ Example 3 (Hard)

Solve $3x^2 – 4x + 5 = 0$

Step 1: $a = 3, b = -4, c = 5$

Step 2: $D = (-4)^2 – 4(3)(5) = 16 – 60 = -44$

Step 3:
Since $D < 0$, roots are imaginary.

$$x = \frac{-(-4) \pm \sqrt{-44}}{2(3)} = \frac{4 \pm i\sqrt{44}}{6}$$

Simplify:

$$x = \frac{2 \pm i\sqrt{11}}{3}$$

✅ Nature of Roots: No real roots (imaginary).

6️⃣ Nature of Roots

Let $D = b^2 – 4ac$

🔹 7️⃣ Summary of Formulas

1. Standard Form: $ax^2 + bx + c = 0$
   

2. Product of roots: $\alpha \beta = \frac{c}{a}$

3. Sum of roots: $\alpha + \beta = -\frac{b}{a}$

4. Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

5. Discriminant: $D = b^2 – 4ac$
   

🔹 8️⃣ Practice Questions (Worksheet)

Each question has 3 variants (a, b, c).

Q1. Solve by factorisation:

(a) $x^2 + 7x + 10 = 0$
(b) $x^2 + 9x + 14 = 0$
(c) $x^2 + 11x + 24 = 0$

Q2. Solve by splitting the middle term:

(a) $2x^2 + 7x + 3 = 0$
(b) $3x^2 + 8x + 4 = 0$
(c) $4x^2 + 9x + 2 = 0$

Q3. Find the discriminant and nature of roots:

(a) $x^2 – 4x + 4 = 0$
(b) $x^2 – 6x + 9 = 0$
(c) $x^2 – 10x + 16 = 0$

Q4. Find the sum and product of the roots:

(a) $2x^2 – 5x + 3 = 0$
(b) $3x^2 + 7x + 2 = 0$
(c) $5x^2 – 9x – 2 = 0$

1. Introduction to Geometry

• Geometry is the branch of mathematics that deals with shapes, sizes, relative positions of figures, and properties of space.

• The word Geometry comes from Greek words: Geo = Earth and Metron = Measurement.

2. Fundamental Terms

2.1 Point

• A point is a precise location in space.

• It has no length, breadth, or thickness.

• Notation: Usually denoted by a capital letter, e.g., $A, B, C$.

2.2 Line

• A line is a straight path of points extending infinitely in both directions.

• Notation: Line passing through points $A$ and $B$ is written as $AB$ or $\overleftrightarrow{AB}$.

2.3 Line Segment

• A line segment is part of a line with two endpoints.

• Notation: Line segment joining points $A$ and $B$ is $\overline{AB}$.

2.4 Ray

• A ray starts at one point and extends infinitely in one direction.

• Notation: Ray starting at $A$ passing through $B$ is $\overrightarrow{AB}$.

2.5 Plane

• A plane is a flat surface that extends infinitely in all directions.

• Represented by a quadrilateral figure in diagrams.

3. Angle And Its Types

• An angle is formed by two rays with a common endpoint called the vertex.

3.1 Classification

1. Acute Angle: $0^\circ < \theta < 90^\circ$
   

2. Right Angle: $\theta = 90^\circ$
   

3. Obtuse Angle: $90^\circ < \theta < 180^\circ$
   

4. Straight Angle: $\theta = 180^\circ$
   

5. Reflex Angle: $180^\circ < \theta < 360^\circ$
   

Formula:

$$\text{Sum of angles on a straight line} = 180^\circ$$$$\text{Sum of angles around a point} = 360^\circ$$

4. Triangle And Its Types

• A triangle has 3 sides and 3 angles.

4.1 Based on Sides

1. Equilateral: All sides equal

2. Isosceles: Two sides equal

3. Scalene: All sides unequal

4.2 Based on Angles

1. Acute-angled: All angles < $90^\circ$
   

2. Right-angled: One angle = $90^\circ$
   

3. Obtuse-angled: One angle > $90^\circ$
   

Important Property (Triangle Sum Theorem):

$$\text{Sum of angles in a triangle} = 180^\circ$$

5. Quadrilateral And Its Types

• A quadrilateral has 4 sides and 4 angles.

5.1 Types

1. Square: All sides equal, all angles $90^\circ$
   

2. Rectangle: Opposite sides equal, all angles $90^\circ$
   

3. Parallelogram: Opposite sides parallel and equal

4. Rhombus: All sides equal, opposite angles equal

5. Trapezium: One pair of opposite sides parallel

Property: Sum of angles in a quadrilateral:

$$\text{Sum of angles} = 360^\circ$$

6. Circles And Its Parts.

• A circle is a set of points equidistant from a fixed point called the centre.

• Radius (r): Distance from centre to any point on the circle

• Diameter (d): Twice the radius, $d = 2r$
  

• Circumference (C): $C = 2 \pi r$
  

• Area (A): $A = \pi r^2$
  

7. Basic Geometrical Constructions

1. Constructing a bisector of a line segment

2. Constructing an angle bisector

3. Constructing perpendiculars from a point on a line or outside a line

4. Constructing triangles using SSS, SAS, ASA, RHS criteria

8. Important Theorems

1. Pythagoras Theorem:

$$\text{In a right triangle: } AB^2 + BC^2 = AC^2$$

2. Triangle Sum Theorem:

$$\angle A + \angle B + \angle C = 180^\circ$$

3. Exterior Angle Theorem:

$$\text{Exterior angle of a triangle} = \text{Sum of the two opposite interior angles}$$

9. Tips & Tricks

• Always label points clearly in diagrams.

• Use a protractor for accurate angle measurement.

• Remember the sum of angles for triangle = 180°, quadrilateral = 360°.

• Practice constructing triangles using different combinations of sides and angles.

Diagram Placeholders:

• [Point, Line, Line Segment, Ray]

• [Triangle with labeled angles]

• [Quadrilateral types]

• [Circle with radius and diameter]

Worksheet on Basics of Geometry (Math Olympiad Preparation)

Topic: Basics of Geometry
Class: 6–9 (CBSE & Olympiad Level)
Marks: Practice Worksheet
Time: 45–60 minutes

Section A — Very Short Answer Questions (1 Mark Each)

(Concept Recall & Definitions)

1. Define a point and a line in geometry.

2. How many endpoints does a line segment have?

3. What is the measure of a straight angle?

4. Name the instrument used to draw circles.

5. Write the sum of all angles around a point.

6. How many vertices does a triangle have?

7. What is the sum of the angles of a quadrilateral?

8. Which type of angle is greater than 180° but less than 360°?

9. Write the relationship between radius and diameter.

10. Give one example of a real-life object in the shape of a circle.

Section B — Short Answer Questions (2 Marks Each)

(Understanding & Application)

11. Draw and name the following:
     (a) Line segment
     (b) Ray
     (c) Line

12. If one angle of a triangle is $90^\circ$ and another is $45^\circ$, find the third angle.

13. The sum of two angles is $130^\circ$. Find the measure of their supplementary angles.

14. In a quadrilateral, three angles are $80^\circ, 90^\circ, 75^\circ$. Find the fourth angle.

15. A circle has a radius of 7 cm. Find its circumference using $\pi = \frac{22}{7}$.
     $C=2\pi r=2\times \frac{22}{7}\times 7=?$

Section C — Application / Reasoning (3 Marks Each)

16. The sum of two adjacent angles on a straight line is always $180^\circ$. Prove this statement using a neat diagram.

17. A triangle has sides 6 cm, 8 cm, and 10 cm. Verify whether it is a right-angled triangle using the Pythagoras Theorem.
     $a^2 + b^2 = c^2$
    

18. In a circle with diameter 14 cm, find:
     (a) Radius
     (b) Circumference
     (c) Area
     Use $\pi = \frac{22}{7}$.

19. The exterior angle of a triangle is $120^\circ$ and one of the interior opposite angles is $40^\circ$.
     Find the other interior opposite angle.

20. In parallelogram ABCD, $\angle A = 70^\circ$. Find all other angles.
     (Hint: Opposite angles are equal and adjacent angles are supplementary.)

Section D — Higher Order Thinking (HOTS) Questions (4–5 Marks Each)

21. A triangle has two equal angles and the third angle is $96^\circ$.
     Find each of the equal angles and name the triangle.

22. The sum of the interior angles of an $n$-sided polygon is $1440^\circ$.
     Find the value of $n$.

23. Draw a circle of radius 4 cm, mark points:
     (a) Inside the circle
     (b) On the circle
     (c) Outside the circle

24. A quadrilateral has three angles measuring $110^\circ, 95^\circ, 70^\circ$. Find the fourth angle and classify the quadrilateral.

25. A line segment $AB = 8 \text{ cm}$ is bisected at $M$.
     Find $AM$ and $MB$, and justify your answer using geometric reasoning.

Section E — Multiple Choice Questions (MCQs)

(For Quick Revision)

26. The total number of right angles in a rectangle is:
     A. 1 B. 2 C. 3 D. 4

27. The angle formed by two perpendicular lines is:
     A. Acute B. Right C. Obtuse D. Straight

28. The sum of all angles in a triangle is: 
     A. $90^\circ$ B. $180^\circ$ C. $270^\circ$ D. $360^\circ$
    

29. A line segment joining the centre of a circle to a point on its circumference is called:
     A. Chord B. Diameter C. Radius D. Tangent

30. Which of the following statements is false?
     A. Every square is a rectangle.
     B. Every rectangle is a square.
     C. Every square is a rhombus.
     D. Every rhombus is a parallelogram.

Worksheet Hints, Solutions & Answers

Section A – Very Short Answer (1 Mark Each)

Q1. Define a point and a line.

Hint: Recall definitions from basic geometry.
Solution:

• A point has position but no size.

• A line extends endlessly in both directions, made of infinite points.
  Answer: Point – no dimensions; Line – extends infinitely both ways.

Q2. How many endpoints does a line segment have?

Hint: Think of a line with fixed ends.
Solution: A line segment has two endpoints.
Answer: 2 endpoints.

Q3. Measure of a straight angle?

Hint: It lies on a straight line.
Solution: $180^\circ$.
Answer: $180^\circ$

Q4. Instrument to draw circles?

Hint: Used with pencil and pin.
Answer: Compass.

Q5. Sum of all angles around a point?

Solution:

$$\text{Sum of all angles around a point} = 360^\circ$$

Answer: $360^\circ$

Q6. Vertices of a triangle?

Answer: 3 vertices.

Q7. Sum of angles of a quadrilateral?

Solution:

$$\text{Sum} = 360^\circ$$

Answer: $360^\circ$

Q8. Type of angle greater than 180° but less than 360°?

Answer: Reflex angle.

Q9. Relationship between radius (r) and diameter (d)?

$$d = 2r$$Answer: Diameter = 2 × Radius.

Q10. Example of circle shape?

Answer: Clock, coin, wheel.

Section B – Short Answer (2 Marks Each)

Q11. Draw and name: line segment, ray, line.

Hint: Use ruler and pencil.
Solution:

• $\overline{AB}$ → line segment

• $\stackrel{}{}$$\overrightarrow{AB}$ → ray

• $\overleftrightarrow{AB}$ → line
  Answer: As shown by symbols above.

Q12. Triangle with angles 90° and 45° → find third.

Solution:

$$\text{Sum}={180}^{\circ }\Rightarrow 90+45+x=180\Rightarrow x=45$$

Answer: Third angle = $45^\circ$.

Q13. Two angles sum 130° → find supplementary angles.

Hint: Supplementary ⇒ sum 180°.
Solution:

$$180 – 130 = 50$$

Answer: $50^\circ$.

Q14. Quadrilateral angles 80°, 90°, 75° → fourth?

$$80+90+75+x=360$$

$$80 + 90 + 75 + x = 360 \Rightarrow x = 115$$Answer: Fourth angle = $115^\circ$.

Q15. Circle radius 7 cm → circumference.

$$C = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44$$Answer: $C = 44\ \text{cm}$.

Section C – Application (3 Marks Each)

Q16. Prove: Adjacent angles on a straight line = 180°.

Hint: Angles on straight line form linear pair.
Solution:
Let ∠1 + ∠2 form straight line.
By linear-pair axiom:

$$\angle1 + \angle2 = 180^\circ$$

Answer: Hence proved.

Q17. Triangle sides 6 cm, 8 cm, 10 cm → right triangle?

$$6^2 + 8^2 = 36 + 64 = 100 = 10^2$$

Answer: Yes, right-angled at side 6 and 8 cm.

Q18. Circle diameter 14 cm → radius, C, A.

$$r = 7,\quad C = 2\pi r = 44,\quad A = \pi r^2 = \frac{22}{7}\times7\times7 = 154$$Answer: Radius 7 cm, Circumference 44 cm, Area 154 cm².

Q19. Exterior angle 120°, interior opposite 40° → other?

$$120 = 40 + x \Rightarrow x = 80$$

Answer: $80^\circ$.

Q20. In parallelogram ABCD, ∠A = 70° → others?

$$\angle A = \angle C = 70,\quad \angle B = \angle D = 110$$

Answer: 70°, 110°, 70°, 110°.

Section D – HOTS (4–5 Marks Each)

Q21. Two equal angles, third 96°.

$$x + x + 96 = 180 \Rightarrow 2x = 84 \Rightarrow x = 42$$

Answer: Equal angles = 42° each; Isosceles triangle.

Q22. Sum of interior angles = 1440°.

Formula → $(n-2) × 180 = 1440$

$$n-2 = 8 \Rightarrow n = 10$$

Answer: 10-sided polygon (decagon).

Q23. Circle radius 4 cm → mark points.

Hint: Use compass radius 4 cm.
Answer: One inside, one on, one outside — as constructed.

Q24. Quadrilateral angles 110°, 95°, 70° → fourth?

$$110 + 95 + 70 + x = 360 \Rightarrow x = 85$$

Answer: Fourth angle = 85°; Irregular quadrilateral.

Q25. $AB = 8$ cm bisected at M.$$AM = MB = \frac{8}{2} = 4$$

Answer: AM = MB = 4 cm; M is midpoint.

Section E – MCQs

Bonus Challenge – Clock Problem

At 3:00 the clock hands are at a right angle (90°). Find the next time between 3:00 and 4:00 when they again form a 90° angle.

1) Write angles of hour and minute hands (in degrees) at time $3{:}t$ minutes

• Hour hand: at 3:00 it is at $90^\circ$. It moves $0.5^\circ$ per minute, so at $t$ minutes past 3:

$$\text{Hour angle} = 90 + 0.5t$$

• Minute hand: moves $6^\circ$per minute, so at $t$ minutes:

$$\text{Minute angle} = 6t$$

2) Angle between the hands

The (smaller) angle between them is the absolute difference:

$$\text{Angle} = \big|6t – (90 + 0.5t)\big| = \left|\frac{11}{2}t – 90\right|$$(we used $6t – 0.5t = 5.5t = \tfrac{11}{2}t$).

3) Set the angle equal to $90^\circ$ and solve

We need

$$\left|\frac{11}{2}t – 90\right| = 90.$$

This gives two cases.

Case A: $\dfrac{11}{2}t – 90 = 90$

• Add $90$ to both sides:

  $$\dfrac{11}{2}t = 180.$$

• Multiply both sides by $2$:

  $$11t = 360.$$

• Divide by $11$. Do the division digit-by-digit:

  $$t = \frac{360}{11}.$$

  Long division: $11$ goes into $360$ exactly $32$ times (because $11\times32=352$), remainder $360-352=8$.
  So

  $$t=32\frac{8}{11}\text{ minutes}\mathrm{.}$$

• Convert the fractional minute $\dfrac{8}{11}$ to seconds:

  $$\frac{8}{11}\times 60 = \frac{480}{11}\ \text{seconds}.$$
  

  Divide $480$ by $11$ : $11\times43=473$, remainder $480-473=7$.
  So $\dfrac{480}{11}=43\;\frac{7}{11}$ seconds.

  Therefore

  $$t = 32\ \text{minutes}\;43\;\frac{7}{11}\ \text{seconds}.$$

  As a decimal, $\dfrac{7}{11}$ second $\approx 0.63636$ s, so

  $$t\approx 32\text{ minutes }43.636\text{ seconds}\mathrm{.}$$

Case B: $\dfrac{11}{2}t – 90 = -90$

• Add $90$ to both sides:

  $$\frac{11}{2}t=0\Rightarrow t=0.$$

This is the starting time $3{:}00$ (the initial right angle).

We want the next time after 3:00, so we take the positive solution from Case A.

Final answer (exact and approximate)

• Exact: $t = \dfrac{360}{11}$ minutes after 3:00, i.e. $3{:}32\!:\!43\ \dfrac{7}{11}$ seconds.

• Approximate: 3:32:43.636… (3 hours, 32 minutes, 43.636 seconds).

So the hands next form a right angle at about 3:32:43.64.

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Chapter: Coordinate Geometry — Class 9 Maths (NCERT)

🔹 1. Introduction

Coordinate Geometry (also called Cartesian Geometry) is the branch of mathematics in which the position of a point, line, or shape is described using ordered pairs (x, y) on a plane.

It forms a bridge between Algebra and Geometry.

🔹 2. Cartesian System

(a) Coordinate Axes

• Two perpendicular lines intersecting at a point O (origin) form the coordinate axes.

• The horizontal line is called the x-axis.

• The vertical line is called the y-axis.

(b) Quadrants

• The plane is divided into four quadrants by the axes.

Diagram: Cartesian plane showing the four quadrants and origin.

🔹 3. Coordinates of a Point

Each point on the plane is represented by an ordered pair $(x, y)$:

• $x$: the abscissa (distance along the x-axis)

• $y$: the ordinate (distance along the y-axis)

Example:
Point $A(3, 2)$ means $x = 3$, $y = 2$.

🔹 4. Signs of Coordinates

🔹 5. Coordinates on the Axes

• On x-axis: $y = 0$→ Points like $(2, 0)$
  

• On y-axis: $x = 0$→ Points like $(0, -3)$
  

• At origin: $(0, 0)$
  

🔹 6. Distance Formula

To find the distance between two points
$A(x_1, y_1)$ and $B(x_2, y_2)$:

$$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^{\mathrm{2<span\; style="font-size:\; revert;\; font-family:\; Manrope,\; sans-serif;\; letter-spacing:\; -0.1px;"></span>}}}$$

Example:

Find the distance between $P(2, 3)$ and $Q(5, 7)$.

$$AB = \sqrt{(5 – 2)^2 + (7 – 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

🔹 7. Section Formula

If a point $P(x, y)$ divides the line joining $A(x_1, y_1)$ and $B(x_2, y_2)$ in the ratio $m : n$, then:

$$x = \frac{m x_2 + n x_1}{m + n}, \quad y = \frac{m y_2 + n y_1}{m + n}$$