CBSE 9TH MATHS PROBLEMSOLVING Archives

CBSE style practice questions for Heron’s Formula based on past exam patterns

February 5, 2025 by Yash

CBSE style practice questions for Heron’s Formula based on past exam patterns

### **📌 Section A: Basic Questions (1-2 Marks Each)**
1️⃣ Find the area of a triangle whose sides are **9 cm, 12 cm, and 15 cm** using Heron’s formula.
2️⃣ A triangle has sides **5 cm, 12 cm, and 13 cm**. Calculate its area.
3️⃣ Find the semi-perimeter of a triangle with sides **18 cm, 24 cm, and 30 cm**.
4️⃣ A triangle has a perimeter of **40 cm**. Two of its sides are **15 cm and 12 cm**. Find the third side if the area is **80 cm²**.
5️⃣ If the sides of a triangle are in the ratio **3:4:5** and its perimeter is **36 cm**, find its area.

—

### **📌 Section B: Intermediate-Level Questions (3-4 Marks Each)**
6️⃣ A triangular park has sides **40 m, 50 m, and 60 m**. Find the area of the park and the cost of laying grass at ₹5 per square meter.
7️⃣ Find the height of a triangle with **base = 14 cm** and **area = 84 cm²** using Heron’s formula.
8️⃣ A triangle has sides **7 cm, 8 cm, and 9 cm**. Find the **difference** between its area using Heron’s formula and using the base-height formula.
9️⃣ Find the area of a **right-angled triangle** whose hypotenuse is **13 cm** and one of the sides is **5 cm**, using Heron’s formula.
🔟 A farmer has a **triangular field** with sides **26 m, 28 m, and 30 m**. He wants to fence it with three rounds of wire. If the cost of fencing is ₹15 per meter, find the total fencing cost.

—

### **📌 Section C: Higher Order Thinking (HOTs) Questions (5-6 Marks Each)**
1️⃣1️⃣ A **quadrilateral** is divided into **two triangles** with sides **6 cm, 8 cm, 10 cm** and **5 cm, 12 cm, 13 cm**. Find its total area using Heron’s formula.
1️⃣2️⃣ The sides of a triangular garden are **80 m, 90 m, and 100 m**. Find the cost of **paving** the garden with tiles at ₹20 per square meter.
1️⃣3️⃣ A **trapezium** is divided into two triangles of sides **10 cm, 12 cm, 14 cm** and **8 cm, 15 cm, 17 cm**. Find the total area of the trapezium.
1️⃣4️⃣ Find the area of an **isosceles triangle** with base **24 cm** and equal sides **26 cm** using Heron’s formula.
1️⃣5️⃣ A **triangle-shaped flag** has sides **20 cm, 21 cm, and 29 cm**. Find the **cost of coloring** it at ₹8 per cm².

—

### **📌 Bonus Challenge Questions**
🎯 A **triangular swimming pool** has sides **50 m, 80 m, and 100 m**. The **depth is 2.5 m**. Find the **total water capacity** in liters. (1 m³ = 1000 liters)
🎯 A **triangular plot** has sides **25 m, 39 m, and 50 m**. The government wants to **build a circular park inside** that covers the maximum possible area. Find the **radius of the largest inscribed circle** in the triangle.

—

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Critical Evaluation of Chapter 10: Circles (CBSE Class 9 Mathematics)

February 1, 2025 by Yash

# **Critical Evaluation of Chapter 10: Circles (CBSE Class 9 Mathematics)**

## **1. Key Concepts in Circles**
1. **Definition of a Circle:** A set of all points in a plane that are equidistant from a fixed point called the center.
2. **Basic Terms in Circles:**
– **Radius (r):** Distance from the center to any point on the circle.
– **Diameter (d):** Twice the radius. \(d = 2r\).
– **Chord:** A line segment joining two points on a circle.
– **Secant:** A line that intersects a circle at two points.
– **Tangent:** A line that touches the circle at exactly one point.
– **Arc:** A portion of the circle’s circumference.
– **Sector:** Region enclosed by two radii and an arc.
– **Segment:** Region enclosed by a chord and an arc.

3. **Important Theorems:**
– **Theorem 1:** Equal chords of a circle subtend equal angles at the center.
– **Theorem 2:** The perpendicular from the center to a chord bisects the chord.
– **Theorem 3:** Equal chords are equidistant from the center.
– **Theorem 4:** The angle subtended by an arc at the center is twice the angle subtended at any point on the remaining part of the circle.
– **Theorem 5:** The sum of opposite angles of a cyclic quadrilateral is 180°.
– **Theorem 6:** The length of the tangent from an external point to a circle is equal.

—

## **2. Critical Evaluation of the Chapter**

### **(a) Conceptual Challenges Faced by Students**
1. **Difficulty Understanding Theorems and Proofs**
– Many students struggle to visualize circle theorems and their proofs.
– Example: Proving that tangents from an external point are equal requires a clear understanding of perpendicular bisectors.

2. **Confusion in Applying Theorems**
– Students often misapply angle properties, especially in cyclic quadrilaterals.
– Example: Misinterpreting that the angle at the center is **twice** the angle on the circumference.

3. **Understanding Tangents and Chords**
– Tangents and their properties, such as equal lengths, are often misused in solving numerical problems.

4. **Visualizing Geometrical Relationships**
– Many students struggle with diagrammatic interpretations, especially when multiple circles or chords are involved.

—

## **3. Evaluation of Past 10 Years’ CBSE Questions**

### **1. Definition-Based Questions**
#### **Question (2023, 2019, 2017, 2015)**
– Define a circle and explain the terms **radius, chord, secant, tangent, and arc**.

#### **Solution:**
– A **circle** is the set of all points equidistant from a fixed point (center).
– **Radius:** A segment from the center to the circumference.
– **Chord:** A segment joining two points on the circle.
– **Secant:** A line cutting the circle at two points.
– **Tangent:** A line touching the circle at exactly one point.
– **Arc:** A curved portion of the circumference.

—

### **2. Theorem-Based Questions**
#### **Question (2022, 2018, 2016, 2013)**
– State and prove: *The perpendicular from the center to a chord bisects the chord.*

#### **Solution:**
– Given: A circle with center \(O\), and a chord \(AB\).
– To Prove: \(OD \perp AB\) bisects \(AB\) at \(D\).
– Proof:
– Join \(OA\) and \(OB\).
– In \(\triangle OAD\) and \(\triangle OBD\):
– \(OA = OB\) (radii of the same circle).
– \(OD = OD\) (common side).
– \(\angle ODA = \angle ODB = 90^\circ\) (given).
– By **RHS Congruence Theorem**, \(\triangle OAD \cong \triangle OBD\).
– \(AD = BD\).

✅ **Hence proved**.

—

### **3. Numerical Questions on Chords & Tangents**
#### **Question (2021, 2019, 2014)**
– The radius of a circle is **10 cm**. A chord is **16 cm** long. Find the perpendicular distance of the chord from the center.

#### **Solution:**
Using the theorem: *The perpendicular from the center bisects the chord.*

– Given: \(OA = 10 cm\), \(AB = 16 cm\), \(AD = 8 cm\)
– Using Pythagoras theorem in \(\triangle OAD\):
\[
OD^2 + AD^2 = OA^2
\]
\[
OD^2 + 8^2 = 10^2
\]
\[
OD^2 + 64 = 100
\]
\[
OD^2 = 36
\]
\[
OD = 6 cm
\]

✅ **Answer: 6 cm**

—

### **4. Angle Property Questions**
#### **Question (2020, 2018, 2016)**
– In a circle, an arc subtends an angle **70°** at the center. Find the angle subtended at the circumference.

#### **Solution:**
Using the theorem: *The angle at the center is twice the angle at the circumference.*
\[
\theta = \frac{1}{2} \times 70^\circ = 35^\circ
\]
✅ **Answer: 35°**

—

### **5. Application-Based Questions**
#### **Question (2019, 2015, 2012)**
– Two tangents are drawn from an external point **P** to a circle with center **O**. Prove that these tangents are equal in length.

#### **Solution:**
– Given: \(PA\) and \(PB\) are tangents from \(P\) to the circle.
– To Prove: \(PA = PB\).
– Proof:
– Join \(OA, OB, OP\).
– \(OA = OB\) (radii of the same circle).
– \(\angle OAP = \angle OBP = 90^\circ\) (tangent is perpendicular to radius).
– In \(\triangle OAP\) and \(\triangle OBP\):
– \(OA = OB\)
– \(OP = OP\) (common side)
– \(\angle OAP = \angle OBP = 90^\circ\)
– By **RHS Congruence Theorem**, \(\triangle OAP \cong \triangle OBP\).
– \(PA = PB\) (corresponding sides).

✅ **Hence proved**.

—

## **4. Exam Preparation Tips**
1. **Memorize All Theorems:** Understanding and proving theorems is critical.
2. **Practice Diagram-Based Questions:** Many CBSE questions are based on geometrical constructions.
3. **Work on Angle and Chord Problems:** Learn how to apply theorems logically in problem-solving.
4. **Solve Past Year Papers:** Focus on numerical and proof-based questions.

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Detailed Solutions for Chapter 9 Practice Problems

January 28, 2025 by Yash

Detailed Solutions for Chapter 9 Practice Problems

1. Prove that if two triangles are on the same base and between the same parallels, they have equal areas.**

Given
– Two triangles \( \triangle ABC \) and \( \triangle DBC \) share the same base \( BC \).
– Both triangles lie between the same parallels \( BC \) and \( AD \).

**To Prove**:
\[
\text{Area of } \triangle ABC = \text{Area of } \triangle DBC
\]

**Proof**:
1. Draw a perpendicular from \( A \) to \( BC \), meeting \( BC \) at \( P \). Let the height be \( h_1 \).
2. Similarly, draw a perpendicular from \( D \) to \( BC \), meeting \( BC \) at \( Q \). Let the height be \( h_2 \).
3. Since both \( A \) and \( D \) lie on the same parallel \( AD \), we know \( h_1 = h_2 \).

The area of \( \triangle ABC \):
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BC \times h_1
\]

The area of \( \triangle DBC \):
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times BC \times h_2
\]

Since \( h_1 = h_2 \):
\[
\text{Area of } \triangle ABC = \text{Area of } \triangle DBC
\]

**Conclusion**:
Triangles on the same base and between the same parallels have equal areas.

—

**2. A parallelogram has sides \( 12 \, \text{cm} \) and \( 15 \, \text{cm} \). Its height corresponding to the base \( 12 \, \text{cm} \) is \( 8 \, \text{cm} \). Find its area.**

**Solution**:
– **Base** \( = 12 \, \text{cm} \)
– **Height** \( = 8 \, \text{cm} \)

Area of a parallelogram:
\[
\text{Area} = \text{Base} \times \text{Height}
\]

Substitute the values:
\[
\text{Area} = 12 \times 8 = 96 \, \text{cm}^2
\]

#### **Answer**:
The area of the parallelogram is \( 96 \, \text{cm}^2 \).

—

**3. The vertices of a triangle are \( A(-2, -3), B(3, 5), \text{and } C(5, -1) \). Calculate its area using coordinate geometry.**

**Solution**:
The formula for the area of a triangle with vertices \( (x_1, y_1), (x_2, y_2), (x_3, y_3) \) is:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|
\]

Substitute \( A(-2, -3), B(3, 5), C(5, -1) \):
\[
\text{Area} = \frac{1}{2} \left| -2(5 – (-1)) + 3(-1 – (-3)) + 5((-3) – 5) \right|
\]
\[
= \frac{1}{2} \left| -2(6) + 3(2) + 5(-8) \right|
\]
\[
= \frac{1}{2} \left| -12 + 6 – 40 \right|
\]
\[
= \frac{1}{2} \left| -46 \right|
\]
\[
= \frac{1}{2} \times 46 = 23 \, \text{sq. units}
\]

#### **Answer**:
The area of the triangle is \( 23 \, \text{sq. units} \).

—

### **4. A quadrilateral \( ABCD \) is divided into two triangles by a diagonal. Prove that the sum of their areas equals the area of the quadrilateral.**

#### **Given**:
Quadrilateral \( ABCD \) with diagonal \( AC \) dividing it into \( \triangle ABC \) and \( \triangle ACD \).

#### **To Prove**:
\[
\text{Area of } \triangle ABC + \text{Area of } \triangle ACD = \text{Area of Quadrilateral } ABCD
\]

#### **Proof**:
1. Draw diagonal \( AC \).
2. The area of \( \triangle ABC \) is given by:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
3. The area of \( \triangle ACD \) is similarly calculated using \( AC \) as the base.

Adding the areas of the two triangles:
\[
\text{Area of } \triangle ABC + \text{Area of } \triangle ACD = \text{Area of Quadrilateral } ABCD
\]

#### **Conclusion**:
The total area of the two triangles equals the area of the quadrilateral.

—

### **5. Prove that the diagonals of a parallelogram divide it into four triangles of equal areas.**

#### **Given**:
A parallelogram \( ABCD \) with diagonals \( AC \) and \( BD \) intersecting at \( O \).

#### **To Prove**:
\[
\text{Area of } \triangle AOB = \text{Area of } \triangle BOC = \text{Area of } \triangle COD = \text{Area of } \triangle DOA
\]

#### **Proof**:
1. In \( \triangle AOB \) and \( \triangle COD \):
– \( AC \) is the diagonal, and it divides the parallelogram into two equal parts.
– Therefore, \( \text{Area of } \triangle AOB = \text{Area of } \triangle COD. \)

2. Similarly, in \( \triangle BOC \) and \( \triangle DOA \):
– \( BD \) divides the parallelogram into two equal parts.
– Therefore, \( \text{Area of } \triangle BOC = \text{Area of } \triangle DOA. \)

3. Since \( AC \) and \( BD \) intersect at \( O \), the four triangles are of equal area.

#### **Conclusion**:
The diagonals of a parallelogram divide it into four triangles of equal areas.

—

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Critical Analysis of Chapter 9: Areas of Parallelograms and Triangles

January 28, 2025 by Yash

Critical Analysis of Chapter 9: Areas of Parallelograms and Triangles
Detailed Analysis with Examples (Based on CBSE Exam Trends)

—

### **Key Concepts and Their Critical Analysis**

1. **Derivation of Formulas for the Area of a Parallelogram and Triangle Using Base and Height**
– **Critical Analysis**:
Understanding that the area depends on the perpendicular distance (height) drawn from a vertex to the base is crucial. Many students confuse height with any side of the figure. Proper visualization of perpendicular height relative to the base is emphasized in CBSE.

– **Example from CBSE Exams**:
**Q1 (2017)**: Prove that the area of a parallelogram is the product of its base and corresponding height.
**Q2 (2015)**: The base of a triangle is 8 cm, and its height is 5 cm. Find the area.

– **Example for Practice**:
Prove that if the diagonals of a parallelogram bisect each other, the parallelogram divides into two triangles of equal areas.

—

2. **Application of Coordinate Geometry to Calculate Areas**
– **Critical Analysis**:
Application of the formula:
\[
\text{Area of a Triangle} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
\]
helps in handling irregular figures and shapes plotted on the Cartesian plane. This concept often confuses students due to sign errors and incorrect substitutions. CBSE examiners focus on step-by-step calculations.

– **Example from CBSE Exams**:
**Q1 (2019)**: Calculate the area of a triangle with vertices \( A(1, 2), B(4, 6), C(7, 2) \).
**Q2 (2022)**: A triangle is formed by the points \( (0, 0), (3, 4), (6, 0) \). Find its area using coordinate geometry.

– **Example for Practice**:
Determine the area of a quadrilateral with vertices \( (1, 2), (3, 4), (5, 6), (1, 6) \) by dividing it into two triangles.

—

3. **Relationship Between Parallelograms and Triangles Sharing the Same Base and Between the Same Parallels**
– **Critical Analysis**:
Students often struggle with visualizing why triangles within parallelograms have equal areas when sharing a base and height. This concept is critical in solving both proof-based and numerical problems in CBSE. Emphasis is on diagrammatic representation in solutions.

– **Example from CBSE Exams**:
**Q1 (2018)**: Prove that triangles on the same base and between the same parallels are equal in area.
**Q2 (2021)**: Two triangles \( ABC \) and \( DEF \) share the same base and are between the same parallels. Show that their areas are equal.

– **Example for Practice**:
A parallelogram \( ABCD \) has diagonals intersecting at \( O \). Prove that triangles \( AOB \) and \( COD \) have equal areas.

—

### **Challenges and Solutions with Examples**

1. **Understanding the Concept of Height Relative to the Base**
– **Challenge**: Students often fail to correctly identify the height of a shape relative to its base, especially in irregular figures.
– **Solution**: Use multiple examples with diagrams to illustrate perpendicular heights for various cases.

– **Example for Practice**:
A parallelogram has a base of 12 cm and a height of 5 cm. Find the area when the base changes to 10 cm but remains within the same parallels.

—

2. **Confusion While Applying Area Formulas in Coordinate Geometry**
– **Challenge**: Substitution of incorrect signs or coordinates during calculations.
– **Solution**: Encourage writing the coordinate formula and substituting values systematically with clear steps.

– **Example for Practice**:
A triangle has vertices at \( (2, 3), (4, 8), (6, 3) \). Calculate its area and verify the result by splitting the triangle into two right triangles.

—

3. **Difficulty Visualizing Relationships Between Shapes**
– **Challenge**: Proving equal areas or relationships between triangles and parallelograms is difficult without proper diagrammatic representation.
– **Solution**: Focus on step-by-step geometric proofs supported by labeled diagrams.

– **Example for Practice**:
Prove that a triangle formed by the midpoints of the sides of a parallelogram has one-fourth the area of the parallelogram.

—

### **CBSE Exam Trends (Last 10 Years)**

– Proof-based questions from this chapter are common, accounting for 3–4 marks.
– Coordinate geometry-based area calculations appear as numerical problems, often worth 3 marks.
– Diagram-based problems related to parallelograms and triangles sharing the same base are frequently tested.

—

### **Practice Problems for CBSE Exam Preparation**

1. Prove that if two triangles are on the same base and between the same parallels, they have equal areas.
2. A parallelogram has sides \( 12 \, \text{cm} \) and \( 15 \, \text{cm} \). Its height corresponding to the base \( 12 \, \text{cm} \) is \( 8 \, \text{cm} \). Find its area.
3. The vertices of a triangle are \( A(-2, -3), B(3, 5), \text{and } C(5, -1) \). Calculate its area using coordinate geometry.
4. A quadrilateral \( ABCD \) is divided into two triangles by a diagonal. Prove that the sum of their areas equals the area of the quadrilateral.
5. Prove that the diagonals of a parallelogram divide it into four triangles of equal areas.

—

Would you like me to provide detailed solutions for these practice problems or additional examples? 😊

Introduction to Euclid’s Geometry with detailed explanations, examples, and solutions to help you excel..

January 13, 2025 by Yash

Example 1: Understanding Postulates

Question

Using Euclid’s postulates, prove that:
“A straight line can be drawn from any one point to any other point.”

Solution

Postulate Referenced:
- Euclid’s First Postulate states, “A straight line can be drawn from any one point to any other point.”
Explanation:
- Consider two points AA and BB in a plane.
- According to the postulate, there exists one and only one straight line that passes through both AA and BB.
Diagram:
Draw a plane with two points labeled AA and BB, and connect them with a straight line.
This represents the unique straight line as per the postulate.
Conclusion:
The statement is a direct application of Euclid’s First Postulate.

Example 2: Understanding Axioms

Question

Explain the axiom: “Things that are equal to the same thing are equal to one another.” Illustrate with an example.

Solution

Axiom Statement:
- This is Euclid’s First Axiom, which states that if two quantities are each equal to a third quantity, they are equal to each other.
Explanation:
- Let a=ba = b and b=cb = c.
- Since both aa and cc are equal to bb, it follows that a=ca = c.
Example:
- Suppose AB=5 cmAB = 5 \, \text{cm}, BC=5 cmBC = 5 \, \text{cm}.
- Since ABAB and BCBC are both equal to 5 cm, we conclude AB=BCAB = BC.
Diagram:
Draw a line segment divided into two parts ABAB and BCBC, both labeled 5 cm5 \, \text{cm}.

Example 3: Application of Euclid’s Fifth Postulate

Question

Explain the significance of Euclid’s Fifth Postulate and provide an example.

Solution

Postulate Statement:
- Euclid’s Fifth Postulate states, “If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side.”
Explanation:
- This is the basis for parallel and non-parallel lines.
- If the sum of the interior angles is less than 180∘180^\circ, the lines will meet on that side.
Example:
- Consider two lines l1l_1 and l2l_2, and a transversal tt.
- Measure the interior angles on the same side of the transversal.
  - If ∠1+∠2<180∘\angle 1 + \angle 2 < 180^\circ, the lines l1l_1 and l2l_2 will meet.
Diagram:
Draw two lines and a transversal, showing interior angles ∠1\angle 1 and ∠2\angle 2.

Example 4: Exam-Style Proof

Question

Prove that:
The sum of the angles in a triangle is 180∘180^\circ.

Solution

Given: A triangle ABCABC.
To Prove: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ.
Construction:
- Draw a line DEDE parallel to BCBC through AA.
- Extend ABAB and ACAC to meet DEDE.
Proof:
- By the Alternate Interior Angle Theorem, ∠CAB=∠DAB,∠ACB=∠EAC.\angle CAB = \angle DAB, \quad \angle ACB = \angle EAC.
- Along the straight line DEDE, ∠DAB+∠BAC+∠EAC=180∘.\angle DAB + \angle BAC + \angle EAC = 180^\circ.
- Substituting, ∠CAB+∠ABC+∠BCA=180∘.\angle CAB + \angle ABC + \angle BCA = 180^\circ.
Conclusion: ∠A+∠B+∠C=180∘.\angle A + \angle B + \angle C = 180^\circ.
Diagram:
Draw triangle ABCABC with the parallel line DEDE and the angles marked.

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Let’s explore Coordinate Geometry even further with challenging problems, advanced applications, and step-by-step solutions to deepen understanding

December 31, 2024 by Yash

Advanced Problem-Solving in Coordinate Geometry

1. Intersection of Lines

Problem: Find the point of intersection of the lines represented by:

2x+3y=132x + 3y = 13
x−2y=−5x – 2y = -5.

Solution:
To find the intersection, solve the two equations simultaneously.

Equation 1: 2x+3y=132x + 3y = 13
Equation 2: x−2y=−5x – 2y = -5.

From Equation 2: x=2y−5.x = 2y – 5.

Substitute x=2y−5x = 2y – 5 into Equation 1: 2(2y−5)+3y=13.2(2y – 5) + 3y = 13. 4y−10+3y=13.4y – 10 + 3y = 13. 7y=23.7y = 23. y=237.y = \frac{23}{7}.

Substitute y=237y = \frac{23}{7} into x=2y−5x = 2y – 5: x=2(237)−5.x = 2\left(\frac{23}{7}\right) – 5. x=467−357=117.x = \frac{46}{7} – \frac{35}{7} = \frac{11}{7}.

Answer: The lines intersect at (117,237)\left(\frac{11}{7}, \frac{23}{7}\right).

2. Verifying Collinearity

Problem: Check if the points A(1,2)A(1, 2), B(3,6)B(3, 6), and C(5,10)C(5, 10) are collinear.

Solution:
To verify collinearity, calculate the area of the triangle formed by these points using the formula: Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.

Substitute A(1,2)A(1, 2), B(3,6)B(3, 6), C(5,10)C(5, 10): Area=12∣1(6−10)+3(10−2)+5(2−6)∣.\text{Area} = \frac{1}{2} \left| 1(6 – 10) + 3(10 – 2) + 5(2 – 6) \right|. =12∣1(−4)+3(8)+5(−4)∣.= \frac{1}{2} \left| 1(-4) + 3(8) + 5(-4) \right|. =12∣−4+24−20∣.= \frac{1}{2} \left| -4 + 24 – 20 \right|. =12∣0∣=0.= \frac{1}{2} \left| 0 \right| = 0.

Since the area is 00, the points are collinear.

Answer: AA, BB, and CC are collinear.

3. Shortest Distance from a Point to a Line

Problem: Find the shortest distance from the point P(4,1)P(4, 1) to the line 3x−4y+5=03x – 4y + 5 = 0.

Solution:
The formula for the shortest distance from a point (x1,y1)(x_1, y_1) to a line Ax+By+C=0Ax + By + C = 0 is: Distance=∣Ax1+By1+C∣A2+B2.\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}.

Here, A=3A = 3, B=−4B = -4, C=5C = 5, x1=4x_1 = 4, y1=1y_1 = 1: Distance=∣3(4)−4(1)+5∣32+(−4)2.\text{Distance} = \frac{|3(4) – 4(1) + 5|}{\sqrt{3^2 + (-4)^2}}. =∣12−4+5∣9+16.= \frac{|12 – 4 + 5|}{\sqrt{9 + 16}}. =∣13∣25=135.= \frac{|13|}{\sqrt{25}} = \frac{13}{5}.

Answer: The shortest distance is 135\frac{13}{5} or 2.6 units2.6 \, \text{units}.

4. Equation of a Line

Problem: Find the equation of a line passing through (2,3)(2, 3) and perpendicular to the line 4x−3y+7=04x – 3y + 7 = 0.

Solution:
For a line perpendicular to another line, the slope of the new line is the negative reciprocal of the given line’s slope.

Find the slope of 4x−3y+7=04x – 3y + 7 = 0:
Rewrite in slope-intercept form (y=mx+cy = mx + c): 3y=4x+7⇒y=43x+73.3y = 4x + 7 \quad \Rightarrow \quad y = \frac{4}{3}x + \frac{7}{3}. Slope (mm) of the given line = 43\frac{4}{3}.
Slope of the required line = −1m=−34-\frac{1}{m} = -\frac{3}{4}.
Use the point-slope form to find the equation of the line: y−y1=m(x−x1).y – y_1 = m(x – x_1). Substitute (x1,y1)=(2,3)(x_1, y_1) = (2, 3), m=−34m = -\frac{3}{4}: y−3=−34(x−2).y – 3 = -\frac{3}{4}(x – 2). y−3=−34x+64.y – 3 = -\frac{3}{4}x + \frac{6}{4}. y=−34x+32+3.y = -\frac{3}{4}x + \frac{3}{2} + 3. y=−34x+92.y = -\frac{3}{4}x + \frac{9}{2}.

Answer: Equation of the line: y=−34x+92y = -\frac{3}{4}x + \frac{9}{2}.

Applications in Depth

Astronomy:
Coordinate geometry is used to calculate distances between planets and map the universe.
Mobile App Development:
Apps like Uber and Google Maps use coordinate systems to plot locations and calculate distances.
Sports and Gaming:
Coordinate geometry helps in analyzing player movements, designing game levels, and calculating trajectories in sports.
Engineering:
Engineers use coordinate geometry in structural design, such as calculating the alignment of beams and supports.

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“Coordinate Geometry”, which introduces the fundamental concepts of representing and analyzing geometric shapes in a two-dimensional plane using the Cartesian system

December 31, 2024 by Yash

Key Concepts

1. Cartesian System

The Cartesian system uses two perpendicular axes:

The horizontal axis (xx-axis).
The vertical axis (yy-axis).

Origin (OO): Intersection point of the axes (0,0)(0, 0).
Points are represented as ordered pairs (x,y)(x, y), where:

xx is the abscissa (horizontal distance from the origin).
yy is the ordinate (vertical distance from the origin).

2. Quadrants

The plane is divided into four quadrants:

Quadrant I: x>0,y>0x > 0, y > 0.
Quadrant II: x<0,y>0x < 0, y > 0.
Quadrant III: x<0,y<0x < 0, y < 0.
Quadrant IV: x>0,y<0x > 0, y < 0.

3. Plotting Points

Start at the origin.
Move horizontally to the xx-coordinate.
Move vertically to the yy-coordinate.

Example: Plot (2,3)(2, 3).

Move 2 units right (positive xx).
Move 3 units up (positive yy).

4. Distance Formula

The distance between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}.

Example: Find the distance between A(1,2)A(1, 2) and B(4,6)B(4, 6). d=(4−1)2+(6−2)2=32+42=9+16=25=5.d = \sqrt{(4 – 1)^2 + (6 – 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

5. Section Formula

A point P(x,y)P(x, y) dividing a line segment joining (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) in the ratio m:nm:n is: P(x,y)=(mx2+nx1m+n,my2+ny1m+n).P\left(x, y\right) = \left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right).

Example: Find the point dividing the line joining A(1,2)A(1, 2) and B(4,6)B(4, 6) in the ratio 2:12:1. x=2(4)+1(1)2+1=8+13=3.x = \frac{2(4) + 1(1)}{2 + 1} = \frac{8 + 1}{3} = 3. y=2(6)+1(2)2+1=12+23=4.67.y = \frac{2(6) + 1(2)}{2 + 1} = \frac{12 + 2}{3} = 4.67.

Point P=(3,4.67)P = (3, 4.67).

6. Midpoint Formula

The midpoint of a line segment joining (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is: M(x,y)=(x1+x22,y1+y22).M\left(x, y\right) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).

Example: Find the midpoint of A(1,2)A(1, 2) and B(4,6)B(4, 6). x=1+42=2.5, y=2+62=4.x = \frac{1 + 4}{2} = 2.5, \, y = \frac{2 + 6}{2} = 4.

Midpoint M=(2.5,4)M = (2.5, 4).

7. Area of a Triangle

The area of a triangle with vertices (x1,y1)(x_1, y_1), (x2,y2)(x_2, y_2), and (x3,y3)(x_3, y_3) is: Area=12∣x1(y2−y3)+x2(y3−y1)+x3(y1−y2)∣.\text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right|.

Example: Find the area of a triangle with vertices A(1,2)A(1, 2), B(4,6)B(4, 6), and C(3,5)C(3, 5). Area=12∣1(6−5)+4(5−2)+3(2−6)∣.\text{Area} = \frac{1}{2} \left| 1(6 – 5) + 4(5 – 2) + 3(2 – 6) \right|. =12∣1(1)+4(3)+3(−4)∣.= \frac{1}{2} \left| 1(1) + 4(3) + 3(-4) \right|. =12∣1+12−12∣=12∣1∣=0.5.= \frac{1}{2} \left| 1 + 12 – 12 \right| = \frac{1}{2} \left| 1 \right| = 0.5.

Answer: Area = 0.5 square units0.5 \, \text{square units}.

Critical Analysis

Conceptual Clarity:
- Understanding quadrants is essential for correctly identifying the position of points.
- Mastery of formulas like distance, section, and midpoint helps solve geometry problems efficiently.
Applications:
- Coordinate geometry has real-world applications in navigation, engineering, and computer graphics.
- It lays the foundation for advanced topics like vectors and 3D geometry.
Common Mistakes:
- Confusing the signs of coordinates in different quadrants.
- Misapplying formulas, especially in the section formula with incorrect ratios.
Visualization:
- Drawing rough sketches improves accuracy.
- Graphical representation helps in better understanding spatial relationships.

Applications of Coordinate Geometry

1. Map Reading

Identifying locations on a map using coordinates.

2. Navigation

GPS systems use coordinate geometry to calculate routes and distances.

3. Design and Animation

Used in computer-aided design (CAD) and animation software to position objects.

4. Physics

Analyzing motion in two dimensions.

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Dive deeper into problem-solving techniques in polynomials, We’ll explore…

December 31, 2024 by Yash

Factoring Techniques
Solving Quadratic Polynomials
Graphical Analysis
Application-based Problems

1. Factoring Techniques

Factoring is crucial for simplifying and solving polynomial equations.

Case 1: Common Factor

Find the greatest common factor (GCF) and factor it out.
Example: Factor 6×3+9x26x^3 + 9x^2. =3×2(2x+3).= 3x^2(2x + 3).

Case 2: Grouping

Group terms to create common factors.
Example: Factor x3+3×2+x+3x^3 + 3x^2 + x + 3. =(x3+3×2)+(x+3).= (x^3 + 3x^2) + (x + 3). =x2(x+3)+1(x+3).= x^2(x + 3) + 1(x + 3). =(x+3)(x2+1).= (x + 3)(x^2 + 1).

Case 3: Splitting the Middle Term

Split the middle term in quadratic polynomials.
Example: Factor x2−5x+6x^2 – 5x + 6.
Find two numbers that multiply to 66 and add to −5-5 (i.e., −2-2 and −3-3): =x2−2x−3x+6.= x^2 – 2x – 3x + 6. =x(x−2)−3(x−2).= x(x – 2) – 3(x – 2). =(x−2)(x−3).= (x – 2)(x – 3).

2. Solving Quadratic Polynomials

To solve ax2+bx+c=0ax^2 + bx + c = 0, use:

Method 1: Factoring

Solve x2−5x+6=0x^2 – 5x + 6 = 0: x2−5x+6=(x−2)(x−3)=0.x^2 – 5x + 6 = (x – 2)(x – 3) = 0. x=2, x=3.x = 2, \, x = 3.

Method 2: Completing the Square

Solve x2−4x+3=0x^2 – 4x + 3 = 0 by completing the square:

Rewrite x2−4x+3x^2 – 4x + 3 as: x2−4x=−3.x^2 – 4x = -3.
Add (−42)2=4\left(\frac{-4}{2}\right)^2 = 4 to both sides: x2−4x+4=−3+4.x^2 – 4x + 4 = -3 + 4.
Simplify: (x−2)2=1.(x – 2)^2 = 1.
Solve: x−2=±1 ⟹ x=3, x=1.x – 2 = \pm 1 \implies x = 3, \, x = 1.

Method 3: Quadratic Formula

For ax2+bx+c=0ax^2 + bx + c = 0: x=−b±b2−4ac2a.x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}.

Example: Solve 2×2−4x−6=02x^2 – 4x – 6 = 0: a=2, b=−4, c=−6.a = 2, \, b = -4, \, c = -6. x=−(−4)±(−4)2−4(2)(−6)2(2).x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(-6)}}{2(2)}. x=4±16+484.x = \frac{4 \pm \sqrt{16 + 48}}{4}. x=4±644.x = \frac{4 \pm \sqrt{64}}{4}. x=4±84.x = \frac{4 \pm 8}{4}. x=3, x=−1.x = 3, \, x = -1.

3. Graphical Analysis

Graphs of polynomials provide visual insights into their behavior and solutions.

Quadratic Polynomials:

Shape: Parabola.
Key points:
- Vertex: Turning point of the parabola.
- Zeros: Points where the parabola intersects the xx-axis.
- Direction: Opens upwards (a>0a > 0) or downwards (a<0a < 0).

Example: y=x2−4x+3y = x^2 – 4x + 3.

Factorize: (x−1)(x−3)(x – 1)(x – 3), so zeros are x=1x = 1, x=3x = 3.
Vertex: Midpoint of zeros: x=1+32=2, y=22−4(2)+3=−1.x = \frac{1 + 3}{2} = 2, \, y = 2^2 – 4(2) + 3 = -1.
Plot points: (1, 0), (3, 0), (2, -1).

4. Application-Based Problems

Problem 1: Cost Optimization

A farmer builds a rectangular pen with xx-meter fencing on one side. The total area is 60 m260 \, \text{m}^2. Find xx.

Area = x⋅widthx \cdot \text{width}: 60=x⋅(x−2) ⟹ x2−2x−60=0.60 = x \cdot (x – 2) \implies x^2 – 2x – 60 = 0.

Solve using the quadratic formula: x=−(−2)±(−2)2−4(1)(−60)2(1).x = \frac{-(-2) \pm \sqrt{(-2)^2 – 4(1)(-60)}}{2(1)}. x=2±4+2402.x = \frac{2 \pm \sqrt{4 + 240}}{2}. x=2±2442.x = \frac{2 \pm \sqrt{244}}{2}. x=2±2612.x = \frac{2 \pm 2\sqrt{61}}{2}. x=1±61.x = 1 \pm \sqrt{61}.

Problem 2: Physics – Motion

The height hh of a ball is modeled by h(t)=−5t2+20t+25h(t) = -5t^2 + 20t + 25. Find the time when the ball hits the ground.
Set h(t)=0h(t) = 0: −5t2+20t+25=0.-5t^2 + 20t + 25 = 0.

Divide through by −5-5: t2−4t−5=0.t^2 – 4t – 5 = 0.

Factorize: (t−5)(t+1)=0.(t – 5)(t + 1) = 0. t=5, t=−1.t = 5, \, t = -1.

Since tt cannot be negative, t=5 secondst = 5 \, \text{seconds}.

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