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9TH CBSE MATHS

Critical Evaluation of Chapter 13: Surface Areas and Volumes (CBSE Class 9 Maths)

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Critical Evaluation of Chapter 13: Surface Areas and Volumes (CBSE Class 9 Maths)

#### **1. Key Concepts Covered in Chapter 13**
This chapter focuses on finding the **surface areas and volumes** of different three-dimensional (3D) geometric shapes, including:

– **Cuboid and Cube** (Surface Area & Volume)
– **Right Circular Cylinder** (Surface Area & Volume)
– **Right Circular Cone** (Surface Area & Volume)
– **Sphere and Hemisphere** (Surface Area & Volume)
– **Frustum of a Cone** (Not in Class 9 CBSE but useful)
– **Combination of Solids** (Applying formulas to composite shapes)
– **Conversion of Solids** (Reshaping one solid into another)

—

### **2. Common Challenges Faced by Students**
✅ **Conceptual Issues**
– Confusing **total surface area** vs. **curved/lateral surface area**
– Understanding when to use **πr²h** vs. **⅓πr²h** for cylinders and cones

✅ **Formula Misapplication**
– Mixing up **sphere’s** surface area formula with **hemisphere’s**
– Forgetting **units conversion** (e.g., cm³ to m³ for volume)

✅ **Difficulty in Composite Figures**
– Combining volumes or areas of different shapes
– Identifying hidden surfaces in diagrams

✅ **Word Problems Misinterpretation**
– Struggling to extract the correct values from problem statements
– Forgetting to apply unit conversions

—

### **3. Analysis of CBSE Exam Trends (Past 10 Years)**

| **Topic** | **Frequently Asked Question Type** | **Marks Distribution** |
|———–|———————————-|———————-|
| **Surface Area of Cube/Cuboid** | Direct formula-based questions | 1 – 2 marks |
| **Surface Area of Cylinder/Cone/Sphere** | Find the total or curved surface area | 2 – 3 marks |
| **Volume of Cube/Cylinder/Cone** | Application-based problem-solving | 2 – 4 marks |
| **Combination of Solids** | Real-life scenario questions | 4 – 6 marks |
| **Conversion of Solids** | Melting and reshaping into a different shape | 3 – 5 marks |

—

### **4. CBSE Exam Questions (Last 10 Years) with Solutions**

#### **Q1: Find the Surface Area of a Sphere (Basic Question)**
📌 **CBSE 2019**: Find the total surface area of a sphere with radius **7 cm**.

✏ **Solution:**
\[
\text{Total Surface Area} = 4\pi r^2
\]
\[
= 4 \times \frac{22}{7} \times 7^2
\]
\[
= 4 \times \frac{22}{7} \times 49 = 616 \text{ cm}^2
\]
✅ **Answer: 616 cm²**

—

#### **Q2: Find the Volume of a Cone**
📌 **CBSE 2020**: Find the volume of a cone with **radius 5 cm** and **height 12 cm**.

✏ **Solution:**
\[
\text{Volume} = \frac{1}{3} \pi r^2 h
\]
\[
= \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12
\]
\[
= \frac{1}{3} \times \frac{22}{7} \times 25 \times 12
\]
\[
= \frac{6600}{21} = 314.29 \text{ cm}^3
\]
✅ **Answer: 314.29 cm³**

—

#### **Q3: Composite Figure (Water Tank Problem)**
📌 **CBSE 2018**: A cylindrical water tank of **radius 3 m** and **height 10 m** is **topped with a hemisphere**. Find the total surface area.

✏ **Solution:**

1️⃣ **Find the Surface Area of Cylinder (excluding top)**
\[
\text{Lateral Surface Area} = 2\pi rh = 2 \times \frac{22}{7} \times 3 \times 10 = 188.57 \text{ m}^2
\]

2️⃣ **Find the Curved Surface Area of Hemisphere**
\[
\text{Curved Surface Area} = 2\pi r^2 = 2 \times \frac{22}{7} \times 3^2 = 56.57 \text{ m}^2
\]

3️⃣ **Total Surface Area:**
\[
\text{Total Surface Area} = 188.57 + 56.57 = 245.14 \text{ m}^2
\]
✅ **Answer: 245.14 m²**

—

#### **Q4: Volume Conversion (CBSE 2017)**
📌 **CBSE 2017**: A solid metallic sphere of **radius 4 cm** is melted and recast into small **cylinders** with radius **1 cm** and height **2 cm**. How many cylinders are formed?

✏ **Solution:**

1️⃣ **Find the Volume of Sphere**
\[
V_{\text{sphere}} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 4^3
\]
\[
= \frac{4}{3} \times \frac{22}{7} \times 64 = 107.43 \text{ cm}^3
\]

2️⃣ **Find the Volume of Cylinder**
\[
V_{\text{cylinder}} = \pi r^2 h = \frac{22}{7} \times 1^2 \times 2 = 6.28 \text{ cm}^3
\]

3️⃣ **Find Number of Cylinders**
\[
\text{Number of cylinders} = \frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{107.43}{6.28} = 17.1 \approx 17
\]
✅ **Answer: 17 cylinders**

—

### **5. Exam Tips for Scoring Full Marks in Chapter 13**

✅ **Memorize All Formulas**
– **Surface Area & Volume of Basic Shapes**
– **Curved vs. Total Surface Area Differences**

✅ **Understand Composite Figures**
– Break them into **basic 3D shapes**
– Solve step-by-step

✅ **Practice Word Problems**
– Focus on **melting and recasting problems**
– Learn **real-life applications**

✅ **Use Approximate Values for Quick Calculation**
– Use **π ≈ 3.14** when needed
– Approximate **square roots** to avoid calculator dependency

—

### **6. Tags for This Chapter**
**CBSE Class 9, Surface Areas and Volumes, Important CBSE Questions, Past 10 Years CBSE Papers, CBSE Exam Preparation, Mensuration, Volume Conversion, Real-Life Applications of Geometry, Mathematics Problem Solving, Word Problems in Math, CBSE Board Exam Tips, Geometry Formulas, CBSE 2024-25**

—

### **🔹 Final Thoughts**
This chapter is **highly scoring** if you practice **past CBSE exam questions**. Focus on **formula application, word problems, and composite shapes**. **Would you like more practice questions or additional explanations on any topic?** 😊

Important CBSE-style practice questions on Heron’s Formula

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Important CBSE-style practice questions on Heron’s Formula

## **Solution 1: Find the Area of a Triangle (Basic Calculation)**
📌 **Question:**
Find the area of a triangle with sides **9 cm, 12 cm, and 15 cm** using Heron’s formula.

### **Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{a + b + c}{2} = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18 \text{ cm}
\]

### **Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]

\[
= \sqrt{18(18-9)(18-12)(18-15)}
\]

\[
= \sqrt{18 \times 9 \times 6 \times 3}
\]

\[
= \sqrt{2916}
\]

\[
= 54 \text{ cm}^2
\]

✅ **Final Answer: 54 cm²**

—

## **Solution 2: Find the Cost of Paving a Triangular Garden**
📌 **Question:**
A **triangular garden** has sides **80 m, 90 m, and 100 m**. Find the cost of paving the garden at ₹20 per square meter.

### **Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{80 + 90 + 100}{2} = \frac{270}{2} = 135 \text{ m}
\]

### **Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{135(135-80)(135-90)(135-100)}
\]

\[
= \sqrt{135 \times 55 \times 45 \times 35}
\]

Using calculations:

\[
= \sqrt{31185000} \approx 5587.5 \text{ m}^2
\]

### **Step 3: Find the Total Cost**
\[
\text{Cost} = \text{Area} \times \text{Rate per m}^2
\]

\[
= 5587.5 \times 20
\]

\[
= 1,11,750
\]

✅ **Final Answer: ₹1,11,750**

—

## **Solution 3: Finding the Height of a Triangle**
📌 **Question:**
A triangle has a **base of 14 cm** and an **area of 84 cm²**. Find its height.

### **Step 1: Use the Basic Area Formula**
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]

\[
84 = \frac{1}{2} \times 14 \times h
\]

### **Step 2: Solve for Height (h)**
\[
h = \frac{84 \times 2}{14} = \frac{168}{14} = 12 \text{ cm}
\]

✅ **Final Answer: Height = 12 cm**

—

## **Solution 4: Finding the Area of a Right-Angled Triangle**
📌 **Question:**
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.

### **Step 1: Use Pythagoras Theorem to Find the Third Side**
\[
\text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2
\]

\[
13^2 = 5^2 + h^2
\]

\[
169 = 25 + h^2
\]

\[
h^2 = 169 – 25 = 144
\]

\[
h = 12 \text{ cm}
\]

✅ So, the sides of the triangle are **5 cm, 12 cm, and 13 cm**.

### **Step 2: Find the Semi-Perimeter (s)**
\[
s = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \text{ cm}
\]

### **Step 3: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{15(15-5)(15-12)(15-13)}
\]

\[
= \sqrt{15 \times 10 \times 3 \times 2}
\]

\[
= \sqrt{900} = 30 \text{ cm}^2
\]

✅ **Final Answer: 30 cm²**

—

## **Solution 5: Quadrilateral Divided into Two Triangles**
📌 **Question:**
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.

### **Step 1: Find the Area of Triangle 1 (6 cm, 8 cm, 10 cm)**
**Semi-Perimeter:**
\[
s = \frac{6+8+10}{2} = \frac{24}{2} = 12 \text{ cm}
\]

\[
\text{Area} = \sqrt{12(12-6)(12-8)(12-10)}
\]

\[
= \sqrt{12 \times 6 \times 4 \times 2}
\]

\[
= \sqrt{576} = 24 \text{ cm}^2
\]

✅ **Triangle 1 Area = 24 cm²**

—

### **Step 2: Find the Area of Triangle 2 (5 cm, 12 cm, 13 cm)**
**Semi-Perimeter:**
\[
s = \frac{5+12+13}{2} = \frac{30}{2} = 15 \text{ cm}
\]

\[
\text{Area} = \sqrt{15(15-5)(15-12)(15-13)}
\]

\[
= \sqrt{15 \times 10 \times 3 \times 2}
\]

\[
= \sqrt{900} = 30 \text{ cm}^2
\]

✅ **Triangle 2 Area = 30 cm²**

—

### **Step 3: Total Area of the Quadrilateral**
\[
= 24 + 30 = 54 \text{ cm}^2
\]

✅ **Final Answer: 54 cm²**

—

## **📌 Summary of Important Exam Tips**
✔ **Always find the semi-perimeter first.**
✔ **Check for right-angled triangles using Pythagoras’ theorem.**
✔ **If given a quadrilateral, split it into two triangles.**
✔ **Always include the correct units (cm², m², etc.) for full marks.**

—

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

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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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Past CBSE exam questions on Heron’s Formula

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Past CBSE exam questions** on **Heron’s Formula

# **Problem 1: Direct Application of Heron’s Formula**
📌 **CBSE 2018 Question:**
Find the area of a triangle with sides **7 cm, 8 cm, and 9 cm** using Heron’s formula.

#### **Solution:**
**Step 1: Find the Semi-Perimeter (s)**
\[
s = \frac{7+8+9}{2} = \frac{24}{2} = 12 \text{ cm}
\]

**Step 2: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
\]
\[
= \sqrt{12(12-7)(12-8)(12-9)}
\]

**Step 3: Simplify the Expression**
\[
= \sqrt{12 \times 5 \times 4 \times 3}
\]

\[
= \sqrt{720}
\]

\[
= 6\sqrt{20} = 6 \times 4.47 \approx 26.82 \text{ cm}^2
\]

✅ **Final Answer: 26.82 cm²**

—

### **Problem 2: Application in Composite Figures**
📌 **CBSE 2022 Question:**
A quadrilateral is divided into two triangles with sides **5 cm, 6 cm, 7 cm** and **8 cm, 10 cm, 6 cm**. Find its total area using Heron’s formula.

#### **Solution:**

✅ **Step 1: Find the Area of Triangle 1 (5 cm, 6 cm, 7 cm)**

**Semi-perimeter:**
\[
s = \frac{5+6+7}{2} = \frac{18}{2} = 9 \text{ cm}
\]

Apply Heron’s Formula:
\[
\text{Area} = \sqrt{9(9-5)(9-6)(9-7)}
\]

\[
= \sqrt{9 \times 4 \times 3 \times 2}
\]

\[
= \sqrt{216} \approx 14.7 \text{ cm}^2
\]

✅ **Step 2: Find the Area of Triangle 2 (8 cm, 10 cm, 6 cm)**

**Semi-perimeter:**
\[
s = \frac{8+10+6}{2} = \frac{24}{2} = 12 \text{ cm}
\]

**Apply Heron’s Formula:**
\[
\text{Area} = \sqrt{12(12-8)(12-10)(12-6)}
\]

\[
= \sqrt{12 \times 4 \times 2 \times 6}
\]

\[
= \sqrt{576} = 24 \text{ cm}^2
\]

✅ **Final Step: Total Area of Quadrilateral**
\[
= 14.7 + 24 = 38.7 \text{ cm}^2
\]

✅ **Final Answer: 38.7 cm²**

—

### **Problem 3: Real-Life Application-Based Question**
📌 **CBSE 2021 Question:**
A triangular plot has sides **50 m, 72 m, and 78 m**. Find the **cost of fencing** it at ₹12 per meter.

#### **Solution:**
✅ **Step 1: Find the Perimeter**
\[
\text{Perimeter} = 50 + 72 + 78 = 200 \text{ m}
\]

✅ **Step 2: Find the Semi-Perimeter (s)**
\[
s = \frac{200}{2} = 100 \text{ m}
\]

✅ **Step 3: Apply Heron’s Formula**
\[
\text{Area} = \sqrt{100(100-50)(100-72)(100-78)}
\]

\[
= \sqrt{100 \times 50 \times 28 \times 22}
\]

\[
= \sqrt{3080000} \approx 1755.69 \text{ m}^2
\]

✅ **Step 4: Find the Cost of Fencing**
Since fencing is charged at ₹12 per meter:

\[
\text{Cost} = \text{Perimeter} \times 12
\]

\[
= 200 \times 12 = 2400
\]

✅ **Final Answer: ₹2400**

—

### **Key Takeaways for Exams:**
✔ **Direct Application** questions appear in every exam.
✔ **Composite Figures** questions are tested in higher-order thinking problems.
✔ **Real-Life Application Questions** appear every **2-3 years**.
✔ **Show Full Steps in Exams** for full marks.

—

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Chapter 11: Constructions (CBSE Class 9 Mathematics)

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Chapter 11: Constructions (CBSE Class 9 Mathematics)

## **1. Key Concepts in Constructions**
Chapter 11 of CBSE Class 9 Mathematics focuses on **geometrical constructions** using a ruler and compass. This chapter is crucial as it builds the foundation for higher-level geometry and practical applications.

### **Topics Covered:**
1. **Basic Constructions**
– Constructing a bisector of a given angle.
– Constructing the perpendicular bisector of a line segment.

2. **Construction of Triangles**
– Given base, base angle, and sum of the other two sides.
– Given base, base angle, and difference of the other two sides.
– Given perimeter and two base angles.

—

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

### **(a) Common Challenges Faced by Students**
1. **Difficulty in Maintaining Precision:**
– Errors in measurements lead to incorrect constructions.
– Slight mistakes in using a compass result in inaccurate figures.

2. **Misinterpretation of Given Data:**
– Confusion between sum/difference of sides in triangle constructions.
– Misplacing the base angle or using the wrong approach.

3. **Errors in Using a Compass and Ruler Together:**
– Students often struggle with drawing perpendicular bisectors accurately.

4. **Forgetting Theoretical Justifications:**
– Many students focus on drawing but fail to justify their steps logically.

—

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

### **1. Basic Construction-Based Questions**

#### **Question (2023, 2020, 2017, 2015)**
– Construct an angle of **75°** using a compass and bisector method.

#### **Solution:**
1. Draw a ray \( OA \).
2. Using a compass, mark an arc from \( O \) that cuts \( OA \) at \( B \).
3. Without changing the compass width, place it at \( B \) and draw another arc to get a point \( C \).
4. Keeping the compass at \( C \), mark another arc to get point \( D \).
5. Bisect \( \angle ABC \) to get \( 75^\circ \).

✅ **Final Answer: Angle of 75° Constructed**

—

### **2. Triangle Construction Questions**

#### **Question (2022, 2018, 2014)**
– Construct a **triangle** where the base is **7 cm**, base angle is **50°**, and the sum of the other two sides is **12 cm**.

#### **Solution:**
1. Draw the base \( BC = 7 cm \).
2. Draw \( \angle B = 50^\circ \).
3. Extend the line beyond \( B \).
4. From \( B \), draw an arc of **12 cm** along the extended line.
5. Connect this new point with \( C \).
6. Bisect this line to locate the third vertex.

✅ **Final Answer: Triangle Constructed Successfully**

—

### **3. Construction with Perpendicular Bisectors**

#### **Question (2021, 2019, 2016)**
– Construct a **perpendicular bisector** for a line segment of **8 cm**.

#### **Solution:**
1. Draw a line \( AB = 8 cm \).
2. Using a compass, place it at \( A \) and draw arcs above and below the line.
3. Without changing the radius, repeat the process from \( B \).
4. Mark the intersection points and draw a perpendicular bisector.

✅ **Final Answer: Perpendicular bisector drawn correctly**

—

### **4. Application-Based Questions**

#### **Question (2020, 2015, 2012)**
– Given the perimeter of a **triangle = 15 cm** and base angles **45° and 60°**, construct the triangle.

#### **Solution:**
1. Draw a line segment equal to **15 cm (perimeter)**.
2. Construct **45°** and **60°** at each end.
3. The intersection of these two rays gives the required triangle.

✅ **Final Answer: Triangle Constructed Using Perimeter**

—

## **4. Exam Preparation Tips**
1. **Practice with Precision:** Small errors can affect the accuracy of the construction.
2. **Revise All Triangle Construction Methods:** Understand when to apply each method.
3. **Use Proper Justifications:** Always explain why a step is performed.
4. **Solve Past Year Papers:** This helps in identifying common patterns.

—

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