The importance of Class 6 Science (Curiosity) Chapter 2 MCQs lies in their ability to reinforce a student's understanding of the chapter's core concepts, which may focus on topics like "Components of Food" or another relevant theme (depending on the exact chapter title). These MCQs help students assRead more
The importance of Class 6 Science (Curiosity) Chapter 2 MCQs lies in their ability to reinforce a student’s understanding of the chapter’s core concepts, which may focus on topics like “Components of Food” or another relevant theme (depending on the exact chapter title). These MCQs help students assess their knowledge of fundamental ideas such as nutrients (carbohydrates, proteins, fats, vitamins, and minerals), their sources and their roles in maintaining good health. Practicing these questions enhances observational skills, logical reasoning, and the ability to apply scientific concepts to real-life scenarios, such as identifying balanced diets, understanding the importance of nutrition and preventing diseases caused by deficiencies. Additionally, MCQs prepare students for exams by improving speed, accuracy and familiarity with question patterns. Mastering this chapter builds a strong foundation for advanced topics in biology and health science, fostering both academic success and a deeper understanding of the importance of proper nutrition for overall well-being.
The importance of Class 6 Science (Curiosity) Chapter 1 MCQs lies in their ability to reinforce a student's understanding of the chapter's core concepts, which may focus on topics like "Food: Where Does It Come From?" or another relevant theme (depending on the exact chapter title). These MCQs helpRead more
The importance of Class 6 Science (Curiosity) Chapter 1 MCQs lies in their ability to reinforce a student’s understanding of the chapter’s core concepts, which may focus on topics like “Food: Where Does It Come From?” or another relevant theme (depending on the exact chapter title). These MCQs help students assess their knowledge of fundamental ideas such as the sources of food (plants and animals), types of food and the process of food production. Practicing these questions enhances observational skills, logical reasoning and the ability to apply scientific concepts to real-life scenarios, such as identifying plant and animal-based food products or understanding the journey of food from farm to table. Additionally, MCQs prepare students for exams by improving speed, accuracy and familiarity with question patterns. Mastering this chapter builds a strong foundation for advanced topics in biology, agriculture, and environmental science, fostering both academic success and a deeper appreciation for the origins and diversity of food sources.
Sides: 3 cm, 4 cm, 5 cm → Right triangle Area = (1/2) × base × height = (1/2) × 3 × 4 = 6 cm² This question is related to Chapter 10, "Heron’s Formula," from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals usingRead more
Sides: 3 cm, 4 cm, 5 cm → Right triangle
Area = (1/2) × base × height
= (1/2) × 3 × 4 = 6 cm²
This question is related to Chapter 10, “Heron’s Formula,” from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals using Heron’s Formula. Provide an answer based on your understanding of the chapter.
Sides: 91 cm, 98 cm, 105 cm Semi-perimeter (s) = 147 cm Area = √[s(s-a)(s-b)(s-c)] = √[147(56)(49)(42)] = 4116 cm² Using Area = (1/2) × base × height: 4116 = (1/2) × 105 × h h = (4116 × 2) / 105 = 78.4 cm This question pertains to Chapter 10, "Heron’s Formula," in the Class 9 NCERT Mathematics textbRead more
Sides: 91 cm, 98 cm, 105 cm
Semi-perimeter (s) = 147 cm
Area = √[s(s-a)(s-b)(s-c)]
= √[147(56)(49)(42)]
= 4116 cm²
Using Area = (1/2) × base × height:
4116 = (1/2) × 105 × h
h = (4116 × 2) / 105 = 78.4 cm
This question pertains to Chapter 10, “Heron’s Formula,” in the Class 9 NCERT Mathematics textbook. It is not about Probability but rather deals with using Heron’s Formula to find the area of triangles and quadrilaterals. Provide an answer based on your understanding of the chapter.
Cost = Rs. 783, Rate = Rs. 58/hectare Area = 783 / 58 = 13.5 hectares = 135,000 m² Base = 3 × altitude (h) Area of triangle = (1/2) × base × height 135,000 = (1/2) × 3h × h 135,000 = (3/2)h² h² = (135,000 × 2) / 3 = 90,000 h = √90,000 = 300 m Base = 3h = 3 × 300 = 900 m This question is from ChapterRead more
Cost = Rs. 783, Rate = Rs. 58/hectare
Area = 783 / 58 = 13.5 hectares = 135,000 m²
Base = 3 × altitude (h)
Area of triangle = (1/2) × base × height
135,000 = (1/2) × 3h × h
135,000 = (3/2)h²
h² = (135,000 × 2) / 3 = 90,000
h = √90,000 = 300 m
Base = 3h = 3 × 300 = 900 m
This question is from Chapter 10, “Heron’s Formula,” in the Class 9 NCERT Mathematics textbook. It focuses on applying Heron’s Formula to calculate the areas of triangles and quadrilaterals, not Probability. Provide an answer based on your understanding of the chapter.
Class 6 Science Curiosity Chapter 2 MCQ?
The importance of Class 6 Science (Curiosity) Chapter 2 MCQs lies in their ability to reinforce a student's understanding of the chapter's core concepts, which may focus on topics like "Components of Food" or another relevant theme (depending on the exact chapter title). These MCQs help students assRead more
The importance of Class 6 Science (Curiosity) Chapter 2 MCQs lies in their ability to reinforce a student’s understanding of the chapter’s core concepts, which may focus on topics like “Components of Food” or another relevant theme (depending on the exact chapter title). These MCQs help students assess their knowledge of fundamental ideas such as nutrients (carbohydrates, proteins, fats, vitamins, and minerals), their sources and their roles in maintaining good health. Practicing these questions enhances observational skills, logical reasoning, and the ability to apply scientific concepts to real-life scenarios, such as identifying balanced diets, understanding the importance of nutrition and preventing diseases caused by deficiencies. Additionally, MCQs prepare students for exams by improving speed, accuracy and familiarity with question patterns. Mastering this chapter builds a strong foundation for advanced topics in biology and health science, fostering both academic success and a deeper understanding of the importance of proper nutrition for overall well-being.
For Practice MCQ visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-6/science/
Class 6 Science Curiosity Chapter 1 MCQ?
The importance of Class 6 Science (Curiosity) Chapter 1 MCQs lies in their ability to reinforce a student's understanding of the chapter's core concepts, which may focus on topics like "Food: Where Does It Come From?" or another relevant theme (depending on the exact chapter title). These MCQs helpRead more
The importance of Class 6 Science (Curiosity) Chapter 1 MCQs lies in their ability to reinforce a student’s understanding of the chapter’s core concepts, which may focus on topics like “Food: Where Does It Come From?” or another relevant theme (depending on the exact chapter title). These MCQs help students assess their knowledge of fundamental ideas such as the sources of food (plants and animals), types of food and the process of food production. Practicing these questions enhances observational skills, logical reasoning and the ability to apply scientific concepts to real-life scenarios, such as identifying plant and animal-based food products or understanding the journey of food from farm to table. Additionally, MCQs prepare students for exams by improving speed, accuracy and familiarity with question patterns. Mastering this chapter builds a strong foundation for advanced topics in biology, agriculture, and environmental science, fostering both academic success and a deeper appreciation for the origins and diversity of food sources.
For Practice MCQ visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-6/science/
The sides of a triangle are 3 cm , 4 cm and 5 cm. Its area is
Sides: 3 cm, 4 cm, 5 cm → Right triangle Area = (1/2) × base × height = (1/2) × 3 × 4 = 6 cm² This question is related to Chapter 10, "Heron’s Formula," from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals usingRead more
Sides: 3 cm, 4 cm, 5 cm → Right triangle
Area = (1/2) × base × height
= (1/2) × 3 × 4 = 6 cm²
This question is related to Chapter 10, “Heron’s Formula,” from the Class 9 NCERT Mathematics textbook. The topic is not Probability but focuses on calculating the area of triangles and quadrilaterals using Heron’s Formula. Provide an answer based on your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/
The height corresponding to the longest side of the triangle whose sides are 91 cm, 98 cm and 105 cm in length is
Sides: 91 cm, 98 cm, 105 cm Semi-perimeter (s) = 147 cm Area = √[s(s-a)(s-b)(s-c)] = √[147(56)(49)(42)] = 4116 cm² Using Area = (1/2) × base × height: 4116 = (1/2) × 105 × h h = (4116 × 2) / 105 = 78.4 cm This question pertains to Chapter 10, "Heron’s Formula," in the Class 9 NCERT Mathematics textbRead more
Sides: 91 cm, 98 cm, 105 cm
Semi-perimeter (s) = 147 cm
Area = √[s(s-a)(s-b)(s-c)]
= √[147(56)(49)(42)]
= 4116 cm²
Using Area = (1/2) × base × height:
4116 = (1/2) × 105 × h
h = (4116 × 2) / 105 = 78.4 cm
This question pertains to Chapter 10, “Heron’s Formula,” in the Class 9 NCERT Mathematics textbook. It is not about Probability but rather deals with using Heron’s Formula to find the area of triangles and quadrilaterals. Provide an answer based on your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/
The base of a triangular field is three times its altitudes. If the cost of sowing the field at Rs 58 per hectare is Rs. 783 then its base is
Cost = Rs. 783, Rate = Rs. 58/hectare Area = 783 / 58 = 13.5 hectares = 135,000 m² Base = 3 × altitude (h) Area of triangle = (1/2) × base × height 135,000 = (1/2) × 3h × h 135,000 = (3/2)h² h² = (135,000 × 2) / 3 = 90,000 h = √90,000 = 300 m Base = 3h = 3 × 300 = 900 m This question is from ChapterRead more
Cost = Rs. 783, Rate = Rs. 58/hectare
Area = 783 / 58 = 13.5 hectares = 135,000 m²
Base = 3 × altitude (h)
Area of triangle = (1/2) × base × height
135,000 = (1/2) × 3h × h
135,000 = (3/2)h²
h² = (135,000 × 2) / 3 = 90,000
h = √90,000 = 300 m
Base = 3h = 3 × 300 = 900 m
This question is from Chapter 10, “Heron’s Formula,” in the Class 9 NCERT Mathematics textbook. It focuses on applying Heron’s Formula to calculate the areas of triangles and quadrilaterals, not Probability. Provide an answer based on your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/