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The height corresponding to the longest side of the triangle whose sides are 42 cm, 34 cm and 20 cm in length is
Longest side = 42 cm Semi-perimeter (s) = 48 cm Area = √[s(s-a)(s-b)(s-c)] = √[48(6)(14)(28)] = 336 cm² Using Area = (1/2) × base × height: 336 = (1/2) × 42 × h h = 336 / 21 = 16 cm This question related to Chapter 10 Mathematics Class 9th NCERT. From the Chapter 10 Heron’s Formula. Probability. GivRead more
Longest side = 42 cm
Semi-perimeter (s) = 48 cm
Area = √[s(s-a)(s-b)(s-c)]
= √[48(6)(14)(28)]
= 336 cm²
Using Area = (1/2) × base × height:
336 = (1/2) × 42 × h
h = 336 / 21 = 16 cm
This question related to Chapter 10 Mathematics Class 9th NCERT. From the Chapter 10 Heron’s Formula. Probability. Give answer according to your understanding.
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Find the area of triangle two sides of which are 18cm and 10ccm and the perimeter is 42 cm.
Using Heron's formula: Sides: a = 18 cm, b = 10 cm, c = 14 cm Semi-perimeter (s) = 21 cm Area = √[s(s-a)(s-b)(s-c)] = √[21(21-18)(21-10)(21-14)] = √[21 × 3 × 11 × 7] = 21√11 cm² This question is connected to Chapter 10, "Heron’s Formula," in the Class 9 NCERT Mathematics textbook. It focuses on applRead more
Using Heron’s formula:
Sides: a = 18 cm, b = 10 cm, c = 14 cm
Semi-perimeter (s) = 21 cm
Area = √[s(s-a)(s-b)(s-c)]
= √[21(21-18)(21-10)(21-14)]
= √[21 × 3 × 11 × 7]
= 21√11 cm²
This question is connected to 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 your response based on the concepts from this chapter.
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Find the area of a triangle, two sides of which are 8 cm and 11 cm and the perimeter is 32 cm
Sides: 8 cm, 11 cm, 13 cm Perimeter = 32 cm → Semi-perimeter (s) = 16 cm Area = √[s(s-a)(s-b)(s-c)] = √[16(16-8)(16-11)(16-13)] = √[16 × 8 × 5 × 3] = √1920 = 8√30 cm² This question is based on Chapter 10, "Heron’s Formula," from the Class 9 NCERT Mathematics textbook. It involves using Heron’s FormuRead more
Sides: 8 cm, 11 cm, 13 cm
Perimeter = 32 cm → Semi-perimeter (s) = 16 cm
Area = √[s(s-a)(s-b)(s-c)]
= √[16(16-8)(16-11)(16-13)]
= √[16 × 8 × 5 × 3]
= √1920
= 8√30 cm²
This question is based on Chapter 10, “Heron’s Formula,” from the Class 9 NCERT Mathematics textbook. It involves using Heron’s Formula to determine the areas of triangles and quadrilaterals, not Probability. Answer according to your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/
The sides of a triangular plot are in the ratio of 3 : 5 : 7 and its perimeter is 300 m. Find its area.
The sides of the triangular plot are in the ratio 3:5:7, and the perimeter is 300 m. Let the sides be 3x, 5x, and 7x. From the perimeter: 3x + 5x + 7x = 300 → 15x = 300 → x = 20. Thus, the sides are: 3x = 60 m, 5x = 100 m, 7x = 140 m. The semi-perimeter (s) is: s = Perimeter / 2 = 300 / 2 = 150 m. URead more
The sides of the triangular plot are in the ratio 3:5:7, and the perimeter is 300 m. Let the sides be 3x, 5x, and 7x.
From the perimeter:
3x + 5x + 7x = 300 → 15x = 300 → x = 20.
Thus, the sides are:
3x = 60 m, 5x = 100 m, 7x = 140 m.
The semi-perimeter (s) is:
s = Perimeter / 2 = 300 / 2 = 150 m.
Using Heron’s formula for the area of a triangle:
Area = √[s(s-a)(s-b)(s-c)],
where a = 60 m, b = 100 m, c = 140 m.
Substitute the values:
Area = √[150(150-60)(150-100)(150-140)]
= √[150 × 90 × 50 × 10]
= √6750000.
Simplify the square root:
√6750000 = √(2² × 3³ × 5⁶) = 2 × 3¹.⁵ × 5³ = 12√30.
Thus, the area of the triangle is 12√30 m².
This question related to Chapter 10 Mathematics Class 9th NCERT. From the Chapter 10 Heron’s Formula. Probability. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/
What is the importance of Class 9 Mathematics Chapter 9 MCQ?
The importance of Class 9 Mathematics Chapter 9 (Circles) MCQs lies in their ability to enhance conceptual clarity, improve exam preparedness, and develop problem-solving skills. These questions test a student's understanding of key concepts like theorems, properties of circles, and their applicatioRead more
The importance of Class 9 Mathematics Chapter 9 (Circles) MCQs lies in their ability to enhance conceptual clarity, improve exam preparedness, and develop problem-solving skills. These questions test a student’s understanding of key concepts like theorems, properties of circles, and their applications, while also helping identify weak areas for improvement. Practicing MCQs boosts confidence by familiarizing students with the exam pattern and improving time management skills. Additionally, mastering this chapter through MCQs builds a strong foundation for advanced geometry topics in higher classes, making it an essential part of effective learning and preparation.
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