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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.
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The base of an isosceles triangle is 24 cm and its area is 192 cm², then its perimeter is
Base = 24 cm, Area = 192 cm² Height = (2 × Area) / Base = (2 × 192) / 24 = 16 cm Using Pythagorean theorem: x² = (24/2)² + 16² = 12² + 16² = 144 + 256 = 400 x = √400 = 20 cm Perimeter = 2x + Base = 2(20) + 24 = 64 cm This question is based on Chapter 10, "Heron’s Formula," from the Class 9 NCERT MatRead more
Base = 24 cm, Area = 192 cm²
Height = (2 × Area) / Base = (2 × 192) / 24 = 16 cm
Using Pythagorean theorem:
x² = (24/2)² + 16² = 12² + 16² = 144 + 256 = 400
x = √400 = 20 cm
Perimeter = 2x + Base = 2(20) + 24 = 64 cm
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 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.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-9/maths/
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/