NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 Solutions

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 Solutions are based on the concepts of SSS congruence rules and RHS congruence rules.

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Exercise Solutions of Class 9 Maths Chapter 7 Triangles

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Important Chapter Related Topics
Mid Point Theorem	Isosceles Triangle	Congruence of triangles
Class 9 Maths Chapter 7 Triangles MCQs	Properties of inequalities and triangles	Sides of a Triangle
Similar Figures	Triangles

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Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . . .
(ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
(iv) – 10, – 6, – 2, 2, . . .
(v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
(vi) 0.2, 0.22, 0.222, 0.2222, . . . .
(vii) 0, – 4, – 8, –12, . . . .
(viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
(ix) 1, 3, 9, 27, . . . .
(x) a, 2a, 3a, 4a, . . . .
(xi) a, \(a^2, a^3, a^4,\) . . . .
(xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
(xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
(xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
(xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

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2.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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3.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm)
Number of leaves
118 - 126
3
127 - 135
5
136 - 144
9
145 - 153
12
154 - 162
5
163 - 171
4
172 - 180
2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

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4.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
  \(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
  \(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

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5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

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6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 7 Triangles

CBSE X Related Questions

2.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 Solutions

Exercise Solutions of Class 9 Maths Chapter 7 Triangles

CBSE X Related Questions

2. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6. If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sin2 A or not

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

6.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not