NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 Solutions

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 Solutions are based on the concepts of Inequalities in a triangle.

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Check out NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 Solutions

Exercise Solutions of Class 9 Maths Chapter 7 Triangles

Also check other 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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Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
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CBSE X Related Questions

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)
View Solution
2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals
- \(2.5\) cm
- \(10.5\) cm
- \(3.5\) cm
- \(\frac{14}{3}\) cm
View Solution
3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).
View Solution
4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.
View Solution
5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is
- \(4 \text{ cm}\)
- \(4\sqrt{2} \text{ cm}\)
- \(8 \text{ cm}\)
- \(2\sqrt{2} \text{ cm}\)
View Solution
6.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :
- 0
- 1
- 3
- 2
View Solution

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 Solutions

Exercise Solutions of Class 9 Maths Chapter 7 Triangles

CBSE X Related Questions

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

Similar Mathematics Concepts

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4 Solutions

Exercise Solutions of Class 9 Maths Chapter 7 Triangles

CBSE X Related Questions

1.Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Prove that : \( \frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\text{cosec}^3 \theta}{\text{cosec}^2 \theta - 1} = \sec \theta \cdot \text{cosec} \theta (\sec \theta + \text{cosec} \theta) \)

2.
In the given figure, \(AB \parallel EF\). If \(AB = 24\) cm, \(EF = 36\) cm and \(DA = 7\) cm, then \(AE\) equals

3.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

4.
PQ and PR are two tangents to a circle with centre O and radius 5 cm. AB is another tangent to the circle at C which lies on OP. If \(OP = 13\) cm, then find the length AB and PA.

5.
If \(PQ\) and \(PR\) are tangents to the circle with centre \(O\) and radius \(4 \text{ cm}\) such that \(\angle QPR = 90^{\circ}\), then the length \(OP\) is

6.
The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :