NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 Solutions

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 Solutions are based on the properties of triangles, including two primary theorems.

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Check out NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 Solutions

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Exercise Solutions of Class 9 Maths Chapter 7 Triangles

Also check other Exercise Solutions of Class 9 Maths Chapter 7 Triangles

Also Check:

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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CBSE X Related Questions

1.
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
2.
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
3.
The natural number 1 is :
- a prime number.
- a composite number.
- prime as well as composite.
- neither prime nor composite.
View Solution
4.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
5.
If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).
View Solution
6.
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

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