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

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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.
If 14th term of an A.P. is 4 and its 15th term is zero, then its first term is
- –48
- –56
- 56
- 48
View Solution
2.
In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).
View Solution
3.
\(17 \times 11 \times 13 + 11\) is
- a prime number.
- multiple of 17.
- a composite number.
- an odd number.
View Solution
4.
In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC.

35(a)(i) Prove that \(\Delta DBA \sim \Delta DAC\)
View Solution
5.
If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :
- \( \sqrt{3} \)
- \( \frac{4}{\sqrt{3}} \)
- \( \frac{\sqrt{3}}{2} \)
- \( 1 \)
View Solution
6.
In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).
View Solution

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