NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4 are provided in this article. Class 10 Maths Chapter 6 Triangles is an important chapter included under the Unit Geometry of class 10 maths syllabus. This chapter covers important concepts related to triangles. Exercise 6.4 mainly includes questions based on areas of similar triangles.

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Check below the NCERT solutions pdf for Class 10 Maths Chapter 6 Exercise 6.4

Find below NCERT solutions of other exercises of class 10 maths chapter 6 Triangles:

Class 10 Chapter 6 Triangles Related Links:

Triangle Theorems	Types of Triangles	Triangles Important Questions
Tangent to a Circle	Important Questions on Circles	Circles Revision Notes
Triangles Revision Notes	MCQ on Triangles	Circles

Class 10 Maths Study Guides:

Class 10 Maths Notes	NCERT Solutions for Class 10 Maths	Mathematics Important Questions
Math Study Material	Important Formulas in Maths	Difference Between in Maths

CBSE X Related Questions

1.
If the zeroes of the polynomial $p(x) = 2x^2 - 7x + 6$ are $\alpha$ and $\beta$, then the value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is
- $\frac{7}{2}$
- $\frac{6}{7}$
- $\frac{-7}{6}$
- $\frac{7}{6}$
View Solution
2.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is
- $6$
- $4$
- $\pm 4$
- $3$
View Solution
3.
For an acute angle $\theta$, if $\sin \theta = \frac{1}{9}$, then value of $\frac{9 \csc \theta + 1}{9 \csc \theta - 1}$ is
- $0$
- $\frac{80}{81}$
- $1$
- $\frac{82}{80}$
View Solution
4.
If roots of the quadratic equation $x^2 - k\sqrt{3}x + 2 = 0$ are real and equal, then value of k is
- $-2$
- $\sqrt{\frac{8}{3}}$
- $1$
- $2$
View Solution
5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is
- $55^\circ$
- $20^\circ$
- $35^\circ$
- $70^\circ$
View Solution
6.
Assertion (A) : H.C.F. $(36 m^{2}, 18 m) = 18 m$, where $m$ is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
View Solution

View All

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Area of a Triangle: Formula, Types & Examples

Triangles Revision Notes

Triangles Important Questions: Important Similarity Concepts

Congruence Of Triangles: Conditions (SSS, SAS, ASA, RHS)

Isosceles Triangle Theorems: Proof and Sample Questions

Triangle Theorems: Properties, Types & Similarity Theorems

NCERT Solutions for Class 10 Maths Chapter 6: Triangles

NCERT Solutions for Class 9 Maths Chapter 7: Triangles

Triangles: Types, Properties, Heron’s Formula and Examples

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Circles Revision Notes: Theorems of Tangent to the Circle

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

CBSE X Related Questions

1.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

2.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles Exercise 6.4

CBSE X Related Questions

1.If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

2.In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

3.For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

5.In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number. Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If the zeroes of the polynomial \(p(x) = 2x^2 - 7x + 6\) are \(\alpha\) and \(\beta\), then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is

2.
In an A.P., if $a_{14} - a_8 = 24$, then the common difference of the A.P. is

3.
For an acute angle \(\theta\), if \(\sin \theta = \frac{1}{9}\), then value of \(\frac{9 \csc \theta + 1}{9 \csc \theta - 1}\) is

4.
If roots of the quadratic equation \(x^2 - k\sqrt{3}x + 2 = 0\) are real and equal, then value of k is

5.
In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

6.
Assertion (A) : H.C.F. \((36 m^{2}, 18 m) = 18 m\), where \(m\) is a prime number.
Reason (R) : H.C.F. of two numbers is always less than or equal to the smaller number.