NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.1 is given in this article. Class 10 Maths Chapter 7 is an important chapter and its Exercise 7.1 has about 10 questions related to the distance formula.

Read More: Coordinate Geometry MCQs

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Check out the solutions of Class 10 Maths NCERT solutions chapter 7 Coordinate Geometry Exercise 7.1

Check out other exercise solutions of Class 10 Maths Chapter 7

Class 10 Chapter 7 Topics:

Section Formula	Cartesian System	Ordinate
Quadrant	Centroid	Midpoint Formula
Centroid Formula	Centroid of a Triangle	Point of Intersection Formula

CBSE Class 10 Maths Study Guides:

Math Formula	Math Study Notes	Trigonometry
Math MCQs	NCERT Solutions for Class 10 Maths	Difference between in Maths
Arithmetic	Mensuration	Calculus
Class 10 Notes	Number System	Geometry

CBSE X Related Questions

1.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).
View Solution
2.
The HCF of 960 and 432 is :
- 48
- 54
- 72
- 36
View Solution
3.
A kite is flying at a height of \(60 \text{ m}\) above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as \(30^{\circ}\). From the bottom of the same building, the angle of elevation of kite is \(45^{\circ}\). Find the length of the string and height of roof from the ground. (Use \(\sqrt{3} = 1.73\))
View Solution
4.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.
View Solution
5.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).
View Solution
6.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals
- \(\frac{21}{4} \text{ cm}\)
- \(\frac{28}{3} \text{ cm}\)
- \(\frac{12}{7} \text{ cm}\)
- \(5.5 \text{ cm}\)
View Solution

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NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

CBSE X Related Questions

1.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

2.
The HCF of 960 and 432 is :

4.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

5.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.1

CBSE X Related Questions

1.If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

2.The HCF of 960 and 432 is :

4.In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

5.Prove that : \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If \(\alpha, \beta\) are the zeroes of the polynomial \(p(x) = x^2 - 3x - 1\), then find the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).

2.
The HCF of 960 and 432 is :

4.
In the given figure, \(AB \parallel DE\) and \(AC \parallel DF\). Show that \(\triangle ABC \sim \triangle DEF\). If \(BC = 10\) cm, \(EB = CF = 5\) cm and \(AB = 7\) cm, then find the length DE.

5.
Prove that :
\(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta\).

6.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals