NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2 Solutions are based on Construction of Tangents to a Circle.

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Exercise Solutions of Class 10 Maths Chapter 11 Constructions

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NCERT Solutions For Class 10 Maths Chapter 11 Exercise 11.1 Solutions

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Chapter Related Topics
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CBSE X Related Questions

1.
Prove that :
$\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta$.
View Solution
2.
In the figure given above, $\triangle ABC \sim \triangle XYZ$, then find the values of $x$ and $y$.
View Solution
3.
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
4.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals
- $x^2 + 5x - 4$
- $(x + 3) (-x + 8)$
- $a(x^2 + 5x - 24)$
- $x^2 - 24$
View Solution
5.
$ABCD$ 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
6.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :
- $q - 9p$
- $p - 9q$
- $p + 9q$
- $2p + 9q$
View Solution

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

Exercise Solutions of Class 10 Maths Chapter 11 Constructions

CBSE X Related Questions

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

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
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.

4.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

6.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.2

Exercise Solutions of Class 10 Maths Chapter 11 Constructions

CBSE X Related Questions

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

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

3.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.

4.If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

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

6.The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

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

2.
In the figure given above, \(\triangle ABC \sim \triangle XYZ\), then find the values of \(x\) and \(y\).

3.
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.

4.
If the zeroes of a polynomial p(x) are $-3$ and 8, then p(x) equals

5.
\(ABCD\) is a parallelogram such that \(AF = 7 \text{ cm}\), \(FB = 3 \text{ cm}\) and \(EF = 4 \text{ cm}\), length \(FD\) equals

6.
The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :