NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1 Solutions are based on Division of a Line Segment and Solved examples based on the concept.

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Exercise Solutions of Class 10 Maths Chapter 11 Constructions

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CBSE X Related Questions

1.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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2.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
  \(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
  \(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

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3.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

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4.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

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5.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

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6.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1

Exercise Solutions of Class 10 Maths Chapter 11 Constructions

CBSE X Related Questions

1.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

4.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

5.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

Similar Mathematics Concepts

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NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1

Exercise Solutions of Class 10 Maths Chapter 11 Constructions

CBSE X Related Questions

1. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

4. Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

5. If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sin2 A or not

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

3.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

4.
Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

5.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not