NCERT Solutions for Class 9 Maths Chapter 11: Constructions

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The NCERT Solutions for Class 9 Maths Constructions are provided in this article. Construction deals with how to accurately draw shapes, angles, and lines. Many geometric figures can be precisely constructed using a variety of construction methods.

Class 9 Maths Chapter 11 Constructions belong to Unit 4 Geometry which has a weightage of 28 marks in the Class 9 Maths Examination. NCERT Solutions for Class 9 Maths for Chapter 11 cover the following important concepts:

Download: NCERT Solutions for Class 9 Mathematics Chapter 11 pdf

NCERT Solutions for Class 9 Mathematics Chapter 11

Important Topics in Class 9 Maths Chapter 11 Constructions

Important Topics in Class 9 Maths Chapter 11 Constructions are elaborated below:

2D and 3D Figures

Geometrically, 2D figures, otherwise known as two-dimensional objects, that can be drawn on a plane surface. 3D-shapes, howerver, are called three-dimensional objects or solids that include three dimensions.

Example: Determine the surface area of a cube, assuming that its edge length is 40 cm.

Solution: The edge of cube is given as = 40cm
As per formula, the surface area of a cube = 6a² (a = the edge length)
Hence, the surface area of the cube is = 6 (40) 2 = 9600 sq.cm

Line Segment

Line segments can be defined as points along a line which are found to be bounded by two distinct points.

Example: What are some properties of a Line Segment?

Solution: The properties of a line segment include:

A line has infinite ends which cannot be measured.
However, a line segment has a start point and an endpoint which means it can be measured.
Line segments normally include a defined length, which lead them to form the sides of any polygon.

Formula of Perimeter Shapes

Perimeter, in geometry, can be defined as the length of a closed figure's boundary. The perimeter formula for regular polygons is usually expressed using algebraic equations.

Perimeter of Different Geometric Figures:

Perimeter of a Hexagon: 6a [where a is the side of a Hexagon]
Perimeter of a Parallelogram: 2(b + h) [where b and h are the base and height respectiveley]
Perimeter of a Trapezoid: a + b + c + d [where a, b, c and d are the respective lengths of the Trapezoid]

NCERT Solutions for Class 9 Maths Chapter 11 Exercises:

The detailed solutions for all the NCERT Solutions for Constructions under different exercises are:

Also Read:

Unit Related Topics
Surface Area of a Right Circular Cylinder	Surface Area of a Right Circular Cone	Surface Area of a Sphere
Volume of a Cuboid	Volume of a Right Circular Cone	Volume of a Sphere
Tangent to a circle	Pascal’s Triangle	Equilateral Triangle
Scalene Triangle	Types Of Triangles	Properties of Triangle
Difference between Area and Volume	Quadrilateral Formula	Class 9 Maths Chapter 11 Constructions MCQs

Also Check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

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CBSE X Related Questions

1.
Prove that $\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2\csc \theta$
View Solution
2.
In a trapezium $ABCD$, $AB \parallel DC$ and its diagonals intersect at $O$. Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]
View Solution
3.
Find the zeroes of the polynomial $2x^2 + 7x + 5$ and verify the relationship between its zeroes and coefficients.
View Solution
4.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
View Solution
5.
In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is
- 4
- $\dfrac{\sqrt{15}}{4}$
- $\sqrt{15}$
- $\dfrac{4}{\sqrt{15}}$
View Solution
6.
Directions: In Question Numbers 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R).
Choose the correct option from the following:
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.
Assertion (A): For any two prime numbers $p$ and $q$, their HCF is 1 and LCM is $p + q$.
Reason (R): For any two natural numbers, HCF × LCM = product of numbers.
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of 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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NCERT Solutions for Class 9 Maths Chapter 11: Constructions

NCERT Solutions for Class 9 Mathematics Chapter 11

Important Topics in Class 9 Maths Chapter 11 Constructions

2D and 3D Figures

Line Segment

Formula of Perimeter Shapes

NCERT Solutions for Class 9 Maths Chapter 11 Exercises:

CBSE X Related Questions

1.
Prove that \(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2\csc \theta\)

2.
In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

3.
Find the zeroes of the polynomial \(2x^2 + 7x + 5\) and verify the relationship between its zeroes and coefficients.

4.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

5.
In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 9 Maths Chapter 11: Constructions

NCERT Solutions for Class 9 Mathematics Chapter 11

Important Topics in Class 9 Maths Chapter 11 Constructions

2D and 3D Figures

Line Segment

Formula of Perimeter Shapes

NCERT Solutions for Class 9 Maths Chapter 11 Exercises:

CBSE X Related Questions

1.Prove that \(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2\csc \theta\)

2.In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

3.Find the zeroes of the polynomial \(2x^2 + 7x + 5\) and verify the relationship between its zeroes and coefficients.

4.From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

5.In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Prove that \(\dfrac{\sin \theta}{1 + \cos \theta} + \dfrac{1 + \cos \theta}{\sin \theta} = 2\csc \theta\)

2.
In a trapezium \(ABCD\), \(AB \parallel DC\) and its diagonals intersect at \(O\). Prove that \[ \frac{OA}{OC} = \frac{OB}{OD} \]

3.
Find the zeroes of the polynomial \(2x^2 + 7x + 5\) and verify the relationship between its zeroes and coefficients.

4.
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$

5.
In a right triangle ABC, right-angled at A, if $\sin B = \dfrac{1}{4}$, then the value of $\sec B$ is