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.
On the day of her examination, Riya sharpened her pencil from both ends as shown below.

The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
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2.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
- 8 cm
- $4\sqrt{5}$ cm
- $2\sqrt{5}$ cm
- 12 cm
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3.
The system of equations $2x + 1 = 0$ and $3y - 5 = 0$ has
- unique solution
- two solutions
- no solution
- infinite number of solutions
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4.
OAB is sector of a circle with centre O and radius 7 cm. If length of arc $ \widehat{AB} = \frac{22}{3} $ cm, then $ \angle AOB $ is equal to
- $ \left(\frac{120}{7}\right)^\circ $
- $ 45^\circ $
- $ 60^\circ $
- $ 30^\circ $
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5.
Prove that: \[ \frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0 \]
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6.
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.
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