NCERT Solutions for Class 7 Mathematics Chapter 10: Practical Geometry

Content Curator

NCERT Solutions for Class 7 Mathematics Chapter 10 Practical Geometry are provided in the article below. Practical geometry is an important branch of geometry which deals with the study of the size, positions, shapes as well as dimensions of objects. Whether you have to draw a line segment or measure it, draw a circle or arcs, draw an angle, etc. it can easily be possible with the help of geometrical tools. Some of the important topics in Practical Geometry chapter are:

Download: NCERT Solutions for Class 7 Mathematics Chapter 10 pdf

NCERT Solutions for Class 7 Mathematics Chapter 10

NCERT Solutions for Class 7 Mathematics Chapter 10 Practical Geometry is given below.

Class 7 Maths Chapter 10 Practical Geometry – Important Topics

Practical Geometry: It is the practical study of constructing geometrical figures.

Perpendicular Bisector: It is the line that is constructed by the use of angles to bisect a line segment into two equal halves.

Angle Bisector: It is a line that is constructed to bisect an angle into two equal halves.

Example: Make a line AB and place a point P on the outside of it. Draw a line CD that runs parallel to AB and passes through point P.

Ans.

Draw an AB line.
Join PQ with a point Q on AB and a point P outside AB.
Draw on the arc to cut AB at X and PQ at Z with Q as the centre and any radius.
Draw an arc cutting QP at Y with P as the centre and the same radius.
Draw an arc to cut the preceding arc at E, with Y as the centre and the radius equal to XZ.
To get the desired line, join PE and produce it on both sides.

NCERT Solutions for Class 7 Maths Chapter 10 Exercises

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercises are given below.

Read More:

Construction of Rhombus	Perpendicular Lines	ASA Triangles Construction
SAS Triangles Construction	RHS Triangles Construction	Perimeter two angles constructing triangles

Class 7 Maths Guides:

NCERT Solutions for Class 7 Maths	Number System	Geometry
Arithmetics	Basic Maths	Maths Formulas

CBSE X Related Questions

1.
If \( 2 \sin A = 1 \), then the value of \( \tan A + \cot A \) is :
- \( \sqrt{3} \)
- \( \frac{4}{\sqrt{3}} \)
- \( \frac{\sqrt{3}}{2} \)
- \( 1 \)
View Solution
2.
In the given figure, PQ is a tangent to a circle with centre \(O(-5, 3)\). If coordinates of P and Q are \((3, 1)\) and \((0, 6)\) respectively, then using distance formula, show that \(PQ \perp OQ\).
View Solution
3.
Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories. Based on above information, answer the following questions:

38(i) Represent the above situation in terms of a pair of linear equations in two variables.
View Solution
4.
In the given figure, two triangles ABC and PQR are shown such that \(\angle A = \angle P\) and \(\angle C = \angle R\). If \(AD \perp BC\) and \(PS \perp QR\), then prove that (i) \(\Delta ADB \sim \Delta PSQ\) (ii) \(AD \times QS = BD \times PS\).
View Solution
5.
In the given figure, \(\Delta ABC\) is a right-angled triangle with \(\angle A = 90^\circ\). AD is perpendicular to BC.

35(a)(i) Prove that \(\Delta DBA \sim \Delta DAC\)
View Solution
6.
There are many varieties of mushrooms available in the world. One such mushroom ‘Amanita muscaria’ has a upper part which is like red cap (hemispherical) and lower part is like white stem (cylindrical). The hemispherical cap’s radius = 3 cm and cylindrical stem is 2 cm high with diameter 1.4 cm. Considering mushroom a solid object, answer the following questions:

36(i) What is the total height of a mushroom ?
View Solution