# NCERT Solutions for Class 7 Maths Chapter 10

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.1, Exercise 10.2, Exercise 10.3, Exercise 10.4, Exercise 10.5 and Miscellaneous Exercise in English Medium free to download in PDF form.

NCERT Solutions 2020-21 are now in updated form following the latest NCERT Books 2020-2021. Videos related to each exercise and NCERT Solutions Offline Apps are also given to free download without any login or registration.

## NCERT Solutions for Class 7 Maths Chapter 10

 Class: 7 Subject: Maths – गणित Chapter 10: Practical Geometry

### 7 Maths Chapter 10 Solutions

Class 7 Maths Chapter 10 all Exercises of Practical Geometry is given below. All the constructions are done step by step. Steps of constructions are also written to understand properly. NCERT Solutions 2020-21 are updated for the this academic session 2020-2021.

• ### 7 Maths Chapter 10 All Exercises in English Medium

#### About NCERT Solutions for Class 7 Maths Chapter 10

In 7 Maths Chapter 10 Practical Geometry, we will learn the construction of a line parallel to a given line, through a point not on the line following the properties regarding the transversal and parallel lines and using ruler and compasses only. During the constructions of a triangle, we should go through the following rules:
1. The exterior angle of a triangle is equal in measure to the sum of interior opposite angles.
2. Sum of the lengths of any two sides of a triangle is greater than the length of the third side.
3. In any right-angled triangle, the square of the length of
4. Hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
5. The total measure of the three angles of a triangle is 180°.

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### Important Questions on Class 7 Maths Chapter 10

Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
To construct: A line, parallel to given line by using ruler and compasses.
Steps of construction:
Draw a line-segment AB and take a point C outside AB.
Take any point D on AB and join C to D.
With D as centre and take convenient radius, draw an arc cutting AB at E and CD at F.
With C as centre and same radius as in step 3, draw an arc GH cutting CD at I.
With the same arc EF, draw the equal arc cutting GH at J.
Join JC to draw a line l.
This the required line AB∥l.
Construct XYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
To construct: XYZ, where XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.
Steps of construction:
(a) Draw a line segment YZ = 5 cm.
(b) Taking Z as centre and radius 6 cm, draw an arc.
(c) Similarly, taking Y as centre and radius 4.5 cm, draw another arc which intersects first arc at point X.
(d) Join XY and XZ.
It is the required XYZ.
Construct DEF such that DE = 5 cm, DF = 3 cm and EDF = 90.
To construct: DEF where DE = 5 cm, DF = 3 cm and EDF = 90
Steps of construction:
(a) Draw a line segment DF = 3 cm.
(b) At point D, draw an angle of 90 with the help of compass i.e., XDF = 90.
(c) Taking D as centre, draw an arc of radius 5 cm, which cuts DX at the point E.
(d) Join EF.
(e) It is the required right angled triangle DEF.
Examine whether you can construct DEF such that EF = 7.2 cm, E = 110 and F = 80. Justify your answer.
Given: In DEF, E = and F =
Using angle sum property of triangle, ∠D + ∠E + ∠F= 180°
⟹∠D+〖110° +〖80〗° =〖180〗°
⟹∠D + 190° = 180°
⟹∠D = 180° – 190° = -10°
Which is not possible.       