NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry - Download PDF

Edited By Ravindra Pindel | Updated on Feb 07, 2024 06:08 PM IST

NCERT Practical Geometry class 7 - The word geometry comprises of two greek words geo and metron. Geo means earth and metron means measurements. Geometry is applied in everyday life from designing simple things like your furniture, doors and windows, the wall clock to the design of complex structures like the designing of a new house. To design all the above structures you need to know about geometry. In CBSE NCERT solutions for Class 7 Maths chapter 10 Practical Geometry, you will get questions related to applications of geometry in real life. You will study a few rulers and compass constructions in this chapter. You will learn how to draw a parallel line to a given line through a point outside the line, how to construct triangles if three sides of triangle are given (SSS), how to construct a triangle if measure of two sides and the angle between them are given (SAS), how to draw a triangle if value of two angles and the length of side included between them are given (ASA), how to draw a triangle if length of hypotenuse and the measure of one leg of a right-angled triangle are given (RHS). There are practice questions given on all the above topics in the NCERT .

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

This Story also Contains

NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th- Important Formulae
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry - Important Points
NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (Intext Questions and Exercise)
NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th Exercise 10.1
NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th Exercise 10.2
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.5
Practical Geometry Class 7 Maths Chapter 10-Topics
Some applications of geometry

You should try to solve all these questions by yourself. You can take help from NCERT Solutions for Class 7 . While solving the problems learn the steps and draw the figures. Just only going through the steps of drawings may not help in the exams. If you have tried to draw all the figures mentioned in the solutions of NCERT for Class 7 Maths chapter 10 Practical Geometry it will be easy for you solve all the questions in the exam related to this chapter. You can get NCERT Solutions from Classes 6 to 12 for Science and Maths by clicking on the above link. Here you will get solutions to five exercises of this chapter.

NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th- Important Formulae

Criteria for Congruence of Triangles:

SSS Congruence: For triangles ABC and DEF, If AB = DE, BC = EF, and AC = DF, Then △ABC ≅ △DEF.
SAS Congruence: For triangles ABC and DEF, If AB = DE, ∠BAC = ∠EDF, and BC = EF, Then △ABC ≅ △DEF.
ASA Congruence: For triangles ABC and DEF, If ∠BAC = ∠EDF, ∠ABC = ∠DEF, and AC = DF, Then △ABC ≅ △DEF.
AAS Congruence: For triangles ABC and DEF, If ∠BAC = ∠EDF, ∠ACB = ∠EFD, and BC = EF, Then △ABC ≅ △DEF.
RHS Congruence: For right-angled triangles ABC and DEF, If ∠CAB = ∠FDE, AC = DF, and BC = EF, Then △ABC ≅ △DEF.

For more, Download Ebook - NCERT Class 7 Maths: Chapterwise Important Formulas And Points

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry - Important Points

Five Measurements Determine a Quadrilateral:

A quadrilateral can be uniquely determined when five measurements related to its sides and angles are known.

Construction with Four Sides and a Diagonal:

If the lengths of all four sides and one diagonal of a quadrilateral are given, the quadrilateral can be constructed uniquely.

Construction with Diagonals and Three Sides:

When the lengths of both diagonals and three sides of a quadrilateral are known, the quadrilateral can be constructed uniquely.

Construction with Adjacent Sides and Angles:

If the lengths of two adjacent sides and the measures of three angles of a quadrilateral are provided, the quadrilateral can be constructed uniquely.

Construction with Three Sides and Included Angles:

If the lengths of three sides and the measures of two angles included between those sides are given, the quadrilateral can be constructed uniquely.

Free download NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry PDF for CBSE Exam.

NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th

Download PDF

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry (Intext Questions and Exercise)

NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th Exercise 10.1

Question:1 Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

Answer:

1. Draw the line AB

1643173693412

2. Mark any point C outside AB and a point D on AB

3. Join CD

1643173716651

4. Draw an arc with convenient radius with D as centre cutting AB at E and DC at F

5. With the same radius as in step 4 draw an arc IGH with C as a centre and cutting line AC at G.

1643173732111

6 Place one tip of the compass on E and adjust the opening so that pencil tip touches at F.

7. With the same opening in step 6 and G as centre cut an arc at J on IGH.

8. Join CJ and extend the line. This line will be parallel to AB

1643173752124

Question:2 Draw a line $l$ . Draw a perpendicular to $l$ at any point on $l$ . On this perpendicular choose a point X, 4 cm away from l. Through X, draw a line m parallel to $l$ .

Answer:

1. Draw a line l. Mark a point P on the line

2. With p as a centre and with any radius draw an arc which touches the line at A and B

1643173776082

3. Take a length more than AP with A as centre draw an arc. with the same length on compass and B as centre draw an arc to cut the previous arc at Q.

4. Join PQ. PQ is perpendicular to the line l

1643173796267

5. Take 4 cm on the compass and with P as centre cut an arc on line PQ at C. The length PC=4 cm

1643173824705

6. with P as a centre and any radius, draw an arc to cut line l at D and PQ at E.

7. with the same radius as in step 6 draw an arc GF with centre C so that it cuts the line PQ at G.

1643173845847

8. Take the length DE on the compass and with G as centre draw an arc to cut FG at J

9. Join CJ and extend the line to name it as line m. Now the line l is parallel to line m

1643173865978

Question:3 Let $l$ be a line and P be a point not on $l$ . Through P, draw a line m parallel to $l$ . Now join P to any point Q on $l$ . Choose any other point R on m. Through R, draw a line parallel to PQ. Let this meet $l$ at S. What shape do the two sets of parallel lines enclose?

Answer:

1643173890497

1. Draw the line PQ parallel to l using the steps to draw parallel lines

2. Draw the line RS parallel to PQ using the steps to draw parallel lines

the two parallel lines make the shape of a parallelogram PQRS.

1643173923370

NCERT Solutions for Maths Chapter 10 Practical Geometry Class 7th Exercise 10.2

Question:1 Construct $\Delta XYZ$ in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm.

Answer:

$\Delta XYZ$ in which XY= 4.5 cm, YZ = 5 cm and ZX = 6 cm :

1. Draw a line XY = 4.5 cm

2. with x as centre and length= 6cm draw an arc

3. with y as centre and length = 5cm draw another arc to cut the arc drawn in step 2 at Z

4. Join XZ and YZ. XYZ is the required triangle

1643173953089

Question:2 Construct an equilateral triangle of side 5.5 cm.

Answer:

an equilateral triangle of side 5.5 cm :

1. Draw a line AB = 5.5 cm

2. with A as centre and length= 5.5cm draw an arc

3. with B as centre and length = 5.5cm draw another arc to cut the arc drawn in step 2 at C

4. Join AC and BC.ABC is the required triangle

1643174012796

Question:3 Draw ?PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of triangle is this?

Answer:

PQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm:

1.Draw a line QR = 3.5 cm

2. With Q as centre and length = 4 cm draw an arc

3. With R as centre and length = 4cm draw another arc to cut the previous arc at P

4. Join QP and QR. Then PQR is the required triangle

1643174049602 1643174083055

As we can see two sides of the triangle are same so The triangle is an isosceles triangle.

Question:4 Construct $\Delta ABC$ such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure $\angle B$ .

Answer:

$\Delta ABC$ such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure $\angle B$ :

1. Draw a line AB = 2.5 cm

2. with A as centre and length= 6.5cm draw an arc

3. with B as centre and length = 6cm draw another arc to cut the arc drawn in step 2 at C

4. Join AC and BC. ABC is the required triangle

1643174148746 1643174176580

On measuring angle B comes out to be 90 degrees.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.3

Question:1 Construct $\Delta DEF$ such that DE = 5 cm, DF = 3 cm and $m\angle EDF=90^{\circ}$ .

Answer:

$\Delta DEF$ such that DE = 5 cm, DF = 3 cm and $m\angle EDF=90^{\circ}$ :

1. Draw a line DE = 5cm

2. Draw a line DX making 90 degrees with DE.

3. cut the length DF =3cm using the compass on the line DX

4. Join E and F

DEF is the required triangle

1643174218987

Question:2 Construct an isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is $110^{\circ}$ .

Answer:

1. Draw a line DE = 6.5 cm

2. Draw a line DX making 110 degrees with DE.

3. cut the length DF =6.5 cm using the compass on the line DX

4. Join E and F

DEF is the required triangle

1643174261746

Question:3 Construct $\Delta ABC$ with BC = 7.5 cm, AC = 5 cm and $m\angle C=60^{\circ}$ .

Answer:

$\Delta ABC$ with BC = 7.5 cm, AC = 5 cm and $m\angle C=60^{\circ}$ :

1. Draw a line AC = 5 cm

2. Draw a line CX making 60 degrees with AC from C.

3. cut the length BC =7.5 cm using the compass on line CX

4. Join A and B

ABC is the required triangle

1643174300070

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.4

Question:1 Construct $\Delta ABC$ , given $m\angle A=60^{\circ}$ , $m\angle B=30^{\circ}$ and AB = 5.8 cm.

Answer:

$\Delta ABC$ , given $m\angle A=60^{\circ}$ , $m\angle B=30^{\circ}$ and AB = 5.8 cm :

1. Draw the line AB=5.8 cm

2. At A making an angle 60 degrees with AB draw a line AX

3. At B making an angle 30 degrees with AB draw a line BY

4. The lines AX and BY intersect at C ( in case the drawn line AX or BY is small extend it to intersect each other)

ABC is the requires triangle

1643174361140

Question:2 Construct $\Delta PQR$ if PQ = 5 cm, $m\angle PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$ .

Answer:

$\Delta PQR$ if PQ = 5 cm, $m\angle PQR = 105^{\circ}$ and $m\angle QRP = 40^{\circ}$ :

the sum of angles of a triangle is 180 degrees. Given angles PQR and QRP, so the angle QPR= 180-(105+40)=35 degrees

1. Draw the line PQ=5 cm

2. At Q making an angle 105 degrees with PQ draw a line QY

3. At P making an angle 35 degrees with PQ draw a line PX

4. The lines PX and QY intersect at R ( in case the drawn line PX or QY is small extend it to intersect each other)

PQR is the required triangle

1643174404066

Question:3 Examine whether you can construct $\Delta DEF$ such that EF = 7.2 cm, $m\angle E = 110^{\circ}$ and $m\angle F = 80^{\circ}$ . Justify your answer.

Answer:

No, we cannot construct such that EF = 7.2 cm, $m\angle E = 110^{\circ}$ and $m\angle F = 80^{\circ}$ . This is because of the property of the triangle of having a sum of internal angles equal to 180 degrees. if we have $m\angle E = 110^{\circ}$ , then we cannot have $m\angle F = 80^{\circ}$ because then the sum would exceed 180 degree which is impossible.

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry Exercise 10.5

Question:1 Construct the right angled $\Delta PQR$ , where $\angle Q=90^{\circ}$ , QR = 8cm and PR = 10 cm.

Answer:

the right-angled $\Delta PQR$ , where $\angle Q=90^{\circ}$ , QR = 8cm and PR = 10 cm:

1. Draw QR = 8 cm

2. Draw a perpendicular QX to QR at Q

3. From R and with length = 10 cm cut an arc on QX. The arc meet QX at P

4. Join PR

PQR is the required triangle

1643174457997

Question:2 Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

Answer:

a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long:

1. Draw QR = 4 cm

2. Draw a perpendicular QX to QR at Q

3. From R and with length = 6 cm cut an arc on QX. The arc meet QX at P

4. Join PR

Triangle PQR is the required triangle

1643174504847

Question:3 Construct an isosceles right-angled triangle ABC, where $m\angle ACB=90^{\circ}$ and AC = 6 cm.

Answer:

1. Draw CA = 6 cm

2. Draw a perpendicular CX to CA at C

3. From C and with length = 6 cm cut an arc on CX. The arc meet CX at B

4. Join BA

Triangle ABC is the required triangle

Practical Geometry Class 7 Maths Chapter 10-Topics

Construction Of A Line Parallel To A Given Line, Through A Point Not On The Line
Construction Of Triangles
Constructing A Triangle When The Lengths Of Its Three Sides Are Known (Sss Criterion)
Constructing A Triangle When The Lengths Of Two Sides And The Measure Of The Angle Between Them Are Known. (Sas Criterion)
Constructing A Triangle When The Measures Of Two Of Its Angles And The Length Of The Side Included Between Them Is Given. (Asa Criterion)
Constructing A Right-angled Triangle When The Length Of One Leg And Its Hypotenuse Are Given (Rhs Criterion )

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.	Chapter Name
Chapter 1	Integers
Chapter 2	Fractions and Decimals
Chapter 3	Data Handling
Chapter 4	Simple Equations
Chapter 5	Lines and Angles
Chapter 6	The Triangle and its Properties
Chapter 7	Congruence of Triangles
Chapter 8	Comparing quantities
Chapter 9	Rational Numbers
Chapter 10	Practical Geometry
Chapter 11	Perimeter and Area
Chapter 12	Algebraic Expressions
Chapter 13	Exponents and Powers
Chapter 14	Symmetry
Chapter 15	Visualising Solid Shapes

NCERT Solutions for Class 7 Subject Wise

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Science

Some applications of geometry

Construction works:- For the construction of houses, bridges, buildings, roads, tanks, dams, etc you need to have knowledge of geometry otherwise you cant do these things with better accuracy. In NCERT solutions for class 7 maths chapter 10 practical geometry, you have learnt some basics of geometry which will help you in higher study.
Games:- In all the computer games or mobile games, you will see particular patterns, maps, 2D and 3D images etc. To designing all these thing you need to have knowledge of geometry.
Art:- Artist uses the geometry to organise the arrangement of space in a picture.
Architecture:- Architect use knowledge of geometry to transform the build the material of architectural design.
Sports:- From designing the pitch, the boundary in cricket to draw the lines in tennis court you need to have knowledge of geometry.

In CBSE NCERT solutions for Class 7 Maths chapter 10 Practical Geometry, you have learnt some fundamental of geometry for simple mathematical geometry. There are many other applications of geometry in the field of designing, civil engineering, CAD(computer-aided design), mapping, GPS and many more which you will learn in higher classes.

Happy Reading!!!

Also Check NCERT Books and NCERT Syllabus here:

Also Read

NCERT Solutions for Class 7 Science Chapter 18 Wastewater Story

NCERT Solutions for Class 7 Science Chapter 17 Forests Our Lifeline

NCERT Solutions for Class 7 Science Chapter 16 Water A Precious Resource

NCERT Solutions for Class 7 Science Chapter 14 Electric Current and Its Effects

NCERT Solutions for Class 7 Science Chapter 13 Motion and Time

NCERT Solutions for Class 7 Science Chapter 12 Reproduction in Plants

Articles

How to Make APAAR ID via DigiLocker and School: A Complete Guide

Dec 17, 2024

Sainik School Entrance Exam 2025: Application, Exam Date, How to Apply, Syllabus

Dec 17, 2024

MVPP Scholarship Exam 2024: Mukhyamantri Vigyan Pratibha Pariksha Registration, Exam Date

Dec 16, 2024

Silverzone ISSO Answer Key 2024-25: Download PDF

Dec 16, 2024

Silverzone IOEL Answer Key 2024 – Download International Olympiad of English Language Solutions

Dec 16, 2024

Silverzone Olympiad IOM Answer Key 2024-25: Qualifying Marks, Score

Dec 16, 2024

Silverzone STEM Olympiad Answer Key 2024-25: Check Answer Key for all Classes

Dec 16, 2024

Silverzone ABHO Answer Key 2024 | Download Akhil Bhartiya Hindi Olympiad Solutions

Dec 16, 2024

Upcoming School Exams

Bihar Open Board 10th Examination

Application Date:11 November,2024 - 10 January,2025

Admit Card

Result

Bihar Open Board 12th Examination

Application Date:11 November,2024 - 10 January,2025

Admit Card

Result

National Means Cum-Merit Scholarship

Admit Card Date:27 November,2024 - 22 December,2024

Eligibility Criteria

Application Process

Exam Pattern

Gujarat Board Higher Secondary School Certificate

Late Fee Application Date:13 December,2024 - 22 December,2024

Exam Pattern

Result

Admit Card

Madhya Pradesh State Open School 10th Examination

Admit Card Date:13 December,2024 - 31 December,2024

Admit Card

Result

View All School Exams

Get answers from students and experts

Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is