Careers360 Logo
NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry - Download PDF

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 .

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

16431740496021643174083055

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

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

NCERT Solutions for Class 7 Subject Wise

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:

Frequently Asked Questions (FAQs)

1. What is practical geometry?

Practical geometry Class 7th is a branch of geometry in which we learn the method of constructing various geometrical shapes like line segments, angles, triangles, and different polygons.  We use various geometrical tools to construct the required shapes with the exact measurements given.

2. How many exercises in NCERT solution Class 7 Maths Chapter 10?

There are 5 exercises in NCERT chapter 10 maths class 7th. 

Practical Geometry Class 7 NCERT Exercise 10.1- 3 questions

Practical Geometry Class 7th NCERT Exercise 10.2- 4 questions

Practical Geometry Class 7th NCERT Exercise Exercise 10.3 - 3 questions

Practical Geometry Class 7 NCERT Exercise 10.4 - 3 questions

Practical Geometry Class 7th NCERT Exercise Exercise 10.5 - 3 questions

3. Who is the father of geometry?

 The Father of Geometry is  Euclid. He was a Greek Mathematician. He discussed many concepts related to Geometry, students can get brief idea from class 7 chapter 10 maths about these concepts. Also you can practice practical geometry class 7 pdf to get deeper understanding of concepts.

Articles

Get answers from students and experts

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 3.8×107 J 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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top