NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry

 

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry: 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. You should try to solve all these questions by yourself. You can take help from CBSE NCERT solutions for class 7 maths chapter 10 practical geometry. 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 class 6 to 12 for science and maths by clicking on the above link. Here you will get solutions to five exercises of this chapter.

Exercise:10.1

Exercise:10.2

Exercise:10.3

Exercise:10.4

Exercise:10.5

This chapter of NCERT Solutions for Class 7 Maths contains the following topics

10.1  Introduction

10.2  Construction Of A Line Parallel To A Given Line, Through A Point Not On The Line

10.3  Construction Of Triangles

10.4  Constructing A Triangle When The Lengths Of Its Three Sides Are Known (Sss Criterion)

10.5  Constructing A Triangle When The Lengths Of Two Sides And The Measure Of The Angle Between Them Are Known. (Sas Criterion)

10.6  Constructing A Triangle When The Measures Of Two Of Its Angles And The Length Of The Side Included Between Them Is Given. (Asa Criterion)

10.7  Constructing A Right-angled Triangle When The Length Of One Leg And Its Hypotenuse Are Given (Rhs Criterion)

The NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry is given Below:

 

Solutions of NCERT for class 7 maths chapter 10 practical geometry-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

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

3. Join CD

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.

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

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

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

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

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.

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

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:

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.

 

CBSE NCERT solutions for class 7 maths chapter 10 practical geometry-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

 

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

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 

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

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

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

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

 

Solutions of NCERT 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

 

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

 

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.

 

CBSE 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

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

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

NCERT Solutions for Class 7 Maths- Chapter-wise 

Chapter No.

Chapter Name

Chapter 1

Solutions of NCERT for class 7 maths chapter 1 Integers

Chapter 2

CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals

Chapter 3

NCERT solutions for class 7 maths chapter 3 Data Handling

Chapter 4

Solutions of NCERT for class 7 maths chapter 4 Simple Equations

Chapter 5

CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles

Chapter 6

NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties

Chapter 7

Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles

Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities

Chapter 9

CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers

Chapter 10

NCERT solutions for class 7 maths chapter 10 Practical Geometry

Chapter 11

Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area

Chapter 12

CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions

Chapter 13

NCERT solutions for class 7 maths chapter 13 Exponents and Powers

Chapter 14

Solutions of NCERT for class 7 maths chapter 14 Symmetry

 NCERT Solutions for Class 7- Subject-wise 

Solutions of NCERT for class 7 maths

CBSE 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!!!  

 

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