NCERT Solutions For Class 7 Maths Chapter 10

NCERT Solutions For Class 7 Maths Chapter 10 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry are provided here so that students can check for the solutions whenever they are facing difficulty while solving the questions from NCERT class 7 maths book. These solutions for chapter 10, are available in PDF format so that students can download it and learn offline as well, where they can use it as a worksheet also.

Maths is a subject which requires understanding and reasoning skills with logic. Along with this, it also requires students to practice maths on a regular basis. Students of Class 7 are suggested to solve NCERT questions in order to strengthen the fundamentals and be able to solve questions that are usually asked in the examination. From all the chapters, Practical Geometry is an important topic for the students, and one needs to practice thoroughly to score well in the examination.

Class 7 Maths NCERT Solutions – Practical Geometry

NCERT Solutions Class 7 Maths Practical Geometry PDF provided here are in a detailed manner, where one can find a step-by-step solution to all the questions of this chapter. These PDF solutions are prepared by our subject experts under the guidelines of the NCERT curriculum and as per CBSE syllabus (2018-2019), to assist students in their exam preparations.

The topics covered in chapter 10 of class 7 Maths are;

  • Introduction to Construction of a line parallel to a given line,
  • Construction of triangles,
  • Constructing a triangle when the lengths of its three sides are known (SSS criterion),
  • Constructing a triangle when 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 its length of the side included between them is given (ASA criterion),
  • Constructing a right-angled triangle when the length of one leg and when its hypotenuse is given (RHS criterion), etc.

Apart from practising solutions for chapter 10, students are advised to solve sample papers and previous year question papers, to get an idea to types of questions asked from Practical Geometry and also marks contained by it. It will also help students to score well in the final exam of class 7.

NCERT Class 7 Maths Solutions For Maths Chapter 10 Exercises

Exercise 10.1


Q1:

Draw a line and name it PQ, take a point R outside it. Through R, draw a line parallel to PQ using the ruler and compass only.

Solution:

To construct: A line, parallel to the given line by using ruler and compass.

Construction:

(i) Draw a line segment PQ and take a point R outside PQ.

(ii) Take any point X on PQ and join R to X.

(iii) With X as the center and take convenient radius, draw an arc cutting PQ at Y and RX at Z.

(iv) With R as center and the same radius as in step (iii), draw an arc AB cutting PX at O

(v) With the same arc YZ, draw the equal arc cutting AB at M.

(vi) Join MR to draw a line N.

This the required line PQ ǁ N

1.1

 

Q2 :

Draw a line N. Draw a perpendicular to N at any point on N. On this perpendicular choose a point P, 4cm away from N. Through A, draw a line P parallel to N.

Solution:

To construct: A line parallel to given line when perpendicular line is also given.

Construction:

(i) Draw a line N and take a point O on it.

(ii) At point O, draw a perpendicular line Q

(iii) Take OA = 4 cm on line Q.

(iv) At point A again draw a perpendicular line P

It is the required construction.

1.2

 

Q3:

Let X be a line and A be a point not on X. Through A, draw a line Y parallel to X. Now join A to any point B on X. Choose any point C on Y. Through C, draw a line parallel to AB. Let this meet X at P. What shape do the two sets of parallel lines enclose?

Solution:

To construct: A pair of parallel lines intersecting other part of parallel lines.

Construction:

(i) Draw a line X and take a point A outside of X

(ii) Take point B on line X and join AB

(iii) Make equal angle at point A such that Ð B = Ð A

(iv) Extend line at A to get line Y

(v) Similarly, take a point C which intersects at P on line X, draw line PC

Thus, we get parallelogram ABCP

 1.3

 

Exercise 10.2


Q1 :

Construct \(\Delta\) PQR in which PQ = 4.5 cm , QR = 5 cm and RP = 6cm.

Solution:

Construct: \(\Delta\) PQR, where PQ = 4.5 cm, QR = 5 cm, RP = 6 cm

Construction:

(i) Draw a line segment QR = 5 cm.

(ii) Taking R as center and radius 6cm, draw arc.

(iii) Similarly, taking Q as center and radius 4.5 cm  draw another arc which intersects the first arc at point P

(iv) Join PR and PR

It is the required \(\Delta\) PQR

 2.1

 

Q2:

Construct an equilateral triangle of side 5.5 cm

Solution:

To construct: A \(\Delta\) PQR where PQ = QR = QP = 5.5 cm

Construction :

(i) Draw a line segment BC = 5.5 cm

(ii) Taking points Q and R as centers and radius 5.5 cm , draw arcs which intersect at point P.

(iii) Join PQ and PR

It is the required \(\Delta\) PQR

 2.2

 

Q3:

Draw \(\Delta\) ABC with AB = 4cm, BC = 3.5 and AC = 4cm. What type of triangle is this?

Solution:

To construct: \(\Delta\) ABC, in which AB = 4cm, BC = 3.5 and AC = 4 cm

Construction:

(i) draw a line segment BC = 3.5 cm

(ii) Taking B as center and radius 4 cm, draw an arc.

(iii) Similarly, taking C as center and radius 4 cm, draw another arc which intersects first arc at A.

(iv) Join AB and AC.

It is the required isosceles \(\Delta\) ABC

2.3

 

 Q4:

Construct \(\Delta\) PQR such that PQ = 2.5 cm , QR = 6 cm and PR = 6.5 cm. Measure \(\angle\) Q

Solution:

To Construct:

\(\Delta\) PQR such that PQ = 2.5 cm, QR = 6cm and PR = 6 cm and PR = 6.5 cm

Construction :

(i) Draw a line segment QR = 6cm

(ii) Taking B as center and radius 2.5 cm, draw an arc.

(iii) Similarly, taking C as center and radius 6.5 cm , draw another arc which intersects first arc at point P

(iv) Join PQ and PR

(v) Measure angle Q with the help of protractor

It is the required \(\Delta\) PQR where \(\angle\) Q = 80o

2.4

 

 

Exercise 10.3


Q1: Construct   a \(\Delta\)  ABC so that AC = 5 cm, AB = 3 cm and \(\angle\) CAB = 90O .

Solution:

Constructing a \(\Delta\) DEF where DE = 5 cm, DF = 3 cm and m\(\angle\) EDF = 90O .

Steps of construction:

(a) Draw a line segment AB = 3 cm.

(b) At point A, draw an angle of 90O with the help of compass i.e., \(\angle\) XAB = 900.

(c) Taking A as the center, draw an arc of radius 5 cm, which cuts AX at the point E.

(d)Connect BC.

This  is the required right angled triangle ACB:

3.1

 

Q2: Construct an isosceles triangle which has equal sides of length  6.5 cm and with a angle of 110 ° between them.

Solution:

Constructing an isosceles triangle where AC = AB = 6.5 cm and \(\angle\) A = 110 ° .

Steps of construction:

(a) Make a line segment AB= 6.5 cm.

(b) At  A, make a  \(\angle\) 110° with the help of protractor, i.e., \(\angle\) XAB = 110 ° .

(c) Take A as the  center, and draw an arc of radius 6.5 cm, which cuts AX at point  C.

(d) Connect BC

This is the required isosceles \(\Delta\) ABC:

3.2

 

Q3:

Construct a \(\Delta\) ZXC with XC = 7.5 cm, ZC = 5 cm and \(\angle\) C = 60O .

Solution:

Constructing a  \(\Delta\) ZXC where XC = 7.5 cm, ZC = 5 cm and \(\angle\) C = 60O.

Steps of construction:

(a) Make a line segment XC = 7.5 cm.

(b) At C, make an angle of 600 with the help of protractor, i.e., \(\angle\) XCB = 60O .

(c) Take C as the center  and draw an arc of 5cm, which cuts the line YC at the point Z.

(d)Connect  AB

Thus, this is the required \(\Delta\) ZXC:

3.3

 

Exercise 10.4


Q1:

Construct a triangle ZXC, where\(\angle\) Z = 60O , \(\angle\) X = 30O and ZX = 5.8 cm.

Solution:

Now, constructing a \(\Delta\) ZXC where \(\angle\) Z = 60O , \(\angle\) X = 300 and ZX = 5.8 cm.

Steps for construction:

(a) Make a line segment ZX = 5.8 cm.

(b) At  Z, make an angle \(\angle\) XZC = 60O  with a compass.

(c) At X, draw \(\angle\) QXZ = 300  with  a compass.

(d) ZP and XQ intersect at the point C.

Thus we have the required triangle ZXC:

4.1

 

Q2:

Construct a triangle ABC if AB = 5 cm, \(\angle\) ABC = 105° and \(\angle\) BCA = 40°.

Solution:

Given:

\(\angle\)ABC  = 105° and \(\angle\) BAC = 40°

We know ,sum of the angles of a triangle = 180 .°

\(\angle\) ABC + \(\angle\) BCA + \(\angle\) CAB = 180°

105. 40 °+ 40°+\(\angle\)BCA = 180°

145° + \(\angle\) BCA = 180°

\(\angle\)BCA = 180° – 145°

\(\angle\)BCA = 35°

Constructing a triangle ABC  whose\(\angle\) A = 35°, \(\angle\) B = 105° and AB = 5 cm.

Steps of construction:

(a) Make a line segment PQ = 5 cm.

(b) At  A, make an angle  \(\angle\) QAB = 35° with the help of a protractor.

(c) At  B, make an angle \(\angle\) ABP = 105° with the help of a protractor.

(d) QA and BP intersect at point C.

Thus, we have the required triangle ABC.

4.2

 

Q3:

Can you construct  a \(\Delta\) ABC such that AB = 4 cm, \(\angle\) B = 120° and \(\angle\) C= 70 °. Explain your answer.

Solution:

Given:

In \(\Delta\) ABC, \(\angle\) A = 120° and m\(\angle\) F = 70 °

We know the angles of a triangle sum up to 180o

Thus, we have

\(\angle\)A + \(\angle\)C+ \(\angle\)B= 180°

⟹\(\angle\)A + 120° + 70° = 180°

⟹\(\angle\)A + 190° = 180°

 

⟹\(\angle\)A = 180° − 190° = −10°

Since the sum of the angles is coming more than 180o, this triangle construction is not possible.

 

Exercise 10.5


Q1:

Construct a right angled \(\Delta\) ABC, where \(\angle\) B = 90o  BC = 8 cm and AC = 10 cm

Solution:

To construct:

A right angled triangle ABC where \(\angle\) B = 90o BC = 8cm and AC = 10cm

Construction:

(i) Draw a line segment BC = 8cm

(ii) At point B, draw BP  \(\perp\)  BC

(iii) taking C as the center, draw an arc of radius 10cm

(iv) this arc cuts BP at point X

(V) join XB

It is the required right angled triangle ABC.

5.1

 

Q2 :

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

Solution:

TO construct:

A right angled triangle ABC where AC = 6cm and BC = 4cm

Construction:

(i) Draw a line segment BC = 4cm

(ii) At point X, draw BX  \(\perp\)  BC

(iii) Taking C as the center and radius 6cm, draw an arc. (Hypotenuse)

(iv) This arc cuts the BX at point A

(v) Join AC

It is the required right angled triangle ABC.

5.2

 

Q3 :

Construct an isosceles right angled triangle PQR, where \(\angle\)PQR = 90o and PR = 6cm.

Solution:

To construct:

An isosceles right triangle PQR where \(\angle\)R = 90o, PR = QR = 6cm

Construction:

(i) Draw a line segment PR = 6cm

(ii) At point R, draw XR  \(\perp\)  RP

(iii) Taking R as the center and radius 6 cm, draw an arc

(iv) This arc cuts RX at point Q

(v) Join QP

It is the required isosceles right angled triangle PQR

5.3

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