NCERT Solutions for Class 7 Maths Chapter 10 - Practical Geometry

NCERT Maths Class 7 Chapter 10 is on Practical Geometry. This chapter is an essential one for students since it encapsulates the basic knowledge and practice that is needed to be able to put geometry to real-world use. This chapter is covered in the Class 7 maths syllabus and has 7 major sections or topics. To be able to gather the information from and engage with the curated contents of this chapter, students must go through these individual topics and concepts very carefully. This will ensure easy understanding and practice of the solutions that have been provided for this chapter on Practical Geometry.

• Introduction

• 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)

• Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle between them are known (SAS)

• Constructing a Triangle When the Measures of Two of Its Angles and the Length of the Side included between them are given (ASA)

• Constructing a Right-Angled Triangle When the Length of One Leg and Its Hypotenuse are Given (RHS Criterion)

Our advice to the students is that they learn each of these topics carefully and then get on to attempt and practise the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry on Vedantu.

Exercise 10.1

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.

Ans: Construct: A line, parallel to given line by using ruler and compasses.

Steps:

(a) Draw a line-segment \[AB\] and take a point \[C\] outside \[AB\].

(b) Take any point \[D\] on \[AB\] and join \[C\] to \[D\].

(c) With \[D\] as centre and take a convenient radius, draw an arc cutting \[AB\] at \[E\] and \[CD\] at \[F\].

(d) With \[C\] as centre and same radius as in step c, draw an arc \[GH\] cutting \[CD\] at \[P\].

(e) With the same arc \[EF\], draw the equal arc cutting \[GH\] at \[J\] with P as centre.

(f) Join \[JC\] to draw a line \[l\]. This the required line \[AB||l\].

2. Draw a line \[{l}\]. Draw a perpendicular to \[{l}\] at any point on \[{l}\]. On this perpendicular choose a point \[{x}\], \[{4cm}\]  away from \[{l}\]. Through \[{x}\], draw a line m parallel to \[{l}\].

Ans: Construct: A line parallel to given line when perpendicular line is also given.

Steps:

(a) Draw a line \[l\] and take a point \[P\] on it.

(b) At point \[P\], draw a perpendicular line \[n\].

(c) Take \[PX=4cm\]  on line \[n\].

(d) At point \[X\] , again draw a perpendicular line \[m\]. It is the required construction.

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?

Ans: Construct: A pair of parallel lines intersecting other part

of parallel lines.

Steps:

(a) Draw a line \[l\] and take a point \[P\] outside of \[l\].

(b) Take point \[Q\] on line \[l\] and join \[PQ\].

(c) Make an equal angle at point \[P\] such that \[\angle Q=\angle P\].

(d) Extend line at \[P\] to get line \[m\].

(e) Similarly, take a point \[R\] on line \[m\], at point \[R\], draw angles such that \[\angle P=\angle R\].

(f) Extended line at \[R\] which intersects at \[S\] on line \[l\]. Draw line \[RS\].

Exercise 10.2

1. Construct \[{ }\!\!\Delta\!\!\text{ XYZ}\] in which \[{XY=4}{.5cm}\], \[{YZ}={5cm}\] and \[{ZX}={6cm}\].

Ans: Construct: \[\Delta XYZ\], where \[XY=4.5cm\], \[YZ=5cm\] and \[ZX=6cm\].

Steps:

(a) Draw a line segment \[YZ=5cm\].

(b) Taking  \[Z\] as centre and radius \[6cm\], draw an arc.

(c) Similarly, taking \[Y\] as centre and radius \[4.5cm\], draw another arc which intersects the first arc at point \[X\].

(d) Join \[XY\] and \[ZX\].

It is the required \[\Delta XYZ\].

2. Construct an equilateral triangle of side \[{5}.{5cm}\].

Ans: Construct: A \[\Delta ABC\] where \[AB=BC=CA=5.5cm\]

Steps:

(a) Draw a line segment \[BC=5.5cm\]

(b) Taking points \[B\] and \[C\] as centres and radius \[5.5cm\], draw arcs which intersect at point \[A\].

(c) Join \[AB\] and \[CA\].

It is required \[\Delta ABC\].

3. Draw  \[{ }\!\!\Delta\!\!\text{ PQR}\] with \[{PQ}={4cm}\], \[{QR}={3}.{5cm}\] and \[{PR}={4cm}\]. What type of triangle is this?

Ans: Construct: \[\Delta PQR\], in which \[PQ=4cm\], \[QR=3.5cm\] and \[PR=4cm\].

Steps:

(a) Draw a line segment \[QR=3.5cm\].

(b) Taking \[Q\] as centre and radius \[4cm\], draw an arc.

(c) Similarly, taking \[R\] as centre and radius \[4cm\], draw another arc which intersects the first arc at \[P\].

(d) Join \[PQ\] and \[PR\].

It is the required isosceles \[\Delta PQR\].

4. Construct  \[{ }\!\!\Delta\!\!\text{ }{ABC}\] such that \[{AB}={2}.{5cm}\], \[{BC}={6cm}\] and \[{AC}={6}.{5cm}\]. Measure \[\angle {B}\] .

Ans: Construct: \[\Delta ABC\]in which \[AB=2.5cm\], \[BC=6cm\] and \[AC=6.5cm\].

Steps:

(a) Draw a line segment \[BC=6cm\].

(b) Taking \[B\] as centre and radius \[2.5cm\], draw an arc.

(c) Similarly, taking \[C\] as centre and radius \[6.5cm\], draw another arc which intersects the first arc at point \[A\].

(d) Join \[AB\] and \[AC\].

(e) Measure angle \[B\] with the help of a protractor.

It is the required \[\Delta ABC\]  where  \[\angle B=90\].

Exercise 10.3

1. Construct  \[{ }\!\!\Delta\!\!\text{ DEF}\] such that \[{DE}={5cm}\], \[{DF}={3cm}\] and \[{m}\angle {EDF}={90}\].

Ans: Construction: \[\Delta DEF\] where \[DE=5cm\], \[DF=3cm\] and \[m\angle EDF={{90}^{\circ }}\].

Steps:

(a) Draw a line segment \[DF=3cm\].

(b) At point \[D\], draw an angle of \[{{90}^{\circ }}\] with the help of protractor i.e., \[\angle XDF={{90}^{\circ }}\]

(c) Taking \[D\] as centre, draw an arc of radius \[5cm\], which cuts \[DX\] at the point \[E\].

(d) Join \[EF\].

It is the required right angled triangle \[\Delta DEF\].

2. Construct an isosceles triangle in which the lengths of each of its equal sides is \[{6}.{5cm}\] and the angle between them is \[{110}\text{ }{}^\circ \] .

Ans: Construction: An isosceles triangle \[PQR\] where \[PQ=RQ=6.5cm\] and \[\angle Q={{110}^{\circ }}\] 

Steps:

(a) Draw a line segment \[RQ=6.5cm\].

(b) At point \[Q\], draw an angle of \[{{110}^{\circ }}\] with the help of protractor, i.e., \[\angle YQR={{110}^{\circ }}\].

(c) Taking \[Q\] as centre, draw an arc with radius \[6.5cm\], which cuts \[QY\]at point \[P\].

(d) Join \[PR\]

It is the required isosceles triangle \[PQR\].

3. Construct  \[{ }\!\!\Delta\!\!\text{ }{ABC}\] with \[{BC}={7}.{5cm}\], \[{AC}={5cm}\] and \[{m}\angle {C}={60}\].

Ans: Construction: \[\Delta ABC\] where \[BC=7.5cm\], \[AC=5cm\] and \[m\angle C={{60}^{\circ }}\].

Steps:

(a) Draw a line segment \[BC=7.5cm\].

(b) At point \[C\], draw an angle of \[{{60}^{\circ }}\] with the help of protractor, i.e., \[\angle XCB={{60}^{\circ }}\] 

(c) Taking \[C\] as centre and radius \[5cm\], draw an arc, which cuts \[XC\] at the point \[A\].

(d) Join \[AB\]

It is the required triangle \[ABC\].

Exercise 10.4

1. Construct \[\Delta {ABC}\], given \[{m}\angle {A}={6}{{{0}}^{\circ }}\],  \[{m}\angle {B}={3}{{{0}}^{\circ }}\] and \[{AB}={5}.{8cm}\].

Ans: Construction: \[\Delta ABC\] where  \[m\angle A={{60}^{\circ }}\],  \[m\angle B={{30}^{\circ }}\] and \[AB=5.8cm\].

Steps:

(a) Draw a line segment  \[AB=5.8cm\].

(b) At point \[A\], draw an angle \[\angle YAB={{60}^{\circ }}\] with the help of a protractor.

(c) At point \[B\], draw \[\angle XBA={{30}^{\circ }}\] with the help of a protractor.

(d) \[AY\text{ } and \text{ }BX\]  intersect at the point \[C\].

It is the required triangle \[ABC\].

2. Construct \[\Delta {PQR}\] if \[{PQ}={5cm}\], \[{m}\angle {PQR}={10}{{{5}}^{\circ }}\]and \[{m}\angle {QRP}={4}{{{0}}^{\circ }}\] .

Ans: Construction: \[\Delta PQR\] where \[m\angle P=35{}^\circ \], \[m\angle Q=105{}^\circ \]and \[PQ=5\text{ }cm\].

Explanation:

\[\angle RPQ=180-\left( \angle RQP+\angle PRQ \right)\]

From angle sum property

$ \angle RPQ=180-\left( 105+40 \right) $

$ \angle RPQ=180-\left( 145 \right) $

$ \angle RPQ={{35}^{\circ }}$

Steps:

(a) Draw a line segment \[PQ=5\text{ }cm\].

(b) At point \[P\], draw  \[\angle XPQ=35{}^\circ \]  with the help of a protractor.

(c) At point \[Q\], draw  \[\angle YQP=105{}^\circ \] with the help of a protractor.

(d) \[XP\text{ }and\text{ }YQ\] intersect at point \[R\].

It is the required triangle \[PQR\].

3. Examine whether you can construct \[{ }\!\!\Delta\!\!\text{ DEF}\] such that \[{EF}={7}.{2}\text{ }{cm}\], \[{m}\angle {E}={110}{}^\circ \text{ }{and}\text{ }{m}\angle {F}={80}{}^\circ \] .

Justify your answer.

Ans: In \[\Delta DEF\], \[m\angle E=110{}^\circ \],  \[m\angle F=80{}^\circ \],  \[EF=7.2\text{ }cm\]

This \[\Delta DEF\] is not possible to construct because

$ m\angle E+m\angle F={{80}^{\circ }}+{{110}^{\circ }} $ 

$ m\angle E+m\angle F={{190}^{\circ }} $ 

Which is more than the total interior angle of the triangle.

Exercise 10.5

1. Construct the right angled \[\Delta {PQR}\], where \[{m}\angle {Q}={9}{{{0}}^{\circ }}\], \[{QR}={8cm}\] and \[{PR}={10cm}\].

Ans: Construction: A right angled triangle \[\Delta PQR\] where \[m\angle Q={{90}^{o}}\] ,  \[QR=8cm\]and \[PR=10cm\].

Steps:

(a) Draw a line segment \[QR=8cm\].

(b) At point \[Q\], draw \[QX\bot QR\].

(c) Taking \[R\] as centre, draw an arc of radius \[10cm\].

(d) This arc cuts \[QX\] at point \[P\].

(e) Join \[PR\].

It is the required right angled triangle \[PQR\].

2. Construct a right angled triangle whose hypotenuse is \[{6}\text{ }{cm}\] long and one the legs is \[{4}\text{ }{cm}\] long.

Ans: Construction: A right angled triangle \[DEF\] where \[DF=6cm\] and \[EF=4cm\]

Steps:

(a) Draw a line segment \[EF=4cm\].

(b) At point \[E\], draw \[EX\bot EF\].

(c) Taking \[F\] as centre and radius \[6\text{ }cm\], draw an arc. (Hypotenuse)

(d) This arc cuts the \[EX\] at point \[D\].

(e) Join \[DF\].

It is the required right angled triangle \[DEF\].

3. Construct an isosceles right angled triangle \[{ABC}\], where \[{m}\angle {ACB}={9}{{{0}}^{\circ }}\] and \[{AC}={6cm}\].

Ans: Construction: An isosceles right angled triangle \[ABC\] where \[m\angle C={{90}^{o}}\],  \[AC=BC=6cm\].

Steps:

(a) Draw a line segment \[AC=6cm\].

(b) At point \[C\], draw \[XC\bot CA\].

(c) Taking \[C\] as centre and radius \[6cm\], draw an arc.

(d) This arc cuts \[CX\] at point \[B\].

(e) Join \[BA\].

It is the required isosceles right angled triangle \[ABC\].

Miscellaneous Questions 

From 1 - 8 Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and, say why you cannot construct them. Construct the rest of the triangle.

Ans: In \[\Delta ABC\], \[m\angle A={{85}^{\circ }}\], \[m\angle B={{115}^{\circ }}\], \[AB=5\text{ }cm\]

Construction of ∆ ABC is not possible because \[m\angle A +  {m} \angle {B} ={{200}^{\circ }}\],  and we know that the sum of angles of a triangle should be \[{{180}^{\circ }}\] . 

Ans: Construct: \[\Delta PQR\]where \[m\angle Q={{30}^{\circ }}\] ,  \[m\angle R={{60}^{\circ }}\] and \[QR=4.7cm\].

Steps:

(a) Draw a line segment \[QR=4.7cm\].

(b) At point \[Q\], draw \[\angle XQR=30\] with the help of a compass.

(c) At point \[R\], draw \[\angle YRQ=60\] with the help of a compass.

(d) \[XQ\] and \[YR\] intersect at point \[P\].

It is the required triangle \[PQR\].

Ans: We know that the sum of angles of a triangle is 180 Degrees.

 $ m\angle A+m\angle B+m\angle C=180 $

 $ {{70}^{\circ }}+{{50}^{\circ }}+m\angle C={{180}^{\circ }} $ 

 $ {{120}^{\circ }}+m\angle C={{180}^{\circ }} $ 

 $ m\angle C={{180}^{\circ }}-{{120}^{\circ }} $ 

 $ m\angle C={{60}^{\circ }} $

Construct: \[\Delta ABC\] where \[m\angle A={{70}^{\circ }}\] , \[m\angle C={{60}^{\circ }}\] and \[AC=3cm\].

Steps:

(a) Draw a line segment \[AC=3cm\].

(b) At point \[C\], draw \[\angle YCA={{60}^{\circ }}\].

(c) At point \[A\], draw \[\angle XAC={{70}^{\circ }}\]. °

(d) Rays \[XA\] and \[YC\] intersect at point \[B\]

It is the required triangle \[ABC\].

Ans: In \[\Delta LMN\], \[m\angle L={{60}^{\circ }}\],  \[m\angle N={{120}^{\circ }}\],  \[LM=5\text{ }cm\]

This \[\Delta LMN\] is not possible to construct because

  $ m\angle L+m\angle N={{60}^{\circ }}+{{120}^{\circ }} $ 

$ m\angle L+m\angle N={{180}^{\circ }} $ 

Which forms a linear pair.

Ans: \[\Delta ABC\], \[BC=2cm,AB=4cm\text{ } and \text{ }AC=2\text{ }cm\]

This \[\Delta ABC\] is not possible to construct because the condition is

Sum of lengths of two sides of a triangle should be greater than the third side.

Ans: Construction: \[\Delta PQR\] where \[PQ=3.5cm,QR=4cm\text{ } and \text{ }PR=3.5cm\]

Steps:

(a) Draw a line segment \[QR=4cm\].

(b) Taking \[Q\] as centre and radius \[3.5cm\], draw an arc.

(c) Similarly, taking \[R\] as centre and radius \[3.5cm\], draw another arc which intersects the first arc at point \[P\].

It is the required triangle \[PQR\].

Ans 7: Construction: A triangle whose sides are \[XY=3cm,YZ=4cm\text{ }andXZ=5cm\].

Steps:

(a) Draw a line segment \[ZY=4cm\].

(b) Taking \[Z\] as centre and radius \[5cm\], draw an arc.

(c) Taking \[Y\] as centre and radius \[3cm\], draw another arc.

(d) Both arcs intersect at point \[X\].

It is the required triangle \[XYZ\].

Ans: Construction: A triangle \[DEF\] whose sides are \[DE=4.5cm,EF=5.5cm\text{ }and\text{ }DF=4cm\]

Steps:

(a) Draw a line segment \[EF=5.5cm\].

(b) Taking \[E\] as centre and radius \[4.5cm\], draw an arc.

(c) Taking \[F\] as centre and radius \[4cm\], draw another arc which intersects the first arc at point \[D\].

It is the required triangle \[DEF\].

NCERT Solutions for Class 7 Maths – Free PDF Download

Students can download the NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry that are put together by knowledgeable teachers. These solutions give the students a good grasp on the topic. It also helps to enhance the students' learning, which lets the students perform well in their examination.

Class 7 Maths Chapter 10 Includes:

NCERT Solutions for Class 7 Maths Chapter 10 – Practical Geometry

10.1 Introduction

This NCERT Solutions for Class 7th Maths Chapter 10 section explains the concept of Practical Geometry which is about constructing various figures in geometry. Now that a learner knows how to draw a line segment and a perpendicular line, this chapter will add to what they have already leant. Through several activities, the Ch 10 Maths Class 7 NCERT Solutions will teach ways to draw parallel lines and triangles.

10.2 Construction of a Line Parallel to a Given Line, Through a Point Not on The Line

This NCERT Class 7 Maths Chapter 10 section will teach how to draw a pair of lines that are parallel to each other. Parallel lines are those lines that do not intersect and always stay at the same distance from each other. A student should know how to construct parallel lines. It is imperative, especially in the field of architecture where the architects need to draw the blueprints of buildings.

So here is how two exact parallel lines can be drawn. 

• Two lines that are parallel in a plane should not intersect even if those lines are extended till infinity in any of the directions. 

• The distance between the two lines stays the same throughout.

To construct a parallel line, a student needs specific tools like a ruler and a compass.

You also need a line segment AB and a point P outside the line from where the parallel lines need to be drawn.

• Choose any point on the line segment AB and then join this point to the outside point P.

• Use X as the centre and with any radius that is suitable, draw an arc that cuts the line segment PX. Let this point be M on the line PX, and N be the point that cuts AB.

• Now use P as the centre and using the same radius used in the first case draw an arc. Let’s call this EF that cuts the line segment PX on the point say Q.

• Use Q as the centre and with the same radius used before drawing an arc that cuts the arc EF at the point R.

• Now join R and P, and this gives the line segment CD.

The line segment is the parallel line that is required, and it passes through the point P.

10.3 Construction of Triangles

Before you proceed to this Class 7 Chapter 10 Maths section, you should first recall the properties of triangles and the congruence of triangles. Triangles are classified on the bases of sides or their angles. Here are some essential features of triangles.

• The exterior angle of any triangle is the sum of the interior and the opposite angles.

• The three angles of any triangle total to 180 degrees.

• The sum of any two sides of the triangle is greater than the length of the third side.

• If the triangle is a right-angled triangle, then the hypotenuse square is equal to the sum of the length square and the breadth square.

The chapter on the Congruence of Triangles shows how a triangle can be drawn if any of these measurements are provided:

• All three sides.

• Two sides and the angles that lie between them.

• Value of two angles and side that stays between them.

• The hypotenuse and the size of the length of a right-angled triangle.
  
  

10.4 Constructing a Triangle When the Lengths of its Three Sides are Known (SSS Criterion)

It is possible to construct a triangle when all the three sides of the triangle are known. The tools required for this are a ruler and a compass.

• When all three sides have been provided, the students first need to identify the longest measure of the side, which is given. 

• Make this as the base of the triangle. 

• Use the other measurement and mark the arcs by using the endpoints of the base as the triangle’s vertices. 

• Then join the arc intersection with the endpoints of the base and get the triangle required.
  
  

10.5 Constructing a Triangle When the Lengths of Two Sides and the Measure of the Angle Between Them are Known. (SAS Criterion)

This Practical Geometry Class 7 PDF section explains how to construct a triangle when the two sides and the angle are provided. The tools needed for this are a ruler and a compass.

To construct a triangle with the SAS provided, here is what needs to be done. 

Here, the two sides and the angle enclosed between them have been provided. This construction will not be possible if any of the angles are given. Only the enclosed angles should be given.

• Draw a straight line and make the left endpoint as point A. 

• Set the compass to the width of one of the sides. Place the compass on point A and cut an arc on the line drawn. 

• The point B is the point where the arc cuts the line. 

• Now construct an angle with the angle degree provided on line AB at the point A. 

• Now set the compass to the second length provided. 

• Then place the pointer head of the compass on the point A and cut an arc on the angle degree line drawn. 

• The point where the arc crosses the line should be marked as say C. 

• The points B and C should now be joined with a ruler to obtain the triangle ABC.
  
  

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)

This Class 7 Maths Practical Geometry section teaches how to draw a triangle where two angles and a side have been provided.

To satisfy the ASA condition, the given side has to be necessarily the one that is enclosed by the angles known. If any other sides of the triangle have been given, this will not help in drawing a triangle with the ASA method. A protractor and a ruler will be needed to draw this triangle. 

• Use the ruler and draw the line segment with the length that is provided.

• Now use the protractor and then draw a ray at point B, making one of the angles provided.

• Similarly, draw the ray of the other angle provided at point A using a protractor.

• The point where both the rays meet should be marked as C.

This is how you can draw the triangle ABC by this method.

10.7 Constructing a Right-Angled Triangle When the Length of One Leg and its Hypotenuse are Given (RHS Criterion)

This Ch 10 Maths Class 7 section explains how to construct a right-angled triangle where the hypotenuse and one of the sides are given. To construct this triangle, we need a compass and a ruler.  Let the right angle be at point C.

• To construct this triangle first draw a horizontal line of any length and then mark a point C on it.

• Now set the compass to the length of the side provided. Place the head of the compass on point C and mark an arc on both sides of C. Mark the points P and A where the arc crosses the line.

• Now set the compass to the length of the hypotenuse and place the head of the compass on the point P. Then use this to mark an arc above C. 

• Do the same step from point A as well and mark this point like B, which is where both the arcs cross each other. 

• Join A to C and simultaneously B to C, and this gives you the right-angled triangle ACB.