NCERT Solutions Class 8 Maths Chapter 4
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NCERT Solutions for Class 8 Mathematics Chapter 4: Practical Geometry
Mathematics brings curiosity to many students in Class 8. As the difficulty level increases, it introduces theorems and formulas. For students, there are a lot of opportunities in the field of engineering and science. However, it also requires regular problemsolving and a comprehensive strategy to understand the complex theories. Class 8 is an important stage for the students, as it prepares them for the upcoming board classes.
Class 8 Mathematics Chapter 4: Practical Geometry has no mathematical formulas for solving questions. However, it introduces the properties of quadrilaterals and teaches children steps that need to be followed for construction of such geometry figures. There are various exceptional cases where students will discover differently constructed quadrilaterals such as four sides and one diagonal, three sides and two diagonals, and two adjacent sides and three angles, etc.. It also discusses other essential facts to which are provided in the NCERT Solutions for Class 8 Mathematics Chapter 4 in an easy language
The NCERT Solutions for Class 8 Mathematics Chapter 4 are helpful for students as it helps them in scoring high marks in the examination by providing detailed stepbystep explanations of all the problems given in the Chapter 4, Practical Geometry Class 8. Through such extensive explanations, students can clarify all their doubts and leverage them to excel in their exams.
Extramarks is one of the leading online learning platforms trusted by many students. Students pursuing primary and secondary education trust our study materials and NCERT solutions. Mostly all the questions are provided by considering the new pattern of CBSE. Thus, students get thorough knowledge and practise necessary for their exams.
For the latest updates and examrelated news, students can visit our Extramarks’ website. In addition, students can refer to other class solutions as well such as NCERT Solutions Class 9, NCERT Solutions Class 10, and NCERT Solutions Class 11.
Key Topics Covered in NCERT Solutions for Class 8 Mathematics Chapter 4:
NCERT Class 8 Mathematics Chapter 4 introduces different patterns for constructions. In addition, it also recalls the properties of quadrilaterals. This chapter is the perfect amalgamation of both the topics mentioned above. It helps to build on the concepts that students have learned before. It shows them new ways in which they can implement these ideas. NCERT Solutions for Class 8 Mathematics Chapter 4 which covers all the essential chapter related questions and their extensive answers.
Some of the key topics featured in NCERT Solutions for Class 8 Mathematics Chapter 4 are:
Introduction
In Class 7, students learnt the basics of the quadrilateral. Now , in Chapter 4, Class 8, students will learn how we require three measurements to draw a unique triangle. Since three measurements were enough to draw a triangle, a natural question arises whether four measurements would be sufficient to draw a unique foursided closed figure, namely a quadrilateral.
A quadrilateral, a closed shape, is a polygon with four sides, four vertices and four angles. It is made by joining four points that are not collinear. The sum interior tips are always equal to 360 degrees. The Latin words Quadra, which means four, and Latus, which means’ sides’, are the origins of the quadrilateral. Not all four sides of a quadrilateral must be equal in length. Therefore, we can have many quadrilaterals based on angles and sides.
To avail accurate answers of the chapter related questions , students can refer to NCERT Solutions for Class 8 Mathematics Chapter 4.
Constructing a Quadrilateral
Before we build quadrilaterals, let’s first remember what a quadrilateral is. A quadrilateral is a triangular polygon with four vertices and sides surrounding four angles. 360 degrees is the sum of all its interior angles.
A quadrilateral generally has sides with different lengths and angles at other measurements. However, squares, rectangles, etc. are some special quadrilaterals that have equal sides and angles. Students will discover the following quadrilateral constructions discussed in NCERT Solutions for Class 8 Mathematics Chapter 4:
When four sides and one diagonal are given
When the measurements are given, let’s say we are required to construct a quadrilateral PQRS.
 PQ = w cm
 QR = x cm
 RS = y cm
 PS = z cm
Diagonal = u cm
To construct the quadrilaterals with four sides, we must draw a rough figure of the quadrilateral with the provided dimensions. Students can follow the direction explained in NCERT Solutions for Class 8 Mathematics Chapter 4 to remove the quadrilateral:
 Draw a line segment of length y cm and mark the ends as S and R.
 Set your compass to the radius of x cm and make an arc from the point R above the line segment.
 Set the compass to the radius of u cm and make an arc from the point S on the previous arc.
 Mark the point as Q where the two arcs cross each other. Join the points S and Q as well as R and Q.
 Set the compass to the radius of w cm and make an arc from the point Q.
 Set the compass to the radius of z cm and make an arc from the point S on the previous arc.
 Mark the point as P where the two arcs cross each other.
 Join the points P and Q as well as P and S.
When two diagonals and three sides are given
If students are required to construct a quadrilateral ABCD where the measurements are:
 AB = x cm
 BC = y cm
 ∠A = 120 degrees
 ∠B = 110 degrees
 ∠C = 90 degrees
 Mark the ends with A and B.
 Use a protractor to draw a line starting at point A and going 120 degrees. The line segment AB is the result.
 Use the protractor to draw a line starting at point B and ending at 110 degrees. The line segment BA is the other end.
 Your compass should be set to the radius of ycm. Make an arc starting at point B along the 110degree line. The point where the line intersects with the arc is marked C.
 Use the protractor to draw a line starting at point C and going 90 degrees with line segment CB. The point where the arc intersects with the 120degree line is marked D.
 The quadrilateral ABCD is the sum of all the interior angles. You can find the quadrilateral ABCD of required measurements because the sum of all the interior angles is 360 degrees.
Students can refer to our Extramarks NCERT Solutions for Class 8 Mathematics Chapter 4 where we have explained the fundamental properties and step wise methods to calculate the diagonal with two sides.
When three sides and two included angles are given.
If students are required to construct a quadrilateral PQRS where the measurements are:
 QR = x cm
 RS = y cm
 PS = z cm
 ∠S = 100 degrees
 ∠R = 120 degrees
Follow these steps to create quadrilaterals using some of the given measurements:
 Mark the ends with S and R.
 Use a protractor to draw a line starting at point R and ending at point S. Draw a line segment SR.
 Your compass should be set to the radius of ycm. Make an arc starting at point S along the 100degree line. The point where the line intersects with the arc is marked P.
 Similarly, set your compass to a radius of zcm and draw an arc starting at point R along the 120degree line. The point where the line intersects with the arc is marked Q.
 Join the points Q and P.
Calculating the diagonals of the quadrilaterals is easy with the steps mentioned above. However, students can also refer to NCERT Solutions for Class 8 Mathematics Chapter 4.
NCERT Solutions for Class 8 Mathematics Chapter 4: Exercise & Solutions
NCERT Solutions for Class 8 Mathematics Chapter 4, provides chapter end exercises and their solutions. Our subject matter experts prepare it after significant and indepth research and analysis of the syllabus. They have straightforwardly designed the solutions by demonstrating stepwise answers. The solutions are organised based on questions in NCERT books for all the problems. Through these, students will be able to understand each concept which will further help them to compose consice answers in the exams.
Chapter 4 focuses on constructing quadrilaterals when certain constraints are imposed on them, such as forming a quadrilateral when four sides and one diagonal are formed. Students will learn to find a diagonal with four sides and three sides. Further, to clear the doubts and get more information, students can register on our website to get a trial account for students.
Students can refer below for specific questions and solutions:
Along with this, students can also refer to other solutions for primary and secondary classes:
 NCERT Solutions Class 1
 NCERT Solutions Class 2
 NCERT Solutions Class 3
 NCERT Solutions Class 4
 NCERT Solutions Class 5
 NCERT Solutions Class 6
 NCERT Solutions Class 7
 NCERT Solutions Class 8
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NCERT Exemplar for Class 8 Mathematics:
Mathematics is an exciting subject, but some concepts are difficult to comprehend. As a result, students find it difficult to remember them and cannot solve problems in practical geometry. Therefore, solving different questions with a higher difficulty is beneficial. In addition, it helps to remember the formulas and their use at various stages of problems.
NCERT Exemplar consists of all topics and concepts that are explained in easily understandable language. The Class 8 Mathematics exemplar helps build a strong foundation on all the concepts. Most of the questions asked in the annual examination are from the NCERT books. Exemplar is handy to elaborate the solution of some of the interior angles of a hexagon. It also shows how the sum is calculated using a formula and example. It helps students solve problems on a more challenging level and enables them to clear competitions in mathematics.
Some questions are taken from NCERT Solutions for Class 8 Mathematics Chapter 4. It helps students understand how to find the diagonals with different numbers of sides in quadrilaterals.
Key Features of NCERT Solutions for Class 8 Mathematics Chapter 4:
Our NCERT Solutions for Class 8 Mathematics Chapter 4 are available for the students on the website. Some essential topics include quadrilaterals’ sides, constructs, and geometry properties. The solution guide contributes to the preparation of the annual examinations. It covers all essential topics and concepts.
The key concepts from NCERT Solutions for Class 8 Mathematics Chapter 4 are listed here:
 The solutions are wellcurated with examples and exercise questions covering all the essential concepts of the topic.
 Students can solve questions of different difficulty levels and boost their confidence.
 Solutions are clear and concise, with every topic explained in detail in the NCERT textbook used by the author.
 Read the NCERT Solutions for Class 8 Mathematics Chapter 4 available online and conserve your time during examinations by going through precise answers to all the chapterrelated questions. Moreover, one doesn’t have to search for any other reliable study material elsewhere. The solutions guide helps one understand all the related concepts in Mathematics that they may have been taught till date at the school level.
 The solutions cover all the chapters of the Class 8 Mathematics textbook.
 The solutions to Chapter 4 help the students to prepare for their exams by covering all topics with examples and answers to NCERT exercise questions which include all the essential mathematics concepts.
 Topics include sides of quadrilateral polygons, their construction, properties of a polyhedron, equivalence relation between sets, connectivity property of polyhedron and polygon, etc.
Q.1 Construct the following quadrilaterals.
(i) Quadrilateral ABCD  (ii) Quadrilateral JUMP 
AB = 4.5 cm  JU = 3.5 cm 
BC = 5.5 cm  UM = 4 cm 
CD = 4 cm  MP = 5 cm 
AD = 6 cm  PJ = 4.5 cm 
AC = 7 cm  PU = 6.5 cm 
(iii) Parallelogram MORE  (iv) Rhombus BEST 
OR = 6 cm  BE = 4.5 cm 
RE = 4.5 cm  ET = 6 cm 
EO = 7.5 cm 
Ans
Steps of construction:
 Draw a line segment BC of length 5.5 cm.
 With B as centre, draw an arc of radius 4.5 cm.
 With C as centre, draw an arc of radius 7 cm intersecting the previous arc at point A.
 Join AB and AC.
 With C as centre, draw an arc of radius 4 cm at point D.
 With A as centre, draw an arc of radius 6 cm cuting the previous arc at D.
 Join AD and CD.
Therefore, ABCD is the required quadrilateral.
(ii)
Steps of construction:
 Draw a line segment PU of length 6.5 cm.
 With P as centre, draw an arc of radius 4.5 cm.
 With U as centre, draw an arc of radius 3.5 cm intersecting the previous arc at point J.
 Join PJ and JU.
 With P as centre, draw an arc of radius 5 cm at point M.
 With U as centre, draw an arc of radius 4 cm intersecting the previous arc at M.
 Join PM and MU.
Therefore, JUMP is the required quadrilateral.
(iii)
Steps of construction:
 Draw a line segment OR of length 6 cm.
 With O as centre, draw an arc of radius 7.5 cm.
 With R as centre, draw an arc of radius 4.5 cm intersecting the previous arc at point E.
 Join OE and RE.
 With O as centre, draw an arc of radius 4.5 cm at point M.
 With E as centre, draw an arc of radius 6 cm intersecting the previous arc at M.
 Join OM and EM.
Therefore, MORE is the required parallelogram.
(iv)
Steps of construction:
 Draw a line segment ET of length 6 cm.
 With E as centre, draw an arc of radius 4.5 cm.
 With T as centre, draw an arc of radius 4.5 cm intersecting the previous arc at point B.
 Join BE and BT.
 With E as centre, draw an arc of radius 4.5 cm at point S.
 With T as centre, draw an arc of radius 4.5 cm intersecting the previous arc at S.
 Join ES and TS.
Therefore, BEST is the required rhombus.
Q.2 Construct the following quadrilaterals.
(i)  Quadrilateral LIFT  (ii)  Quadrilateral GOLD 
LI = 4 cm  OL = 7.5 cm  
IF = 3 cm  GL = 6 cm  
TL = 2.5 cm  GD = 6 cm  
LF = 4.5 cm  LD = 5 cm  
IT = 4 cm  OD = 10 cm  
(iii)  Rhombus BEND  
BN = 5.6 cm  
DE = 6.5 cm 
Ans
Steps of construction:
 Draw a line segment LT of length 2.5 cm.
 With L as centre, draw an arc of radius 4 cm.
 With T as centre, draw an arc of radius 4 cm intersecting the previous arc at point I.
 Join IL and IT.
 With L as centre, draw an arc of radius 4.5 cm at point F.
 With I as centre, draw an arc of radius 3 cm intersecting the previous arc at F.
 Join IF,LF and FT.
Therefore, LIFT is the required quadrilateral.
(ii)
Steps of construction:
 Draw a line segment GD of length 6 cm.
 With G as centre, draw an arc of radius 6 cm.
 With D as centre, draw an arc of radius 6 cm intersecting the previous arc at point L.
 Join GL and DL.
 With D as centre, draw an arc of radius 10 cm at point O.
 With L as centre, draw an arc of radius 7.5 cm intersecting the previous arc at O.
 Join LO,DO and GO.
Therefore, GOLD is the required quadrilateral.
(iii)
Steps of construction:
 Draw a line segment ED of length 6.5 cm.
 With E and D as centres draw the perpendicular bisector of ED.
 Let the lines interesect each other at O.
With O as centre, draw an arcs of radius 2.8 cm on both the sides.  Let the arcs intersect the perpendicular bisector at B and N.
 Join NE, ND BD and BE.
Therefore, BEND is the required rhombus.
Q.3
$\begin{array}{l}\mathrm{Construct}\text{}\mathrm{the}\text{}\mathrm{following}\text{}\mathrm{quadrilaterals}.\\ \left(\mathrm{i}\right)\text{}\mathrm{Quadrilateral}\text{}\mathrm{MORE}\text{}\left(\mathrm{ii}\right)\text{}\mathrm{Quadrilateral}\text{}\mathrm{PLAN}\\ \begin{array}{l}\mathrm{MO}\text{}=\text{}6\text{}\mathrm{cm}\text{}\text{\hspace{0.33em}\hspace{0.33em}}\mathrm{PL}\text{}=\text{}4\text{}\mathrm{cm}\\ \mathrm{OR}\text{}=\text{}4.5\text{}\mathrm{cm}\mathrm{LA}\text{}=\text{}6.5\text{}\mathrm{cm}\\ \angle \mathrm{M}\text{}=\text{}60\xb0\text{}\text{}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\angle \mathrm{P}\text{}=\text{}90\xb0\\ \angle \mathrm{O}\text{}=\text{}105\xb0\text{}\text{}\text{}\angle \mathrm{A}\text{}=\text{}110\xb0\\ \begin{array}{l}\angle \mathrm{R}\text{}=\text{}105\xb0\text{}\angle \mathrm{N}\text{}=\text{}85\xb0\\ \end{array}\\ \left(\mathrm{iii}\right)\text{}\mathrm{Parallelogram}\text{}\mathrm{HEAR}\text{}\left(\mathrm{iv}\right)\text{}\mathrm{Rectangle}\text{}\mathrm{OKAY}\\ \mathrm{HE}\text{}=\text{}5\text{}\mathrm{cm}\text{}\mathrm{OK}\text{}=\text{}7\text{}\mathrm{cm}\\ \mathrm{EA}\text{}=\text{}6\text{}\mathrm{cm}\text{}\mathrm{KA}\text{}=\text{}5\text{}\mathrm{cm}\\ \angle \mathrm{R}\text{}=\text{}85\xb0\end{array}\end{array}$
Ans
Steps of construction:
 Draw a line segment MO of length 6 cm.
 With O as centre, draw an angle of 105°.
 With O as centre, draw an arc of radius 4.5 cm at point R.
 Join OR.
 Draw an angle of 105°at point R.
 With M as centre, draw an angle of 60° and let it meet the previous ray drawn from point R at point E.
Therefore, MORE is the required quadrilateral.
Steps of construction:
 Draw a line segment PL of length 4 cm.
 With L as centre, draw an angle of 75°.
 With L as centre, draw an arc of radius 6.5 cm.
 Join LA.
 With A as centre, draw an angle of 110°.
 With P as centre, draw an angle of 90° and let it meet the previously drawn arc from point A at point N.
Therefore, PLAN is the required quadrilateral.
Steps of construction:
 Draw a line segment EH of length 5 cm.
 With E as centre, draw an angle of 85°.
 With E as centre, draw an arc of radius 6 cm intersecting the previous arc at point A.
 With H as centre, draw an arc of radius 6 cm at point R.
 With A as centre, draw an arc of radius 5 cm intersecting the previous arc at point R.
Therefore, HEAR is the required parallelogram.
Steps of construction:
 Draw a line segment KO of length 7 cm.
 With K as centre, draw an angle of 90°.
 With K as centre, draw an arc of radius 5 cm at point A.
 Join KA.
 With O as centre, draw an arc of radius 5 cm at Y.
 With A as centre, draw an arc of radius 7 cm intersecting the previous arc at Y.
 Join AY and OY.
Therefore, OKAY is the required rectangle.
Q.4
$\begin{array}{l}\mathrm{Construct}\text{}\mathrm{the}\text{}\mathrm{following}\text{}\mathrm{quadrilaterals}.\\ \left(\text{i}\right)\text{}\mathrm{Quadrilateral}\text{}\mathrm{DEAR}\text{}\left(\mathrm{ii}\right)\text{}\mathrm{Quadrilateral}\text{}\mathrm{TRUE}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{DE}\text{}=\text{}4\text{}\mathrm{cm}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{TR}\text{}=\text{}3.5\text{}\mathrm{cm}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{EA}\text{}=\text{}5\text{}\mathrm{cm}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{RU}\text{}=\text{}3\text{}\mathrm{cm}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{AR}=\text{}4.5\text{}\mathrm{cm}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{UE}\text{}=\text{}4\text{}\mathrm{cm}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\angle \text{E}=\text{}60\xb0\text{\hspace{0.33em} \hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\angle \text{R}=\text{}75\xb0\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\angle \text{A}=\text{}90\xb0\text{\hspace{0.33em} \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\angle \text{U}=\text{}120\xb0\text{}\end{array}$
Ans
Steps of construction:
 Draw a line segment DE of length 4 cm.
 With E as centre, draw an angle of 60°.
 With E as centre and radius 5 cm, draw an arc intersecting the previous ray at point A.
 Join EA.
 With A as centre, draw an angle of 90°.
 With A as centre and radius 4.5 cm, draw an arc intersecting the previous ray at point cm at R.
 Join AR and RD.
Therefore, DEAR is the required quadrilateral.
(ii)
Steps of construction:
 Draw a line segment RU of length 3 cm.
 With R as centre, draw an angle of 75°.
 With R as centre and radius 3.5 cm, draw an arc intersecting the previous ray at point T.
 Join RT.
 With U as centre, draw an angle of 120°.
 With U as centre and radius 4 cm, draw an arc intersecting the previous ray at point cm at E.
 Join UE and ET.
Therefore, TRUE is the required quadrilateral.
Q.5 Draw the following.
The square READ with RE = 5.1 cm.
Ans
Steps of construction:
 Draw a line segment RU of length 5.1 cm.
 With R as centre, draw an angle of 90°.
 With R as centre and radius 5.1 cm, draw an arc intersecting the previous ray at point D.
 Join RD.
 With E as centre, draw an angle of 90°.
 With E as centre and radius 5.1 cm, draw an arc intersecting the previous ray at point cm at A.
 Join EA and AD.
Therefore, TRUE is the required square.
Q.6 A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
Ans
Steps of construction:
 Draw a line segment AC of length 5.2 cm.
 Draw the perpendicular bisector of AC intersecting it at point O.
 With A and C as centres, draw arcs on both the sides of the perpendicular bisector of radius 3.2 cm.
 Let the arcs intersect at point B and D.
 Join AB,AD,BC and CD.
Therefore, ABCD is the required rhombus.
Q.7 Draw a rectangle with adjacent sides of lengths 5 cm and 4 cm.
Ans
Steps of construction:
 Draw a line segment AB of length 5 cm.
 With A as centre, draw an angle of 90°.
 With B as centre, draw an angle of 90°.
 With A and B as centres, draw arcs of radius 4 cm, intersecting the previous line segments at C and D.
 Join CD.
Therefore, ABCD is the required rectangle.
Q.8 Draw a parallelogram OKAY where OK = 5.5 cm and KA= 4.2 cm. Is it unique?
Ans
Steps of construction:
 Draw a line segment OK of length 5.5 cm.
 With K as centre, draw a ray at a convenient angle.
 With K as centre and radius 4.2 cm, draw an arc intersecting the previous arc at A.
 Join KA
 With O as centre,draw a ray parallel to KA.
 With O as centre and radius 4.2 cm, draw an arc intersecting the previous arc at Y.
 Join OY and AY.
Therefore, OKAY is the required parallelogram.
No, the parallelogram OKAY is not unique.
For constructing any quadrilateral, we should have 5 measurements but here, we are given only 4 measurements. So it can be drawn using various different angles.