NCERT Solutions for Class 8 Maths Chapter 4 – Quadrilaterals

NCERT Solutions for Class 8 Maths Chapter 4 – Quadrilaterals are designed to help students clearly understand four-sided figures, their properties, and angle relationships. This chapter forms an essential part of geometry in Class 8 Mathematics and builds the foundation for advanced geometric concepts taught in higher classes.

Quadrilaterals are widely used in geometry and real-life applications. A thorough understanding of this chapter enhances spatial reasoning, logical thinking, and problem-solving ability.

These solutions are prepared strictly according to the latest NCERT syllabus prescribed by the National Council of Educational Research and Training (NCERT) and follow the CBSE examination pattern. Each concept is explained step by step to support effective learning and confident exam preparation.

1.0NCERT Solutions for Class 8 Maths Chapter 4 PDF Download

Students can download the Quadrilaterals Class 8 PDF with answers from the table below. The NCERT Solutions for Class 8 Maths Chapter 4 – Quadrilaterals PDF are meticulously prepared by ALLEN’s expert faculty members, who possess in-depth subject knowledge and a strong understanding of the NCERT and CBSE curriculum.

These solutions help students understand geometric concepts clearly and apply them accurately in examinations.

3.0NCERT Questions with Solutions Class 8 Maths Chapter 4 – Quadrilaterals Detailed Solutions

1. Find all the other angles inside the following rectangles.
   
   
   Solution (i) The given rectangle is $ABCD$.
   
   We have $\angle 1=30^{\circ }$ $\angle 1+\angle 2=90^{\circ }$ $\therefore \angle 2=90^{\circ }-\angle 1=90^{\circ }-30^{\circ }=60^{\circ }$ $\mathrm{MD}=\mathrm{MA}$ $\Rightarrow \angle 3=\angle 2=60^{\circ }$ $\angle 3+\angle \mathrm{Z}4=90^{\circ }$ $\therefore \angle 4=90^{\circ }-\angle 3=90^{\circ }-60^{\circ }=30^{\circ }$ $\mathrm{MC}=\mathrm{MD}$ $\Rightarrow \angle 5=\angle 4=30^{\circ }$ $\angle 5+\angle 6=90^{\circ }$ $\therefore \angle 6=90^{\circ }-\angle 5=90^{\circ }-30^{\circ }=60^{\circ }$ $\mathrm{MB}=\mathrm{MC}$ $\Rightarrow \angle 7=\angle 6=60^{\circ }$ $\mathrm{MB}=\mathrm{MA}$ $\angle 8=\angle 1=30^{\circ }$ In $△\mathrm{AMB}$, we have $\angle 1+\angle 9+\angle 8=180^{\circ }$ $\therefore 30^{\circ }+\angle 9+30^{\circ }=180^{\circ }$ $\therefore \angle 9+180^{\circ }-60^{\circ }=120^{\circ }$ $\angle 11=\angle 9=120^{\circ }$ (Vertically opposite angles) $\angle 9+\angle 10=180^{\circ }$ (Linear angles) $\therefore \angle 10=180^{\circ }-120^{\circ }=60^{\circ }$ $\angle 12=\angle 10=60^{\circ }$ (Vertically opposite angles) $\therefore \angle 2=60^{\circ },\angle 3=60^{\circ },\angle 4=30^{\circ },\angle 5=30^{\circ }$, $\angle 6=60^{\circ },\angle 7=60^{\circ },\angle 8=30^{\circ },\angle 9=120^{\circ }$, $\angle 10=60^{\circ },\angle 11=120^{\circ }$, and $\angle 12=60^{\circ }$
   (ii) The given rectangle is PSRQ.
   
   We have $\angle 9=110^{\circ }$ $\angle 11+\angle 9=110^{\circ }$ (Vertically opposite angles) $\therefore \angle 9+\angle 10^{\circ }=180^{\circ }$ (Linear angles) $\mathrm{MD}=\mathrm{MA}$ $\Rightarrow \angle 3=\angle 2=60^{\circ }$ $\angle 3+\angle \mathrm{Z}4=90^{\circ }$ $\therefore \angle 4=90^{\circ }-\angle 3=90^{\circ }-60^{\circ }=30^{\circ }$ $\mathrm{MC}=\mathrm{MD}$ $\Rightarrow \angle 5=\angle 4=30^{\circ }$ $\angle 5+\angle 6=90^{\circ }$ $\therefore \angle 6=90^{\circ }-\angle 5=90^{\circ }-30^{\circ }=60^{\circ }$ $\mathrm{MB}=\mathrm{MC}$ $\Rightarrow \angle 7=\angle 6=60^{\circ }$ $\mathrm{MB}=\mathrm{MA}$ $\angle 8=\angle 1=30^{\circ }$ In $△\mathrm{AMB}$, we have $\angle 1+\angle 9+\angle 8=180^{\circ }$ $\therefore 30^{\circ }+\angle 9+30^{\circ }=180^{\circ }$ $\therefore \angle 9+180^{\circ }-60^{\circ }=120^{\circ }$ $\angle 11=\angle 9=120^{\circ }$ (Vertically opposite angles) $\angle 9+\angle 10=180^{\circ }$ (Linear angles) $\therefore \angle 10=180^{\circ }-120^{\circ }=60^{\circ }$ $\angle 12=\angle 10=60^{\circ }$ (Vertically opposite angles) $\therefore \angle 2=60^{\circ },\angle 3=60^{\circ },\angle 4=30^{\circ },\angle 5=30^{\circ }$, $\angle 6=60^{\circ },\angle 7=60^{\circ },\angle 8=30^{\circ },\angle 9=120^{\circ }$, $\angle 10=60^{\circ },\angle 11=120^{\circ }$, and $\angle 12=60^{\circ }$,
2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of (i) $30^{\circ }$ (ii) $40^{\circ }$ (iii) $90^{\circ }$ (iv) $140^{\circ }$
   Solution
   (i) Draw a line AB equal to 8 cm . Take point M on AB such that $\mathrm{AM}=\mathrm{BM}=4\mathrm{cm}$. Using a protractor, draw an angle of $30^{\circ }$ at M on MB . On this line, take points C and D such that $\mathrm{MC}=\mathrm{MD}=4\mathrm{cm}$. Join $\mathrm{AD},\mathrm{DB},\mathrm{BC}$, and CA . ABCD is the required quadrilateral.
   
   Since diagonals AB and CD are equal and are bisecting each other at $\mathrm{M},\mathrm{ACBD}$ is a rectangle.
   (ii) Draw a line AB equal to 8 cm. Take point M on AB such that $\mathrm{AM}=\mathrm{BM}=$ 4 cm . Using a protractor, draw an angle of $40^{\circ }$ at M on MB. On this line, take points C and D such that $\mathrm{MC}=\mathrm{MD}=4\mathrm{cm}$. Join $\mathrm{AD},\mathrm{DB},\mathrm{BC}$, and CA . ABCD is the required quadrilateral.
   
   Since diagonals AB and CD are equal and are bisecting each other at $\mathrm{M},\mathrm{ACBD}$ is a rectangle.
   
   (iii) Draw a line AB equal to 8 cm.
   Take a point M on AB such that $\mathrm{AM}=\mathrm{BM}=4\mathrm{cm}$. Using a protractor, draw an angle of $90^{\circ }$ at M on MB.
   On this line, take points C and D such that $\mathrm{MC}=\mathrm{MD}=4\mathrm{cm}$.
   Join $\mathrm{AD},\mathrm{DB},\mathrm{BC}$, and CA . ABCD is the required square.

• (iv) Draw a line AB equal to 8 cm. Take point M on AB such that $\mathrm{AM}=\mathrm{BM}=$ 4 cm . Using a protractor, draw an angle of $140^{\circ }$ at M on MB. On this line, take points C and D such that $\mathrm{MC}=\mathrm{MD}=4\mathrm{cm}$. Join $\mathrm{AD},\mathrm{DB},\mathrm{BC}$, and CA . ABCD is the required quadrilateral.

• (iv) Let AB and CD be two sticks of equal length, say 6 cm. Mark the midpoints of the sticks using a ruler. Fix a screw to the sticks at their midpoints. Using a thread, measure distances AD and BD. Keep on moving the sticks about the screw, so that the distances AD and BD are equal.
  
  In this position, fix the sticks by tightening the screw. The new positions of the sticks are shown in the figure. Tie pieces of thread along AD and BD .
  
  $\therefore$ By the SSS condition, $△\mathrm{AMD}$ and $△\mathrm{BMD}$ are congruent. $\therefore \angle \mathrm{AMD}=\angle \mathrm{BMD}$ Also $\angle \mathrm{AMD}+\angle \mathrm{BMD}=180^{\circ }$ (Linear angles) $\therefore \angle \mathrm{AMD}+\angle \mathrm{AMD}=180^{\circ }$ $\Rightarrow 2\angle \mathrm{AMD}=180^{\circ }$ $\Rightarrow \angle \mathrm{AMD}=90^{\circ }$ $\therefore \angle \mathrm{AMD}=\angle \mathrm{BMD}=90^{\circ }$ ∴ Angle between the sticks is $90^{\circ }$

• We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle? Solution Let ABCD be a quadrilateral in which opposite sides are parallel and equal. Here $AB\Vert DC$ and $AD\Vert BC$. Also, $\mathrm{AB}=\mathrm{DC}$ and $\mathrm{AD}=\mathrm{BC}$. In the quadrilateral ABCD , opposite sides are equal. For ABCD to be a rectangle, we require each angle to be $90^{\circ }$. Given information $\mathrm{AB}\Vert \mathrm{DC}$ and $\mathrm{AD}\Vert \mathrm{BC}$ can not help us to prove that each angle of ABCD is $90^{\circ }$.
  
  $\therefore \mathrm{ABCD}$ may not be a rectangle. ∴ A rectangle can not be defined as a quadrilateral with equal and parallel opposite sides. 4.4 Quadrilaterals with Equal Sidelengths

Figure it out (Pg 102)

• Find the remaining angles in the following quadrilaterals. (i)
  
  (ii)
  
  (iii)
  
  $\backslash \backslash 172.26.70.208\backslash$ SMD_Data$\PNCF\2026-27\Print Module SET-1 NCERT\Mathematics $\backslash 8$ th\3. A Story of Numbers $\Rightarrow \angle \mathrm{AEP}=180^{\circ }-40^{\circ }=140^{\circ }$ We know that opposite angles of a parallelogram are equal. $\therefore \angle \mathrm{EAR}=\angle \mathrm{EPR}=40^{\circ }$ and $\angle \mathrm{ARP}=\angle \mathrm{AEP}=140^{\circ }$. (ii) Since opposite sides of the quadrilateral PQRS are parallel, this is a parallelogram. In the given parallelogram, PQ is a transversal to the parallel lines PS and QR. $\angle \mathrm{SPQ}$ and $\angle \mathrm{RQP}$ are the internal angles on the same sides of the parallel lines. $\therefore \angle \mathrm{SPQ}+\angle \mathrm{RQP}=180^{\circ }$ $\Rightarrow 110^{\circ }+\angle \mathrm{RQP}=180^{\circ }$ $\Rightarrow \angle \mathrm{RQP}=180^{\circ }-110^{\circ }=70^{\circ }$ We know that the opposite angles of a parallelogram are equal. $\therefore \angle \mathrm{QRS}=\angle \mathrm{QPS}=110^{\circ }$ and $\angle \mathrm{PSR}=\angle \mathrm{RQP}=70^{\circ }$. (iii) Here, quadrilateral UVWX is a rhombus, because all sides are equal. We have $\angle 1=30^{\circ }$. In a rhombus, diagonals bisect the angles of the rhombus. $\therefore \angle 6=30^{\circ }$ In $△UVX,\angle 1=\angle 3(\because UV=UX)$ $\therefore \angle 3=30^{\circ }$ $\therefore \angle 4=30^{\circ }$
  
  In $△\mathrm{UVX},\angle 1+\angle 2+\angle 3=180^{\circ }$. $\therefore 30^{\circ }+\angle 2+30^{\circ }=180^{\circ }$ $\Rightarrow \angle 2=180^{\circ }-60^{\circ }=120^{\circ }$ $\Rightarrow \angle 5=\angle 2=120^{\circ }(\because$ Opposite angles are (i) Since opposite sides of the quadrilateral PEAR are parallel, this is a parallelogram. In the given parallelogram, PE is a transversal to the parallel lines PR and EA. $\angle \mathrm{RPE}$ and $\angle \mathrm{AEP}$ are the internal angles on (iv)
  
  Solution Solution (i) Since opposite sides of the quadrilateral PEAR are parallel, this is a parallelogram. In the given parallelogram, PE is a ransversal to the parallel lines PR and EA. A.CP $\angle \mathrm{RPE}$ and $\angle \mathrm{AEP}$ are the internal angles on $\begin{array}{lccccc}\text{ the }& \text{ same }& \text{ side }& \text{ of parallel }& \text{ lines. }& \\ \therefore & \angle \mathrm{RPE}& +& \angle \mathrm{AEP}& =& 180^{\circ }\\ \Rightarrow & 40^{\circ }& +& \angle \mathrm{AEP}& =& 180^{\circ }\end{array}$ equal) $\therefore \angle 2=120^{\circ },\angle 3=30^{\circ },\angle 4=30^{\circ },\angle 5=120^{\circ }$ and $\angle 6=30^{\circ }$. (iv) Here, quadrilateral UVWX is a rhombus, because all sides are equal. We have $\angle 1=20^{\circ }$. In a rhombus, diagonals bisect the angles of the rhombus. $\therefore \angle 6=20^{\circ }$ In $△UVX,\angle 1=\angle 3(\because UV=UX)\therefore \angle 3=20^{\circ }$ $\therefore \angle 4=20^{\circ }$
  
  In $△UVX,\angle 1+\angle 2+\angle 3=180^{\circ }$. $\therefore 20^{\circ }+\angle 2+20^{\circ }=180^{\circ }$ $\Rightarrow \angle 2=180^{\circ }-40^{\circ }=140^{\circ }$ $\Rightarrow \angle 5=\angle 2=140^{\circ }(\because$ Opposite angles are equal) $\therefore \angle 2=140^{\circ },\angle 3=40^{\circ },\angle 4=40^{\circ },\angle 5=120^{\circ }$ and $\angle 6=40^{\circ }$.
• Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm , and intersect at an angle of $140^{\circ }$. Solution Draw a line AB equal to 7 cm . Take point O on AB such that $\mathrm{AO}=\mathrm{OB}=3.5\mathrm{cm}$.
  
  On OB, draw an angle of $140^{\circ }$ at 0 . Take points C and D on the line of angle so that $\mathrm{OC}=\mathrm{OD}=2.5\mathrm{cm}$. $\therefore \mathrm{CD}=5\mathrm{cm}$, and O is the midpoint of AB and AC . Join $\mathrm{AC},\mathrm{CB},\mathrm{BD}$, and DA . ACBD is a quadrilateral, and its diagonals AB and CD bisect at 0 . $\therefore \mathrm{ACBD}$ is the required parallelogram.
• Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm . Solution Draw a line AB equal to 4 cm . Take a point O on AB such that $\mathrm{AO}=\mathrm{OB}=$ 2 cm . On AB, draw a line perpendicular to it and passing through 0 . Take points C and D on this perpendicular so that $\mathrm{OC}=\mathrm{OD}=2.5\mathrm{cm}$.
  
  $\therefore \mathrm{CD}=5\mathrm{cm}$, and O is the midpoint of AB and CD . Join $\mathrm{AC},\mathrm{CB},\mathrm{BD}$, and DA . ACBD is a quadrilateral, and its diagonals AB and CD are bisecting at O and are also perpendicular to each other. $\therefore \mathrm{ACBD}$ is the required rhombus.

Figure it out (Pg 107)

• Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm . Solution Let two equilateral triangles of sides 4 cm be joined as shown below.
  
  The sides of the quadrilateral ABCD are 4 cm each. The angles of the quadrilateral $ABCD$ are $\angle \mathrm{A}=60^{\circ }+60^{\circ }=120^{\circ },\angle \mathrm{B}=60^{\circ },\angle \mathrm{C}=60^{\circ }+60^{\circ }=120^{\circ }$ and $\angle \mathrm{D}=60^{\circ }$.
• Construct a kite whose diagonals are of lengths 6 cm and 8 cm . Solution
• Find the remaining angles in the following trapeziums-
  
  Solution (i) Let the given trapezium be ABCD . Lines AB and DC are Parallel
  
  $\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$ and $\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$. $\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$ $\Rightarrow 135^{\circ }+\angle \mathrm{D}=180^{\circ }$ $\Rightarrow \angle \mathrm{D}=180^{\circ }-135^{\circ }=45^{\circ }$. $\therefore \angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow 105^{\circ }+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow \angle C=180^{\circ }-105^{\circ }=75^{\circ }$. ∴ The remaining angles are $45^{\circ }$ and $75^{\circ }$. (ii) Let the given trapezium be ABCD . Since $\mathrm{AD}=\mathrm{BC},\mathrm{ABCD}$ is an isosceles trapezium.
  
  ∴ Angles opposite to the equal sides are equal. $\therefore \angle \mathrm{C}=\angle \mathrm{D}=100^{\circ }$ Lines AB and DC are parallel. $\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$ and $\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ $\therefore \angle \mathrm{A}+\angle \mathrm{D}=180^{\circ }$ $\Rightarrow \angle \mathrm{A}+100^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{A}=180^{\circ }-100^{\circ }$ $\Rightarrow \angle \mathrm{A}=80^{\circ }$ $\therefore \angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow \angle \mathrm{B}+100^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{B}=180^{\circ }-100^{\circ }$ $\Rightarrow \angle \mathrm{B}=80^{\circ }$ ∴ Remaining angles are $\angle \mathrm{A}=80^{\circ },\angle \mathrm{B}=80^{\circ }$ and $\angle \mathrm{C}=100^{\circ }$.
• Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions- (i) What is the quadrilateral that is both a kite and a parallelogram? (ii) Can there be a quadrilateral that is both a kite and a rectangle? (iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals? Solution We know the following:

• Every rectangle is a parallelogram.
• Every square is a rectangle.
• Every square is a rhombus.
• Every rhombus is a kite. The following Venn diagram shows the sets of parallelograms, kites, rhombuses, rectangles, and squares
  
  (i) The set of rhombuses is common to both the set of kites and the set of parallelograms. ∴ A rhombus is both a kite and a parallelogram. (ii) A kite is not a rectangle, and a rectangle is not a kite. ∴ There can be no quadrilateral that is both a kite and a rectangle. Also, there is no common portion of the set of kites and the set of rectangles. (iii) Every kite is not a rhombus. In the given figure, the kite ABCD is not a rhombus.
  
  The correct relationship is that every rhombus is a kite.

• If PAIR and RODS are two rectangles, find $\angle \mathrm{IOD}$.
  
  Solution From O, draw a line OK parallel to RI.
  
  $\therefore \angle \mathrm{KOR}=\angle \mathrm{ORI}=30^{\circ }$ (Alternate angles) $\therefore \angle \mathrm{KOR}+\angle \mathrm{ROI}=90^{\circ }$ $\Rightarrow 30^{\circ }+\angle \mathrm{ROI}=90^{\circ }$ $\Rightarrow \angle \mathrm{ROI}=90^{\circ }-30^{\circ }$ $\Rightarrow \angle \mathrm{ROI}=60^{\circ }$ Also, $\angle \mathrm{ROI}+\angle \mathrm{IOD}=90^{\circ }$ $\Rightarrow 60^{\circ }+\angle \mathrm{IOD}=90^{\circ }$ $\Rightarrow \angle \mathrm{IOD}=90^{\circ }-60^{\circ }$ $\Rightarrow \angle \mathrm{IOD}=30^{\circ }$.
• Construct a square with diagonal 6 cm without using a protractor. Solution Draw a line AB equal to 6 cm .
  
  We have $6÷2=3$. With centre at A and B, draw arcs of radius slightly greater than 3 cm , say, 4 cm. Join the points of intersection of the arcs. Let this line intersect AB at 0 . Take points $C$ and $D$ on the perpendicular line so that $\mathrm{OC}=\mathrm{OD}=3$
  
  (a)
  
  (b) cm. Join $\mathrm{AC},\mathrm{CB},\mathrm{BD}$, and DA . ACBD is the required square with diagonals equal to 6 cm .
  
  7. CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b). Solution (a) U, V, W, and X are the midpoints of the sides of the square.
  In $△\mathrm{VCU}$ and $△\mathrm{UAX}$, we have $\mathrm{VC}=\mathrm{UA},\angle \mathrm{VCU}=\angle \mathrm{UAX}=90^{\circ }$, and $\mathrm{CU}=\mathrm{AX}$. ∴ By the SAS condition, $△\mathrm{VCU}$ and $△\mathrm{UAX}$ are congruent. $\therefore \mathrm{VU}=\mathrm{UX}$ Similarly, $\mathrm{VU}=\mathrm{XW},\mathrm{VU}=\mathrm{WV}$. ∴ Sides of the quadrilateral UVWX are equal.
  
  In $△\mathrm{VCU},\mathrm{VC}=\mathrm{CU}$ $\Rightarrow \angle 1=\angle 2$ Also, $\angle 1+\angle \mathrm{C}+\angle 2=180^{\circ }$ $\Rightarrow \angle 1+90^{\circ }+\angle 1=180^{\circ }$ $\Rightarrow 2\angle 1=90^{\circ }$ $\Rightarrow \angle 1=45^{\circ }$ $\therefore \angle 2$ is also $45^{\circ }$. Similarly, $\angle 3=\angle 4=45^{\circ }$ We have $\angle 2+\angle \mathrm{VUX}+\angle 3=180^{\circ }$ $\Rightarrow 45^{\circ }+\angle \mathrm{VUX}+45^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{VUX}=180^{\circ }-90^{\circ }$ $\Rightarrow \angle \mathrm{VUX}=90^{\circ }$ Similarly, $\angle \mathrm{VXW}=90^{\circ },\angle \mathrm{XWV}=90^{\circ }$ and $\angle \mathrm{WVU}=90^{\circ }$. ∴ By definition, the quadrilateral UVWX is a square.
  (b) Let ABCD be a square. Take points $\mathrm{P},\mathrm{Q},\mathrm{R}$, and S such that $\mathrm{AS}=\mathrm{BP}=\mathrm{CQ}=\mathrm{DR}$.
  
  Since the sides of squares are equal, we have $\mathrm{DS}=\mathrm{AP}=\mathrm{BQ}=\mathrm{CR}$. In $△\mathrm{PAS}$ and $△\mathrm{SDR}$, we have $\mathrm{PA}=\mathrm{SD},\angle \mathrm{PAS}=\angle \mathrm{SDR}=90^{\circ }$, and $\mathrm{AS}=\mathrm{DR}$. ∴ By the SAS condition, $△\mathrm{PAS}$ and $△\mathrm{SDR}$ are congruent. $\therefore \mathrm{PS}=\mathrm{SR}$ Similarly, $\mathrm{PS}=\mathrm{RQ},\mathrm{PS}=\mathrm{QP}$. ∴ Sides of the quadrilateral PQRS are equal. In $△\mathrm{PAS},\angle 1+\angle 2+90^{\circ }=180^{\circ }$ $\Rightarrow \angle 1+\angle 2=90^{\circ }$ $\Rightarrow \angle 3+\angle 2=90^{\circ }(\because \angle 1=\angle 3)$ Also, $\angle 2+\angle 4+\angle 3=180^{\circ }$ $\Rightarrow 90^{\circ }+\angle 4=180^{\circ }$ $\Rightarrow \angle 4=180^{\circ }-90^{\circ }$ $\Rightarrow \angle 4=90^{\circ }$ ∴ Similarly, $\angle 5=90^{\circ },\angle 6=90^{\circ }$, and $\angle 7=90^{\circ }$. By definition, the quadrilateral PQRS is a square.
  
  8. If a quadrilateral has four equal sides and one angle of $90^{\circ }$, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement. Solution Let $ABCD$ be a quadrilateral such that $AB=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ and $\angle \mathrm{DAB}=90^{\circ }$. Join BD.
  
  In $△\mathrm{ADB}$ and $△\mathrm{CDB}$, we have $\mathrm{AD}=\mathrm{CD},\mathrm{AB}=\mathrm{CB}$, and DB is a common side. $\therefore △\mathrm{ADB}$ and $△\mathrm{CDB}$ are congruent. $\therefore \angle \mathrm{C}=\angle \mathrm{A}=90^{\circ }$ In $△\mathrm{DAB},\angle 1=\angle 2(\because \mathrm{AB}=\mathrm{AD})$ Also, $\angle 1+90^{\circ }+\angle 2=180^{\circ }$ $\Rightarrow \angle 1+\angle 2=90^{\circ }$ $\Rightarrow \angle 1=45^{\circ }$ and $\angle 2=45^{\circ }(\because \angle 1=\angle 2)$ In $△\mathrm{CDB},\angle 3=\angle 4(\because \mathrm{CD}=\mathrm{CB})$ Also, $\angle 3+90^{\circ }+\angle 4=180^{\circ }$ $\Rightarrow \angle 3+\angle 4=90^{\circ }$ $\Rightarrow \angle 3=\angle 4=45^{\circ }(\because \angle 3=\angle 4)$ $\therefore \angle \mathrm{ABC}=\angle 1+\angle 4=45^{\circ }+45^{\circ }=90^{\circ }$ and $\angle \mathrm{ADC}=\angle 2+\angle 3=45^{\circ }+45^{\circ }=90^{\circ }$. ∴ Each angle of the quadrilateral ABCD is $90^{\circ }$. $\therefore \mathrm{ABCD}$ is a square. Also, by measurement, we find $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$ and $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ }$.
  
  9. What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer. Hint: Draw a diagonal and check for congruent triangles. Solution Let ABCD be a quadrilateral in which opposite sides are equal. Join AC.
  
  In $△\mathrm{ADC}$ and $△\mathrm{CBA}$, we have $\mathrm{AD}=\mathrm{CB},\mathrm{DC}=\mathrm{BA}$, and AC is common. ∴ By the SSS condition, AADC and ACBA are congruent. $\therefore \angle 1=\angle 3$ and $\angle 2=\angle 4$ AC is a transversal of lines AB and DC , and alternate angles $\angle 1$ and $\angle 3$ are equal. ∴ Lines AB and DC are parallel. AC is a transversal of lines AD and BC , and alternate angles $\angle 2$ and $\angle 4$ are equal. ∴ Lines AD and BC are parallel. ∴ By definition, the quadrilateral ABCD is a parallelogram.
  
  10. Will the sum of the angles in a quadrilateral such as the following one also be $360^{\circ }$ ? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
  
  Solution : In the given quadrilateral, join BD . In $△\mathrm{ABD}$, we have $\angle \mathrm{A}+\angle 3+\angle 1=180^{\circ }$ In $△\mathrm{CBD}$, we have $\angle \mathrm{C}+\angle 4+\angle 2=180^{\circ }$
  
  Adding, we get $(\angle \mathrm{A}+\angle 3+\angle 1)+(\angle \mathrm{C}+\angle 4+\angle 2)=180^{\circ }+180^{\circ }$ $\Rightarrow \angle \mathrm{A}+(\angle 3+\angle 4)+\angle \mathrm{C}+(\angle 1+\angle 2)=360^{\circ }$ $\Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ }$ ∴ The sum of the angles of the quadrilateral ABCD is $360^{\circ }$. Also, by using a protractor, we find that the sum of all angles is $360^{\circ }$.
  
  11. State whether the following statements are true or false. Justify your answers. (i) A quadrilateral whose diagonals are equal and bisect each other must be a square. (ii) A quadrilateral having three right angles must be a rectangle. (iii)A quadrilateral whose diagonals bisect each other must be a parallelogram. (iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus. (v) A quadrilateral in which the opposite angles are equal must be a parallelogram. (vi) A quadrilateral in which all the angles are equal is a rectangle. (vii) Isosceles trapeziums are parallelograms. Solution
  (i) A quadrilateral whose diagonals are equal and bisect each other need not be a square. In the figure, diagonals AC and DB are equal and bisect each other. Such a quadrilateral is always a rectangle.
  
  ∴ The given statement is false.
  (ii) Let ABCD be a quadrilateral having three right angles at $\mathrm{A},\mathrm{D}$, and C . We have $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}=360^{\circ }$. $\Rightarrow 90^{\circ }+\angle \mathrm{B}+90^{\circ }+90^{\circ }=360^{\circ }$ $\Rightarrow \angle \mathrm{B}=360^{\circ }-270^{\circ }$ $\Rightarrow \angle \mathrm{B}=90^{\circ }$. ∴ Each angle of ABCD is $90^{\circ }$. ∴ Given quadrilateral is a rectangle.
  
  ∴ The given statement is true
  (iii) In the quadrilateral ABCD , the diagonals AC and BD bisect each other. Here, $△\mathrm{AOD}$ and $△\mathrm{COB}$ are congruent. $\therefore \angle 1=\angle 2$
  
  $\therefore \mathrm{BC}$ is parallel to AD . Also, $△\mathrm{AOB}$ and $△\mathrm{COD}$ are congruent. $\therefore \angle 3=\angle 4$ $\therefore \mathrm{AB}$ is parallel to DC . Since opposite sides of ABCD are parallel, it must be a parallelogram. ∴ The given statement is true.
  (iv) Let ABCD be a quadrilateral whose diagonals AC and BD are perpendicular to each other.
  
  This quadrilateral may not be a rhombus, because the diagonals AC and BC may not bisect each other. ∴ The given statement is false.
  (v) Let ABCD be a quadrilateral in which $\angle 1=\angle 3$ and $\angle 2=\angle 4$. We have, $\angle 1+\angle 2+\angle 3+\angle 4=360^{\circ }$. $\Rightarrow \angle 1+\angle 2+\angle 1+\angle 2=360^{\circ }$ $\Rightarrow 2(\angle 1+\angle 2)=360^{\circ }$ $\Rightarrow \angle 1+\angle 2=180^{\circ }$
  
  AB is a transversal of lines AD and BC , and the sum of internal angles $\angle 1$ and $\angle 2$ on the same side is $180^{\circ }$. ∴ Lines AD and BC are parallel. Again, $\angle 1+\angle 2+\angle 3+\angle 4=360^{\circ }$ $\Rightarrow \angle 3+\angle 2+\angle 3+\angle 2=360^{\circ }$ $\Rightarrow 2(\angle 2+\angle 3)=360^{\circ }$ $\Rightarrow \angle 2+\angle 3=180^{\circ }$ BC is a transversal of lines AB and DC , and the sum of internal angles $\angle 2$ and $\angle 3$ on the same sides is $180^{\circ }$. ∴ Lines AB and DC are parallel. ∴ Opposite sides of quadrilateral ABCD are parallel. $\therefore \mathrm{ABCD}$ is a parallelogram. ∴ The given statement is true.
  (vi) Let ABCD be a quadrilateral, where $\angle 1,\angle 2$, $\angle 3$, and $\angle 4$ are all equal. We have $\angle 1+\angle 2+\angle 3+\angle 4=360^{\circ }$ $\therefore \angle 1+\angle 1+\angle 1+\angle 1=360^{\circ }$ $\Rightarrow 4\angle 1=360^{\circ }$ $\Rightarrow \angle 1=90^{\circ }$
  
  $\therefore \angle 2=90^{\circ },\angle 3=90^{\circ },\angle 4=90^{\circ }$ We have, $\angle 5+\angle 6=90^{\circ }$ and $\angle 6+90^{\circ }+\angle 8=180^{\circ }$ $\Rightarrow \angle 5+\angle 6=\angle 6+\angle 8$ $\Rightarrow \angle 5=\angle 8$ Also, $\angle 7+90^{\circ }+\angle 5=180^{\circ }$ $\Rightarrow \angle 7+\angle 5=90^{\circ }$ $\Rightarrow \angle 5+\angle 6=\angle 7+\angle 5$ $\Rightarrow \angle 6=\angle 7$ In $△\mathrm{DAB}$ and $△\mathrm{BCD}$, we have $\angle 5=\angle 8,\angle 7=\angle 6$, and side BD is common. ∴ By the ASA condition, $△\mathrm{DAB}$ and $△\mathrm{BCD}$ are congruent. $\therefore \mathrm{DA}=\mathrm{BC}$ and $\mathrm{AB}=\mathrm{CD}$ ∴ Opposite sides of ABCD are equal. $\therefore \mathrm{ABCD}$ is a rectangle. ∴ The given statement is true
  (vii) An isosceles trapezium ABCD can not be a parallelogram because it has two nonparallel equal lines AD and BC .
  
  ∴ The given statement is false.

4.0Key Topics Covered in Class 8 Maths Chapter -4 

Chapter 4 on Quadrilaterals in the 8th grade (Class VIII) Mathematics book will explore the characteristics of all types of four-sided shapes and how they relate to one another. The following are the main points of focus for this chapter:

• Definition of Quadrilaterals: This section will provide students with a general definition for quadrilaterals: polygons with four sides, four angles and four vertices.
• Different Types of Quadrilaterals: Students will learn about different types of quadrilaterals: parallelograms, rectangles, squares, rhombuses, trapeziums and kites.
• Angle Sum Property Of Quadrilaterals: Students will be taught that the sum total of the angles in a quadrilateral equals 360°.
• Properties Of Parallelograms: This topic introduces students to properties of parallelograms, including the relationship between opposite sides, opposite angles and diagonals.
• Properties Of Special Quadrilaterals: In this section, students will learn about specific properties of rectangle, square, trapezium, and rhombus.
• Properties Of Diagonals Of Quadrilaterals: In this area, students will be studying the relationships between diagonals and their properties.
• Symmetry Within Quadrilaterals: Teachers are expected to provide students with enough examples of symmetric quadrilaterals so they understand symmetry.
• Application Problems About Quadrilaterals: Students will be asked to do word problems related to quadrilaterals through their application of knowledge about quadrilaterals and problem solving related to quadrilaterals.

By learning these ideas about quadrilaterals, students will develop a strong foundation in geometry that will enable them to take more advanced classes in geometry later in their schooling career.

5.0Benefits of Studying Class 8 Maths Chapter 4 – Quadrilaterals

The NCERT questions in this chapter focus on understanding the properties of quadrilaterals and applying geometric rules to solve problems. These step-by-step explanations help students avoid common errors and improve accuracy in geometry questions along with that :

• Improves understanding of geometric shapes and their properties
• Strengthens logical reasoning and spatial thinking
• Builds a strong foundation for advanced geometry topics
• Enhances problem-solving and diagram-based question skills
• Supports better performance in school examinations