NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

The NCERT Solutions for Class 8 Maths Chapter 3 Exercise 3.3 explains one of the most important parts of the chapter, the types of special quadrilaterals. In this exercise you will learn to classify different types of quadrilaterals such as parallelograms, rhombuses, rectangles, and squares based on their sides and the angles they form. Understanding these shapes will allow you to quickly and easily solve geometry questions and also builds a good foundation for solving problems in higher levels and competitive examination.

These NCERT Solutions provided are according to the latest NCERT syllabus and explain each step clearly to improve your understanding. Download the PDF for practice and quick revision.

1.0Download NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3: Free PDF

The free PDF for Exercise 3.3 of NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals is available for download here. Click the link below to download the PDF.

2.0Key Concepts in Exercise 3.3 of Class 8 Maths Chapter 3

In this exercise, you will learn about the different types of quadrilaterals and their characteristics. Below are some of the key concepts covered in this exercise.

• Definition and characteristics of parallelograms
• Properties of rhombus, rectangle and square
• Identifying special quadrilaterals based on their properties
• Differences of similar looking quadrilaterals

3.0NCERT Class 8 Maths Chapter 3: Other Exercises

4.0NCERT Class 8 Maths Chapter 3 Exercise 3.3: Detailed Solutions

• Given a parallelogram ABCD. Complete each statement along with the definition or property used.
  
  (i) $\mathrm{AD}=$ (ii) $\angle \mathrm{DCB}=$ (iii) $\mathrm{OC}=$ (iv) $\mathrm{m}\angle \mathrm{DAB}+\mathrm{m}\angle \mathrm{CDA}=$ Sol. (i) $\mathrm{AD}=\mathrm{BC}$ : In a parallelogram, opposite sides are equal. (ii) $\angle \mathrm{DCB}=\angle \mathrm{DAB}:$ In a parallelogram, opposite angles are equal. (iii) $\mathrm{OC}=\mathrm{OA}$ : The diagonals of a parallelogram bisect each other. (iv) $\mathrm{m}\angle \mathrm{DAB}+\mathrm{m}\angle \mathrm{CDA}=180^{\circ }$ : in a parallelogram, the sum of any two adjacent angles is $180^{\circ }$.
• Consider the following parallelograms. Find the values of the unknowns $\mathrm{x},\mathrm{y},\mathrm{z}$. (i)
  
  (iii)
  
  (iv)
  
  (v)
  
  Sol. (i) Since the sum of any two adjacent angles of a parallelogram is $180^{\circ }$, therefore, $\mathrm{x}+100^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-100^{\circ }=80^{\circ }$ $x+y=180^{\circ }$ $\Rightarrow y=180^{\circ }-x=180^{\circ }-80^{\circ }=100^{\circ }$ $y+z=180^{\circ }$ $\Rightarrow z=180^{\circ }-y=180^{\circ }-100^{\circ }=80^{\circ }$ Hence, $x=80^{\circ },y=100^{\circ }$ and $z=80^{\circ }$ (ii) Since $ABCD$ is a parallelogram, therefore, $AB\Vert DC$ and $AD\Vert BC$. Now, AB || DC and transversal BC intersects them. $\therefore \mathrm{z}=\mathrm{y}[\because$ Alternate angles are equal] and $\mathrm{AD}\Vert \mathrm{BC}$ and transversal AB intersects them. $\therefore \mathrm{z}=\mathrm{x}[\because$ Corresponding angles are equal] $\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}$ Since the sum of any two consecutive angles of a parallelogram is $180^{\circ }$, therefore, $y+50^{\circ }=180^{\circ }$ $\Rightarrow y=180^{\circ }-50^{\circ }=130^{\circ }$ Thus, $x=y=z=130^{\circ }$ (iii) From the figure, clearly, $x=90^{\circ }$ $[\because$ Vertically opp. $\angle$ s are equal] In $△\mathrm{DOC}$, we have $\angle \mathrm{DOC}+\angle \mathrm{OCD}+\angle \mathrm{CDO}=180^{\circ }$ [Angle sum property of triangle] $\Rightarrow \mathrm{x}+30^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow 90^{\circ }+30^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow y=180^{\circ }-120^{\circ }=60^{\circ }$ In the parallelogram $\mathrm{ABCD},\mathrm{AB}||\mathrm{DC}$. Also, the transversal $BD$ intersects them. $\therefore \mathrm{z}=\mathrm{y}[\because$ Alternate angles are equal $]$ $\Rightarrow \mathrm{z}=60^{\circ }$ $\left[\because y=60^{\circ }\right]$ Hence, $\mathrm{x}=90^{\circ },\mathrm{y}=60^{\circ }$ and $\mathrm{z}=60^{\circ }$. (iv) Since $ABCD$ is a parallelogram, the sum of any two adjacent angles of a parallelogram is $180^{\circ }$, therefore, $\mathrm{x}+80^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{x}=180^{\circ }-80^{\circ }$ $\Rightarrow \mathrm{x}=100^{\circ }$ $x+y=180^{\circ }$ $\Rightarrow 100^{\circ }+\mathrm{y}=180^{\circ }$ $\Rightarrow y=180^{\circ }-100^{\circ }$ $\Rightarrow \mathrm{y}=80^{\circ }$ In the parallelogram opposite angles are equal, therefore, $\angle \mathrm{A}=\angle \mathrm{C}$ $100^{\circ }=\angle \mathrm{C}$ and $\angle \mathrm{C}+\mathrm{z}=180^{\circ }$ [Linear pair] $\Rightarrow 100^{\circ }+\mathrm{z}=180^{\circ }$ $\Rightarrow \mathrm{z}=180^{\circ }-100^{\circ }$ $\Rightarrow \mathrm{z}=80^{\circ }$ Hence, $x=100^{\circ },y=80^{\circ }$ and $z=80^{\circ }$ (v) Since ABCD is a parallelogram, therefore opposite angles are equal in it. $\angle B=y$ [opposite angles] $112^{\circ }=y$ and $ADC$ is a triangle, since the sum of angles of triangle is $180^{\circ }$. In $△\mathrm{ADC}$. $\angle A+y+x=180^{\circ }$ $40^{\circ }+112^{\circ }+\mathrm{x}=180^{\circ }$ $152^{\circ }+\mathrm{x}=180^{\circ }$ $\mathrm{x}=180^{\circ }-152^{\circ }$ $\mathrm{x}=28^{\circ }$ In parallelogram $\mathrm{ABCD},\mathrm{AB}\Vert \mathrm{CD}$. Also, the transversal AC intersects them. $\therefore \mathrm{x}=\mathrm{z}[\because$ Alternate angles are equal $]$ $\Rightarrow \mathrm{z}=28^{\circ }[\because \mathrm{x}=28^{\circ }$ by equation 2$]$ Hence, $x=28^{\circ },y=112^{\circ }$ and $z=28^{\circ }$.
• Can $a$ quadrilateral $ABCD$ be a parallelogram if (i) $\angle \mathrm{D}+\angle \mathrm{B}=180^{\circ }$ ? (ii) $\mathrm{AB}=\mathrm{DC}=8\mathrm{cm},\mathrm{AD}=4\mathrm{cm}$ and $\mathrm{BC}=4.4\mathrm{cm}$ ? (iii) $\angle \mathrm{A}=70^{\circ }$ and $\angle \mathrm{C}=65^{\circ }$ ? Sol. (i) If in a quadrilateral $\mathrm{ABCD},\angle \mathrm{D}+\angle \mathrm{B}=$ $180^{\circ }$, then it is not necessary that ABCD is a parallelogram. (ii) Since $AD\ne BC$, i.e., the opposite sides are unequal, so ABCD is not a parallelogram. (iii) Since $\angle \mathrm{A}\ne \angle \mathrm{C}$, i.e., the opposite angles are unequal, so $ABCD$ is not a parallelogram.
• Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure. Sol.
  
  In kite, $\angle \mathrm{B}=\angle \mathrm{D}$.
• The measures of two adjacent angles of a parallelogram are in the ratio $3:2$. Find the measure of each of the angles of the parallelogram. Sol. Let two adjacent angles A and B of |gm ABCD be $3x$ and $2x$ respectively. Since the adjacent angles of a parallelogram are supplementary. $\therefore \angle \mathrm{A}+\angle \mathrm{B}=180^{\circ }$ $\Rightarrow 3\mathrm{x}+2\mathrm{x}=180^{\circ }$ $\Rightarrow 5\mathrm{x}=180^{\circ }$ $\Rightarrow \mathrm{x}=36^{\circ }$ $\therefore \angle \mathrm{A}=3\times 36^{\circ }=108^{\circ }$ and, $\angle \mathrm{B}=2\times 36^{\circ }=72^{\circ }$ Since the opposite angles are equal in a parallelogram, therefore, $\angle \mathrm{C}=\angle \mathrm{A}=108^{\circ }$ and $\angle \mathrm{D}=\angle \mathrm{B}=72^{\circ }$ Hence, $\angle \mathrm{A}=108^{\circ },\angle \mathrm{B}=72^{\circ },\angle \mathrm{C}=108^{\circ }$ and $\angle \mathrm{D}=72^{\circ }$
• Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram. Sol. We know that rectangle is also a parallelogram.
  
  Since ABCD is a parallelogram, the sum of any two adjacent angles of a parallelogram is $180^{\circ }$, therefore, $\mathrm{x}+\mathrm{x}=180^{\circ }$ $x=90^{\circ }$ In a parallelogram opposite angles are equal. $\therefore$ All angles are $90^{\circ }$ hence it is a rectangle.
• The adjacent figure HOPE is a parallelogram. find the angle measures x , $y$ and $z$. State the properties you use to find them.
  
  Sol. Since HOPE is a parallelogram, therefore, HE | OP and HO || EP. Now, HE || OP and transversal HO intersects them. $\therefore \angle \mathrm{EHO}=\angle \mathrm{POX}[\because$ Corresponding angles are equal] $\Rightarrow 40^{\circ }+\mathrm{z}=70^{\circ }\Rightarrow \mathrm{z}=70^{\circ }-40^{\circ }=30^{\circ }$ Again, HE || OP and transversal HP intersects them. $\therefore \angle \mathrm{OPH}=\angle \mathrm{EHP}[\because$ Alternate angles are equal] $\Rightarrow \mathrm{y}=40^{\circ }$ Since opposite angles are equal in a parallelogram. $\therefore \angle \mathrm{HEP}=\angle \mathrm{HOP}$ $\Rightarrow \mathrm{x}=180^{\circ }-\angle \mathrm{POX}=180^{\circ }-70^{\circ }=110^{\circ }$ Hence, $\mathrm{x}=110^{\circ },\mathrm{y}=40^{\circ }$ and $\mathrm{z}=30^{\circ }$
• The following figures GUNS and RUNS are parallelograms. Find $x$ and $y$. (Lengths are in cm) (i)
  
  (ii)
  
  Sol. (i) Since GUNS is a parallelogram, therefore, its opposite sides are equal. i.e. GS $=UN$ and GU $=$ SN $\Rightarrow 3\mathrm{x}=18$, i.e., $\mathrm{x}=6$ and $3\mathrm{y}-1=26$ $\Rightarrow 3y=26+1=27\Rightarrow y=9$ Hence, $\mathrm{x}=6$ and $\mathrm{y}=9$ (ii) We know that diagonals of parallelogram bisect each other and RN and US are diagonals of given parallelogram RUNS. So, $y+7=20\Rightarrow y=13$ and $\mathrm{x}+\mathrm{y}=16\Rightarrow \mathrm{x}+13=16\Rightarrow \mathrm{x}=3$ $\therefore \mathrm{x}=3,\mathrm{y}=13$
• In the figure both RISK and CLUE are parallelograms. Find the value of x .
  
  Sol. In the |gm RISK, we have $\angle$ RIS $=\angle$ RKS $=120^{\circ }[\because$ opp. $\angle$ s are equal $]$ Also, $\angle \mathrm{OIC}=180^{\circ }-\angle \mathrm{RIS}[$ Linear pair $]$ $\Rightarrow \angle \mathrm{OIC}=180^{\circ }-120^{\circ }=60^{\circ }$ In the |gm CLUE, CE | LU and a transversal ICL intersects them. $\therefore \angle \mathrm{ICO}=\angle \mathrm{CLU}[\because$ Corresponding $\angle \mathrm{s}$ are equal] $\Rightarrow \angle \mathrm{ICO}=70^{\circ }$ In $△$ OIC, by angle sum property, we have $\angle \mathrm{IOC}+\angle \mathrm{OIC}+\angle \mathrm{ICO}=180^{\circ }$ $\Rightarrow \angle \mathrm{ICO}+60^{\circ }+70^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{IOC}=180^{\circ }-60^{\circ }-70^{\circ }=50^{\circ }$ $\therefore \mathrm{x}=\angle \mathrm{IOC}=50^{\circ }[\because$ Vertically opp. $\angle \mathrm{s}$ are equal]
• Explain how this figure is a trapezium. Which of its two sides are parallel?
  
  Sol. Since $\angle \mathrm{KLM}+\angle \mathrm{NML}=180^{\circ }$ i.e., the pair of consecutive interior angles are supplementary. Therefore, KL | NM and so KLMN is a trapezium.
• Find $\angle \mathrm{C}$ in the figure if $\overline{\mathrm{AB}}\Vert \overline{\mathrm{DC}}$.
  
  Sol. In this figure, $\mathrm{AB}\mid \mathrm{CD}$, also, the transversal BC intersects them. $\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ }$ [ $\because$ Sum of interior angles on the same side of transversal is $180^{\circ }$ ] $\Rightarrow 120^{\circ }+\angle \mathrm{C}=180^{\circ }$ $\Rightarrow \angle \mathrm{C}=180^{\circ }-120^{\circ }$ $\Rightarrow \angle \mathrm{C}=60^{\circ }$
• Find the measure of $\angle \mathrm{P}$ and $\angle \mathrm{S}$ if $\overline{\mathrm{SP}}\Vert \overline{\mathrm{RQ}}$. (If you find $\mathrm{m}\angle \mathrm{R}$, is there more than one method to find $\mathrm{m}\angle \mathrm{P}$ ) ? Sol. Since $\overline{\mathrm{SP}}\Vert \overline{\mathrm{RQ}}$
  
  $\therefore \angle \mathrm{R}+\angle \mathrm{S}=180^{\circ }[\because$ Sum of interior angles on the same side of transversal is $180^{\circ }$ ] So $\angle \mathrm{R}=\angle \mathrm{S}=90^{\circ }\{\angle \mathrm{R}=90^{\circ }$ (Given) We know that the sum of the interior angles of a quadrilateral is $360^{\circ }$. $\therefore \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}=360^{\circ }$ $\Rightarrow \angle \mathrm{P}+130^{\circ }+90^{\circ }+90^{\circ }=360^{\circ }$ $\Rightarrow \angle \mathrm{P}+310=360^{\circ }$ $\Rightarrow \angle \mathrm{P}=360^{\circ }-310^{\circ }$ $\Rightarrow \angle \mathrm{P}=50^{\circ }$

5.0Key Features and Benefits of Class 8 Maths Chapter 3 Exercise 3.3

• The exercise displays unique quadrilaterals such as rectangles, rhombuses, and squares.
• The concepts align with the latest NCERT Class 8 Maths syllabus.
• Stepwise solutions provided explains how to understand properties of shapes.
• Regular practice of these solutions strengthens core foundations of geometry that are helpful for various competitive exams like the olympiads.
• Supports thinking clearly while also sharpening shape and angle descriptions.