NCERT Solutions Class 8 Maths Chapter 3 - Understanding Quadrilaterals

The NCERT Solutions for class 8th maths understanding quadrilaterals are elaborated in all the exercises of the textbook with their stepwise detailed answers thereby encouraging the students to practice them effectively. 

In class 8 - quadrilaterals, students are exposed to different types of polygons mostly concentrated on four-sided figures. It is very necessary to comprehend ncert class 8 maths chapter 3 Understanding quadrilaterals because as intriguing as these many sided shapes are, they become even more complicated in the ways that they are formed and the shapes formed from them. The focus on quadrilaterals extends beyond the current level of geometry to other disciplines in the real world such as buildings and designs where shapes and measurements quite clearly show their importance.

ALLEN focuses on students, providing them with well-structured NCERT solutions for class 8 maths chapter 3 aiding the understanding of exterior and interior angles of various quadrilaterals such as trapezium, parallelogram, rhombus etc. Making regular practice of these solutions, students will be able to improve their understanding of the concepts as well as work on their speed and accuracy which are required to perform well in the examinations.

1.0NCERT Solutions for Class 8 Maths Chapter 3 PDF

Students can download the Understanding Quadrilaterals Class 8 PDF with answers from the table below. The NCERT Solutions for Class 8 Maths Understanding Quadrilaterals PDF is meticulously designed by ALLEN's expert faculties, who are subject matter experts and have in-depth knowledge of the curriculum.

2.0NCERT Solutions Class 8 Maths Chapter 3 Understanding Quadrilaterals : All Exercises

By solving all class 8 understanding quadrilateral exercises, students can enhance their problem-solving abilities and increase their accuracy and speed in geometry.

3.0NCERT Questions with Solutions Class 8 Maths Chapter 3 - Detailed Solutions

EXERCISE : 3.1

• Given here are some figures :

• Classify each of them on the basis of the following: (a) Simple curve (b) Simple closed curve (c) Polygon (d)Convex polygon (e) Concave polygon Sol. The classification of the given figures is as under : (a) Simple curve : (1), (2), (5), (6) and (7) (b) Simple closed curve : (1), (2), (5), (6) and (7) (d) Convex polygon : (2) (e) Concave polygon: (1)
• How many diagonals does each of the following have? (a) A convex quadrilateral (b) A regular hexagon (c) A triangle Sol. (a) A convex quadrilateral has two diagonals. (b) A regular hexagon has nine diagonals. (c) A triangle has no diagonal
• What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? Sol. The sum of measures of the angles of a convex quadrilateral is $360^{\circ }$ Yes, this property holds in case if the quadrilateral is not convex.
• Examine the table

• What can you say about the angle sum of a convex polygon with number of sides? (a) 7 (b) 8 (c) 10 (d) $n$ Sol. From the given table, clearly we observe that the sum of angles (interior angles) of a polygon with $n$ sides $=(n-2)\times 180^{\circ }$. (a) Here, $\mathrm{n}=7$ $\therefore$ The sum of the angles of a polygon of 7 sides $=(7-2)\times 180^{\circ }=5\times 180^{\circ }=900^{\circ }$ (b) Here $\mathrm{n}=8$ $\therefore$ The sum of the angles of a polygon of 8 sides $=(8-2)\times 180^{\circ }=6\times 180^{\circ }=1080^{\circ }$ (c) Here $\mathrm{n}=10$ $\therefore$ The sum of the angles of a polygon of 10 sides $=(10-2)\times 180^{\circ }=8\times 180=1440^{\circ }$ (d) The sum of the angles of a polygon of ' $n$ ' sides $=(n-2)\times 180^{\circ }$
• What is a regular polygon? State the name of a regular polygon of (i) 3 sides (ii) 4 sides (iii) 6 sides Sol. A polygon is said to be a regular polygon, if all its (i) interior angles are equal; (ii) sides are equal; and (iii) exterior angles are equal. The name of a regular polygon of (i) 3 sides is equilaterial triangle. (ii) 4 sides is square. (iii) 6 sides is regular hexagon.
• Find the angle measure x in the following figures: (a)

• (b)

• (c)

• (d)

• Sol. (a) Since the sum of angles (interior angles) of a quadrilateral is $360^{\circ }$. $\therefore \mathrm{x}+120^{\circ }+130^{\circ }+50^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}+300^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-300^{\circ }=60^{\circ }$ (b) Since the sum of angles (interior angles) of a quadrilateral is $360^{\circ }$ $\therefore \mathrm{x}+70^{\circ }+60^{\circ }+90^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}+220^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-220^{\circ }=140^{\circ }$ (c) Since the sum of angles (interior angles) of a pentagon is $540^{\circ }$. Let its fourth and fifth angle be y & z. $y=180^{\circ }-70^{\circ }=110^{\circ }(\because$ Linear pair $)$ $\mathrm{z}=180^{\circ }-60^{\circ }=120^{\circ }(\because$ Linear pair $)$ Now, $30^{\circ }+\mathrm{x}+\mathrm{x}+110^{\circ }+120^{\circ }=540^{\circ }$ $\Rightarrow 260+2\mathrm{x}=540^{\circ }$ $\Rightarrow 2\mathrm{x}=540^{\circ }-260^{\circ }$ $\Rightarrow 2\mathrm{x}=280^{\circ }\Rightarrow \mathrm{x}=140^{\circ }$ (d) Since the sum of angles (interior angles) of a pentagon is $540^{\circ }$ Here, its a regular pentagon. So all 5 angles are equal. $\therefore$ each angle $=\frac{540^{\circ }}{5}=108^{\circ }$
• (a) Find $x+y+z$

• (b) Find $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}$

• Sol. (a) Since the sum of angles of a triangle is $180^{\circ }$.

• $\therefore \mathrm{m}\angle 1+30^{\circ }+90^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{m}\angle 1+120^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=180^{\circ }-120^{\circ }=60^{\circ }$ Also, $\mathrm{x}+90^{\circ }=180^{\circ }$ [Linear pair] $\Rightarrow \mathrm{x}=180^{\circ }-90^{\circ }=90^{\circ }$ and, $\mathrm{y}+\mathrm{m}\angle 1=180^{\circ }$ $\left[\because \mathrm{m}\angle 1=60^{\circ }\right]$ $\Rightarrow y+60^{\circ }=180^{\circ }$ $\Rightarrow y=180^{\circ }-60^{\circ }=120^{\circ }$ and $\mathrm{z}+30^{\circ }=180^{\circ }$ [Linear pair] $\Rightarrow \mathrm{z}=180^{\circ }-30^{\circ }=150^{\circ }$ Adding (1), (2) and (3), we get $x+y+z=90^{\circ }+120^{\circ }+150^{\circ }=360^{\circ }$ (b) Since the sum of angles of a quadrilateral is $360^{\circ }$.

• $\because \mathrm{m}\angle 1+120^{\circ }+80^{\circ }+60^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{m}\angle 1+260^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=360^{\circ }-260^{\circ }$ $\Rightarrow \mathrm{m}\angle 1=100^{\circ }$ Also, $x+120^{\circ }=180^{\circ }$ (Linear pair) $\mathrm{x}=180^{\circ }-120^{\circ }$ $\mathrm{x}=60^{\circ }$ and $y+80^{\circ }=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{y}=180^{\circ }-80^{\circ }$ $\Rightarrow \mathrm{y}=100^{\circ }$ and $\mathrm{z}+60^{\circ }=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{z}=180^{\circ }-60^{\circ }$ $\Rightarrow \mathrm{z}=120^{\circ }$ and $\mathrm{w}+\mathrm{m}\angle 1=180^{\circ }$ (Linear pair) $\Rightarrow \mathrm{w}+100^{\circ }=180^{\circ }$ $\Rightarrow \mathrm{w}=180^{\circ }-100^{\circ }$ $\Rightarrow \mathrm{w}=80^{\circ }$ Adding (1), (2), (3) and (4), we get $\mathrm{x}+\mathrm{y}+\mathrm{z}+\mathrm{w}=60^{\circ }+100^{\circ }+120^{\circ }+80^{\circ }=360^{\circ }$

EXERCISE 3.2

• Find x in the following figures : (a)

• (b)

• Sol. We know that the sum of the exterior angles formed by producing the sides of a convex polygon in the same order is equal to $360^{\circ }$. Therefore, (a) $x+125^{\circ }+125^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}+250^{\circ }=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-250^{\circ }=110^{\circ }$ (b) $60^{\circ }+90^{\circ }+70^{\circ }+x+90^{\circ }=360^{\circ }$ $\Rightarrow 310^{\circ }+\mathrm{x}=360^{\circ }$ $\Rightarrow \mathrm{x}=360^{\circ }-310^{\circ }$ $\Rightarrow \mathrm{x}=50^{\circ }$
• Find the measure of each exterior angle of a regular polygon of (i) 9 sides (ii) 15 sides. Sol. (i) Each exterior angle of a regular polygon of 9 sides $=\frac{360^{\circ }}{\mathrm{n}}$, where $\mathrm{n}=9$ $\therefore {\left(\frac{360}{9}\right)}^{\circ }=40^{\circ }$ (ii) 15 sides Each exterior angle of a regular polygon of ' n ' sides $=$ where $\mathrm{n}=15$ $\therefore \frac{360^{\circ }}{15}=24^{\circ }$
• How many sides does a regular polygon have if the measure of an exterior angle is $24^{\circ }$ ? Sol. Since the number of sides of a regular polygon $=\frac{360^{\circ }}{\text{ Exterior angle }}$ Here, exterior angle $=24^{\circ }$ $\therefore$ Number of sides of the given polygon $=\frac{360^{\circ }}{24^{\circ }}=15$
• How many sides does a regular polygon have if each of its interior angles is $165^{\circ }$ ? Sol. Let there be n sides of the polygon. Then, its each interior angle $=$ $\therefore \frac{\mathrm{n}-2}{\mathrm{n}}\times 180^{\circ }=165^{\circ }$ $\Rightarrow 180^{\circ }\times \mathrm{n}-360^{\circ }=165^{\circ }\times \mathrm{n}$ $\Rightarrow 180^{\circ }\times \mathrm{n}-165^{\circ }\mathrm{n}=360^{\circ }$ $\Rightarrow 15^{\circ }\times \mathrm{n}=360^{\circ }$ $\Rightarrow \mathrm{n}=\frac{360^{\circ }}{15^{\circ }}=24$ Thus, there are 24 sides of the polygon.
• (a) Is it possible to have a regular polygon with measure of each exterior angle as $22^{\circ }$ ? (b) Can it be an interior angle of a regular polygon? Why? Sol. (a) Since the number of sides of a regular polygon $=\frac{360^{\circ }}{\text{ Exterior angle }}$ $\therefore$ The number of sides of a regular polygon $=\frac{360^{\circ }}{22}[\because$ Exterior angle $=22^{\circ }$, given $]$ $=\frac{180}{11}$ Which is not a whole number. $\therefore$ A regular polygon with measure of each exterior angle as $22^{\circ }$ is not possible. (b) If interior angle $=22^{\circ }$, then its exterior angle $=180^{\circ }-22^{\circ }=158^{\circ }$. But 158 does not divide $360^{\circ }$ exactly. Hence, the polygon is not possible.
• (a) What is the minimum interior angle possible for a regular polygon? Why? (b) What is the maximum exterior angle possible for a regular polygon? Sol. (a) The equilateral triangle being a regular polygon of 3 sides has the least measure of an interior angle $=60^{\circ }$. (b) Since the minimum interior angle of a regular polygon is equal to $60^{\circ }$, therefore, the maximum exterior angle possible for a regular polygon $=180^{\circ }-60^{\circ }=120^{\circ }$.

EXERCISE 3.3

• 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 $360^{\circ }$. $\therefore \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}+\angle \mathrm{S}=360^{\circ }$ $\Rightarrow \angle \mathrm{P}+130^{\circ }+90^{\circ }+90^{\circ }=360^{\circ }$ $\Rightarrow \angle \mathrm{P}+310=360^{\circ }$ $\Rightarrow \angle \mathrm{P}=360^{\circ }-310^{\circ }$ $\Rightarrow \angle \mathrm{P}=50^{\circ }$

EXERCISE 3.4

• State whether True or False : (a) All rectangles are squares. (b) All rhombuses are parallelograms. (c) All squares are rhombuses and also rectangles. (d) All squares are not parallelograms. (e) All kites are rhombuses. (f) All rhombuses are kites. (g) All parallelograms are trapeziums. (h) All squares are trapeziums. Sol. (a) False (b)True (c) True (d) False (e) False (f) True (g) True (h) True
• Identify all the quadrilaterals that have : (a) Four sides of equal length. (b) Four right angles. Sol. (a) The quadrilaterals having four sides of equal length is either a square or a rhombus. (b) The quadrilaterals having four right angles is either a square or a rectangle.
• Explain how a square is (i) A quadrilateral (ii) A parallelogram (iii) A rhombus (iv) A rectangle Sol. (i) A square is 4 sided, so it is a quadrilateral. (ii) A square has its opposite sides parallel, so it is a parallelogram. (iii) A square is a parallelogram with all the four sides equal, so it is a rhombus. (iv) A square is a parallelogram with each angle a right angle, so it is a rectangle.
• Name the quadrilaterals whose diagonals : (i) bisect each other. (ii) are perpendicular bisectors of each other. (iii) are equal. Sol. (i) The quadrilaterals whose diagonals bisect each other can be a parallelogram or a rhombus or a square or a rectangle. (ii) The quadrilaterals whose diagonals are perpendicular bisectors of each other can be a rhombus or a square. (iii) The quadrilaterals whose diagonals are equal can be a square or a rectangle.
• Explain why a rectangle is a convex quadrilateral. Sol. Since the measure of each angle is less than $180^{\circ }$ and also both the diagonals of a rectangle lie wholly in its interior, so a rectangle is a convex quadrilateral.
• ABC is a right angled triangle and $O$ is the mid point of the side opposite to the right angle. Explain why 0 is equidistant from A, B and C. Sol.

• Produce BO to D such that BO = OD. Join $AD$ and $DC$. Then $ABCD$ is a rectangle. In the rectangle $ABCD$, its diagonals $AC$ and $BD$ are equal and bisect each other at 0 , $\therefore \mathrm{OA}=\mathrm{OC}$ and $\mathrm{OB}=0\mathrm{D}$. But AC = BD Therefore, $\mathrm{OA}=\mathrm{OB}=\mathrm{OC}$ Thus, O is equidistant from $\mathrm{A},\mathrm{B}$ and C . The mid point of diagonal $\overline{\mathrm{AC}}$ is 0 .

4.0What Will Students Learn in Chapter 3: Understanding Quadrilaterals?

This chapter lays down solid basic principles for understanding the advanced level of geometry. 

• Definition of quadrilaterals which are four-sided figures and their properties. 
• Different kinds of quadrilaterals such as parallelogram, rectangle and square, rhombus, trapezium, etc.
• Angle sums and angles within and surrounding a quadrilateral are defined and the angles sums are calculated. 
• Properties of diagonals and their relationships in specific quadrilaterals.
• The significance of symmetry and the importance of the patterns in distinguishing between the types of quadrilaterals will be dealt with. 
• Strategies for solving problems concerning various aspects or characteristics of quadrilaterals, including properties and arithmetic questions dealing with such figures.