NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals

A geometrical closed shape made by using 4 straight lines and having four connecting points is called a Quadrilateral. The sum of the interior angles in a quadrilateral is always 360 degrees. NCERT Solutions for Class 8 Maths chapter 3, Understanding Quadrilaterals, cover the chapter to help you with the answers and conceptual clarity. It carries around 35% of geometry weightage and includes questions based on concepts of rectangles, squares, rhombuses, parallelograms, trapeziums, etc.

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1. Understanding Quadrilaterals Class 8 Questions And Answers PDF Free Download
2. Understanding Quadrilaterals Class 8 Solutions - Important Formulae
3. Understanding Quadrilaterals Class 8 Solutions - Topics
4. NCERT Solutions for Class 8 Maths: Chapter Wise
5. Importance of Solving NCERT Questions of Class 8 Maths Chapter 3
6. NCERT Solutions for Class 8 - Subject Wise
7. NCERT Books and NCERT Syllabus

These NCERT Solutions are created by the expert team at Career360, keeping the latest syllabus and pattern of CBSE 2025-26. Practicing questions is important to score good marks in Mathematics. NCERT solutions for Class 8 Maths discussed all questions and answers of all chapters, including Chapter 3, Understanding Quadrilaterals. The subtopics covered in this chapter are polygons, the angle sum property, and properties of different kinds of quadrilaterals. The NCERT solutions for Class 8 Maths chapter 3, Understanding Quadrilaterals, contain the solution to each question in an easy and understandable manner.

Understanding Quadrilaterals Class 8 Questions And Answers PDF Free Download

Understanding Quadrilaterals Class 8 Solutions - Important Formulae

Polygons are categorised based on the number of sides or vertices they possess, as outlined below:

• Angle Sum Property for Quadrilaterals: The sum of all angles of a quadrilateral is 360°.
• Sum of the Measures of Exterior Angles of a Polygon: Regardless of the number of sides in a polygon, the total measure of the exterior angles equals 360 degrees.

>> Categories of Quadrilaterals: Quadrilaterals are grouped based on the measurements of their angles and side lengths. The area represents the total space enclosed by the figure, while the perimeter is the complete distance around its boundaries. Here, you'll find descriptions, area calculations, and perimeter formulas for different types of quadrilaterals:

Q1 Take any quadrilateral, say ABCD (Fig 3.4). Divide it into two triangles by drawing a diagonal. You get six angles: 1, 2, 3, 4, 5, and 6. Use the angle-sum property of a triangle and argue how the sum of the measures of and amounts to ${180}^{\circ }+{180}^{\circ }={360}^{\circ }$.

Answer:

As shown in $\mathrm{△}$ ACD,

$\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3={180}^{\circ }$

As shown in $\mathrm{△}$ ABC,

$\mathrm{\angle }4+\mathrm{\angle }5+\mathrm{\angle }6={180}^{\circ }$

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D=(\mathrm{\angle }1+\mathrm{\angle }4)+\mathrm{\angle }6+(\mathrm{\angle }2+\mathrm{\angle }5)+\mathrm{\angle }3$ ( Since, $\mathrm{\angle }A=(\mathrm{\angle }1+\mathrm{\angle }4)$ , $\mathrm{\angle }B=\mathrm{\angle }6$ , $\mathrm{\angle }C=(\mathrm{\angle }2+\mathrm{\angle }5)$ ,

$\mathrm{\angle }D=\mathrm{\angle }3$ )

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D=(\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3)+(\mathrm{\angle }4+\mathrm{\angle }5+\mathrm{\angle }6)$

$={180}^{\circ }+{180}^{\circ }$

$={360}^{\circ }$

Hence proved, the sum of the measures of $\mathrm{\angle }A,\mathrm{\angle }B,\mathrm{\angle }C$ and
$\mathrm{\angle }D$ amounts to ${180}^{\circ }+{180}^{\circ }={360}^{\circ }$ .

Q2 Take four congruent cardboard copies of any quadrilateral ABCD, with angles as shown [Fig 3.5 (i)]. Arrange the copies as shown in the figure, where angles ∠1, ∠2, ∠3, ∠4 meet at a point [Fig 3.5 (ii)].

What can you say about the sum of the angles ∠1, ∠2, ∠3, and ∠4?

[Note: We denote the angles by ∠1, ∠2, ∠3, etc., and their respective measures by m∠1, m∠2, m∠3, etc.]

The sum of the measures of the four angles of a quadrilateral is___________.
You may arrive at this result in several other ways, also.

Answer:

As we know, the sum of all four angles of a quadrilateral is ${360}^{\circ }$.

$\mathrm{\angle }1,\mathrm{\angle }2,\mathrm{\angle }3,\mathrm{\angle }4$ are four angles of quadrilateral ABCD.

Hence, the sum of these angles is ${360}^{\circ }$ = $\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3+\mathrm{\angle }4$

Angles ∠1, ∠2, ∠3, ∠4 meet at a point and the sum of these angles is ${360}^{\circ }$ = $\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3+\mathrm{\angle }4$.

Q3 As before, consider quadrilateral ABCD (Fig. 3.6). Let P be any point in its interior. Join P to vertices A, B, C, and D. In the figure, consider $\mathrm{\Delta }PAB$. From this we see $x={180}^{\circ }-m\mathrm{\angle }2-m\mathrm{\angle }3$ ; similarly from $\mathrm{\Delta }PBC$ , $y={180}^{\circ }-m\mathrm{\angle }4-m\mathrm{\angle }5$ , from $\mathrm{\Delta }PCD$ , $z={180}^{\circ }-m\mathrm{\angle }6-m\mathrm{\angle }7$ and from $\mathrm{\Delta }PDA$ , $w={180}^{\circ }-m\mathrm{\angle }8-m\mathrm{\angle }1$ . Use this to find the total measure $m\mathrm{\angle }1+m\mathrm{\angle }2+...+m\mathrm{\angle }8$, does it help you to arrive at the result? Remember
$\mathrm{\angle }x+\mathrm{\angle }y+\mathrm{\angle }z+\mathrm{\angle }w={360}^{\circ }$ .

Answer:

As we know $\mathrm{\angle }x+\mathrm{\angle }y+\mathrm{\angle }z+\mathrm{\angle }w={360}^{\circ }$

$\begin{array}{rl}& ({180}^{\circ }-m\mathrm{\angle }2-m\mathrm{\angle }3)+({180}^{\circ }-m\mathrm{\angle }4-m5)+({180}^{\circ }-m\mathrm{\angle }6-m\mathrm{\angle }7)+({180}^{\circ }-m\mathrm{\angle }8-m\mathrm{\angle }1)={360}^{\circ }\\ & m\mathrm{\angle }1+m\mathrm{\angle }2+\dots +m\mathrm{\angle }8=({180}^{\circ }+{180}^{\circ }+{180}^{\circ }+{180}^{\circ })-{360}^{\circ }\\ & m\mathrm{\angle }1+m\mathrm{\angle }2+\dots +m\mathrm{\angle }8={720}^{\circ }-{360}^{\circ }\\ & m\mathrm{\angle }1+m\mathrm{\angle }2+\dots +m\mathrm{\angle }8={360}^{\circ }\end{array}$

Hence, the sum of

$m\mathrm{\angle }1+m\mathrm{\angle }2+\dots +m\mathrm{\angle }8={360}^{\circ }$

Q4 These quadrilaterals were convex. What would happen if the quadrilateral is not convex? Consider quadrilateral ABCD. Split it into two triangles and find the sum of the interior angles (Fig. 3.7).

Answer:

Draw a line matching points B and D.Line BD divides ABCD into two triangles. $\mathrm{△}BCD$ and $\mathrm{△}ABD$ .

Sum of angles of $\mathrm{△}BCD$ = $\mathrm{\angle }DBC+\mathrm{\angle }BDC+\mathrm{\angle }C={180}^{\circ }$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Sum of angles of $\mathrm{△}ABD$ = $\mathrm{\angle }ADB+\mathrm{\angle }ABD+\mathrm{\angle }A={180}^{\circ }$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Here, $\mathrm{\angle }ADB+\mathrm{\angle }BDC=\mathrm{\angle }D$

$\mathrm{\angle }DBC+\mathrm{\angle }ABD=\mathrm{\angle }B$

Adding equations 1 and equation 2,

$(\mathrm{\angle }DBC+\mathrm{\angle }BDC+\mathrm{\angle }C)+$ $(\mathrm{\angle }ADB+\mathrm{\angle }ABD+\mathrm{\angle }A)$ $=$ ${360}^{\circ }$ { $\mathrm{\angle }ADB+\mathrm{\angle }BDC=\mathrm{\angle }D$ and

i.e. $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D$ = ${360}^{\circ }$ . $\mathrm{\angle }DBC+\mathrm{\angle }ABD=\mathrm{\angle }B$ }

The sum of the interior angles in a quadrilateral,

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D$ = ${360}^{\circ }$

Q1 (a) Given here are some figures.

Classify each of them based on the following.

(a) Simple curve

Answer:

(a) Simple curve: The curve that does not cross itself and has only one curve.

Some simple curves are $:1,2,5,6,7$

Q1 (b) Given here are some figures.

Classify each of them based on the following.

(b) Simple closed curve

Answer:

Simple closed curve: A simple curve that is closed by a line segment or curved line.

Some simple closed curves are $:1,2,5,6,7$

Q1 (c) Given here are some figures.

Classify each of them based on the following.

(c) Polygon

Answer:

A normally closed curve made up of more than 4 line segments is called a polygon.

Some polygons are shown in Figures $1,2.$

Q1 (d) Given here are some figures.

Classify each of them based on the following.

(d) Convex polygon

Answer:

(d) Convex polygon: Convex polygons are polygons having all interior angles less than ${180}^{\circ }$.

Convex polygon = 2.

Q1 (e) Given here are some figures.

Classify each of them based on the following.

(e) Concave polygon

Answer:

(e) Concave polygon: A Concave polygon has one or more interior angles greater than ${180}^{\circ }$.

Concave polygon = 1.

Q2 (i) What is a regular polygon?
State the name of a regular polygon of 3 sides

Answer:

A regular polygon is a polygon that has equal sides and equal angles.

The name of a regular polygon of 3 sides is an equilateral triangle.

All sides of the equilateral triangle are equal, and the angles are also equal.

Each angle = ${60}^{\circ }$

Q2 (ii) What is a regular polygon?
State the name of a regular polygon of 4 sides

Answer:

A regular polygon is a polygon that has equal sides and equal angles.

The name of a regular polygon of 4 sides is a square.

Square has all angles of ${90}^{\circ }$ and all sides are equal.

Q2 (iii) What is a regular polygon?
State the name of a regular polygon of 6 sides

Answer:

A regular polygon is a polygon which have equal sides and equal angles.

The name of a regular polygon of 6 sides is a hexagon.

All angles of the hexagon are ${120}^{\circ }$ each.

Q1 (a) Find $x$ in the following figures.

Answer:

The sum of all exterior angles of a polygon is ${360}^{\circ }.$

$x+{125}^{\circ }+{125}^{\circ }={360}^{\circ }.$

$x={360}^{\circ }-{250}^{\circ }.$

$x={110}^{\circ }.$

Q1 (b) Find $x$ in the following figures.

Answer:

The sum of all exterior angles of a polygon is ${360}^{\circ }$

$x+{90}^{\circ }+{60}^{\circ }+{90}^{\circ }+{70}^{\circ }={360}^{\circ }$

$x={360}^{\circ }-{310}^{\circ }$

$x={50}^{\circ }$

Q2 (i) Find the measure of each exterior angle of a regular polygon of 9 sides.

Answer:

A regular polygon of 9 sides has all sides, interior angles, and exterior angles equal.

Sum of exterior angles of a polygon = ${360}^{\circ }$

Let the interior angle be A.

Sum of exterior angles of 9 sided polygon = $9\ast A=$ ${360}^{\circ }$

Exterior angles of 9 sided polygon $=A={360}^{\circ }÷9$

$A={40}^{\circ }$

Hence, the measure of each exterior angle of a regular polygon of 9 sides is ${40}^{\circ }$

Q2 (ii) Find the measure of each exterior angle of a regular polygon of 15 sides.

Answer:

A regular polygon of 15 sides has all sides, interior angles, and exterior angles equal.

Sum of exterior angles of a polygon = ${360}^{\circ }$

Let the interior angle be A.

Sum of exterior angles of 15 sided polygon = $15\ast A=$ ${360}^{\circ }$

Exterior angles of 15 sided polygon $=A={360}^{\circ }÷15$

$A={24}^{\circ }$

Hence, the measure of each exterior angle of a regular polygon of 15 sides is ${24}^{\circ }$

Q3 How many sides does a regular polygon have if the measure of an exterior angle is 24°?

Answer:

The measure of an exterior angle is 24°

A regular polygon has all exterior angles equal.

Sum of exterior angles of a polygon = ${360}^{\circ }$

Let the number of sides be X.

Sum of exterior angles of a polygon = $X\ast {24}^{\circ }=$ ${360}^{\circ }$

Exterior angles of 15 sided polygon $=X={360}^{\circ }÷{24}^{\circ }$

$X=15$

Hence, a $15$ sided regular polygon has a measure of an exterior angle of 24°

Q4 Hoof w many sides does a regular polygon have if each of its interior angles is 165°?

Answer:

The measure of each interior angle is 165°

So, measure of each exterior angle = 180°-165° = 15°

Regular polygons have regular exterior angles equal.

Let the number of sides the the s of the polygon = n

The sum of the Exterior angles of a polygon = ${360}^{\circ }$

$\left(n\right)\ast {15}^{\circ }={360}^{\circ }$

$n=24$

Hence, a regular polygon having each of its interior angles is 165° has 24 sides.

Q5 (a) Is it possible to have a regular polygon with a measure of each exterior angle of 22°?

Answer:

The measure of an exterior angle is 22°

A regular polygon has all exterior angles equal.

Sum of exterior angles of a polygon = ${360}^{\circ }$

Let the number of sides be X.

Sum of exterior angles of a polygon = $X\ast {22}^{\circ }=$ ${360}^{\circ }$

Exterior angles of 15 sided polygon $=X={360}^{\circ }÷{22}^{\circ }$

$X=16.36$

Hence, the side of a polygon should be an integer but as shown ,above, the side is not an integer. So,it is not poss ible to have a regular polygon with the measure of each exterior angle as 22 o .

Q5 (b) Can it be an interior angle of a regular polygon? Why?

Answer:

The measure of an interior angle is 22°

A regular polygon has all interior angles equal.

Let the number of sides and the number of interior angles be n.

Sum of interior angles of a polygon = $(n-2)\ast {180}^{\circ }$

Sum of interior angles of a polygon = $(n-2)\ast {180}^{\circ }$ = $\left(n\right)\ast {22}^{\circ }$

$\left(n\right)\ast {180}^{\circ }-\left(2\right)\ast {180}^{\circ }=\left(n\right)\ast {22}^{\circ }$

$\left(n\right)\ast {180}^{\circ }-\left(n\right)\ast {22}^{\circ }={360}^{\circ }$

$\left(n\right)\ast {158}^{\circ }={360}^{\circ }$

$n=2.28$

The number of sides of a polygon should be an integer, but it is not an integer. So, it cannot be a regular polygon with an interior angle of 22 o

Q6 (a) What is the minimum interior angle possible for a regular polygon? Why?

Answer:

Consider a polygon with the lowest number of sides, i.e., 3.

Sum of interior angles of 3 sided polygon = $(3-2)\ast {180}^{\circ }={180}^{\circ }$

Interior angles of a regular polygon are equal = A.

$\therefore$ $A+A+A={180}^{\circ }$

$3\ast A={180}^{\circ }$

$A={60}^{\circ }$

Hence, the minimum interior angle possible for a regular polygon is ${60}^{\circ }$.

Q6 (b) What is the maximum exterior angle possible for a regular polygon?

Answer:

Let there be a polygon with the minimum number of sides, i.e., 3.

The exterior angles s an equilateral triangle have a maximum measure.

Sum of exterior angles of polygon = ${360}^{\circ }$

Let the exterior angle be A.

$\therefore$ $A+A+A={360}^{\circ }$

$3\ast A={360}^{\circ }$

$A={120}^{\circ }$

Hence, the maximum exterior angle possible for a regular polygon is ${120}^{\circ }$.

Q1 (i) Given a parallelogram ABCD. Complete each statement along with the definition or property used.

$AD=$ ......

Answer:

In a parallelogram, opposite sides are equal in length.

Hence,

$AD=BC$

Q1 (ii) Given a parallelogram ABCD. Complete each statement along with the definition or property used.

$\mathrm{\angle }DCB=$ ......

Answer:

In a parallelogram, opposite angles are equal.

(ii) $\mathrm{\angle }DCB=$ $\mathrm{\angle }BAD.$

Q1 (iii) Given a parallelogram ABCD. Complete each statement along with the definition or property used.

$OC=$ ......

Answer:

In a parallelogram, both diagonals bisect each other.

$OC=$ $OA$

Q1 (iv) Given a parallelogram ABCD. Complete each statement along with the definition or property used.

$m\mathrm{\angle }DAB+m\mathrm{\angle }CDA=$ .....

Answer:

In a parallelogram, adjacent angles are supplementary to each other.

(iv) $m\mathrm{\angle }DAB+m\mathrm{\angle }CDA=$ ${180}^{\circ }$

Q2 (i) Consider the following parallelograms. Find the values of the unknowns $x,y,z$.

Answer:

In a parallelogram, adjacent angles are supplementary to each other.

$\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

${100}^{\circ }+x={180}^{\circ }$

$x={80}^{\circ }$

Opposite angles are equal.

Hence, z = $x={80}^{\circ }$

and y = 100

Q2 (ii) Consider the following parallelograms. Find the values of the unknowns $x,y,z$.

Answer:

${50}^{\circ }+x={180}^{\circ }$ ( Two adjacent angles are supplementary to each other)

$x={130}^{\circ }$

x=y= ${130}^{\circ }$ (opposite angles are equal)

z=x= ${130}^{\circ }$ ( corresponding angles are equal)

Q2 (iii) Consider the following parallelograms. Find the values of the unknowns $x, y, z.

Answer:

x= ${90}^{\circ }$ (vertically opposite angles)

$x+y+{30}^{\circ }={180}^{\circ }$ (sum of angles of a triangle is ${180}^{\circ }$ )

${90}^{\circ }+y+{30}^{\circ }={180}^{\circ }$

$y={60}^{\circ }$

y=z= ${60}^{\circ }$ (alternate interior angles)

Q2 (iv) Consider the following parallelograms. Find the values of the unknowns $x,y,z$.

Answer:

$x+{80}^{\circ }={180}^{\circ }$ (adjacent angles are supplementary)

$x={100}^{\circ }$

y = ${80}^{\circ }$ ( opposite angles are equal)

z= ${80}^{\circ }$ (corresponding angles are equal)

Q2 (v) Consider the following parallelograms. Find the values of the unknowns $x,y,z$.

Answer:

y = ${112}^{\circ }$ (opposite angles are equal)

$z+{40}^{\circ }+{112}^{\circ }={180}^{\circ }$ (adjacent angles are supplementary)

$z={180}^{\circ }-{152}^{\circ }$

$z={28}^{\circ }$

x = z = ${28}^{\circ }$ (alternate angles are equal)

Q3 (i) Can a quadrilateral ABCD be a parallelogram if $\mathrm{\angle }D+\mathrm{\angle }B={180}^{\circ }$

Answer:

(i) $\mathrm{\angle }D+\mathrm{\angle }B={180}^{\circ }$

Opposite angles should be equal, and adjacent angles should be supplementary to each other.

$\mathrm{\angle }B,\mathrm{\angle }D$ are opposite angles.

Hence, a quadrilateral ABCD can be a parallelogram, but it is not confirmed.

Q3 (ii) Can a quadrilateral ABCD be a parallelogram if $AB=DC=8cm,AD=4cm$ and $BC=4.4cm$ ?

Answer:

Opposite sides of a parallelogram are equal in length.

Sinc, $AD=4cm$ and $BC=4.4cm$ are opposite sides and have different lengths.

No, it is not a parallelogram.

Q3 (iii) Can a quadrilateral ABCD be a parallelogram if $\mathrm{\angle }A={70}^{\circ }$ and $\mathrm{\angle }C={65}^{\circ }$ ?

Answer:

Opposite angles of a parallelogram are equal

Since here $\mathrm{\angle }A={70}^{\circ }$ and $\mathrm{\angle }C={65}^{\circ }$ are different.

So, it is not a parallelogram.

Q4 Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles.
of equal measure.

Answer:

The above-shown figure shows two opposite angles are equal. $\mathrm{\angle }B=\mathrm{\angle }D$ .

But, it's not a parallelogram because the other two angles are different, i.e., $\angle A\neq \angle C.

Q5 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.

Answer:

The measures of two adjacent angles of a parallelogram are in the ratio 2.

The sum of the adjacent angles is ${180}^{\circ }$.

$\therefore$ $3\times x+2\times x={180}^{\circ }$

$5\times x={180}^{\circ }$

$x={36}^{\circ }$

Hence, angles are $2\times {36}^{\circ }={72}^{\circ }$ and $3\times {36}^{\circ }={108}^{\circ }$.

Let there be a parallelogram ABCD, then, $\mathrm{\angle }A=\mathrm{\angle }C={108}^{\circ }$ and $\mathrm{\angle }B=\mathrm{\angle }D={72}^{\circ }$. (Opposite angles are equal.)

Q6 Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.

Answer:

Given: Two adjacent angles of a parallelogram have equal measure = $\mathrm{\angle }A=\mathrm{\angle }B$.

$\mathrm{\angle }A+\mathrm{\angle }B={180}^{\circ }$ ( adjacent angles of a parallelogram are supplementary)

$2\times \mathrm{\angle }A={180}^{\circ }$

$\mathrm{\angle }A={90}^{\circ }$

$\therefore$ $\mathrm{\angle }A=\mathrm{\angle }B={90}^{\circ }$

$\mathrm{\angle }A=\mathrm{\angle }C={90}^{\circ }$ and $\mathrm{\angle }B=\mathrm{\angle }D={90}^{\circ }$ ( Opposite angles of a parallelogram are equal)

Hence, $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C=\mathrm{\angle }D={90}^{\circ }$

Q7 The adjacent figure HOPE is a parallelogram. Find the angle measures $x,y$, and $z$. State the properties you use to find them.

Answer:

The adjacent figure HOPE is a parallelogram.

$\mathrm{\angle }HOP+{70}^{\circ }={180}^{\circ }$ (linear pairs)

$\mathrm{\angle }HOP={110}^{\circ }$

$x=\mathrm{\angle }HOP={110}^{\circ }$ (opposite angles of a parallelogram are equal)

$\mathrm{\angle }HOP+\mathrm{\angle }EHO={180}^{\circ }$ ( adjacent angles are supplementary )

${110}^{\circ }+({40}^{\circ }+z)={180}^{\circ }$

$z={180}^{\circ }-{150}^{\circ }$

$z={30}^{\circ }$

y= ${40}^{\circ }$ (Alternate interior angles are equal)

Q8 (i) The following figures, GUNS and RUNS, are parallelograms. Find $x$ and $y$ . (Lengths are in cm)

Answer:

GUNS is a parallelogram, so opposite sides are equal in length

$SG=UN$

$3\times x=18$

$x=18÷3$

$x=6$

$UG=NS$

$3\times Y-1=26$

$3\times Y=27$

$Y=9$

Hence, $x=6$ cm and $Y=9$ cm.

Q8 (ii) The following figures, GUNS and RUNS, are parallelograms. Find $x$ and $y$ . (Lengths are in cm)

Answer:

Diagonals of a parallelogram intersect each other.

$y+7=20$

$y=20-7$

$y=13$

$x+y=16$

$x=16-13$

$x=3$

Hence, $x=3$ cm and $y=13$ cm.

Q9 In the above figure, both RISK and CLUE are parallelograms. Find the value of $x$.

Answer:

$\mathrm{\angle }SKR+\mathrm{\angle }KSI={180}^{\circ }$ ( adjacent angles are supplemantary)

$\mathrm{\angle }KSI={180}^{\circ }-{120}^{\circ }$

$\mathrm{\angle }KSI={60}^{\circ }$

$\mathrm{\angle }CLU=\mathrm{\angle }UEC={70}^{\circ }$ (oppsite angles are equal)

$x+\mathrm{\angle }UEC+\mathrm{\angle }KSI={180}^{\circ }$ (sum of angles of a triangle is ${180}^{\circ }$ )

$x+{70}^{\circ }+{60}^{\circ }={180}^{\circ }$

$x={180}^{\circ }-{130}^{\circ }$

$x={50}^{\circ }$

Q10 Explain how this figure is a trapezium. Which of its two sides is parallel? (Fig. 3.32)

Answer:

Given, $\mathrm{\angle }M+\mathrm{\angle }L={100}^{\circ }+{80}^{\circ }$ = ${180}^{\circ }$ .

A transverse line is intersecting two lines such that the sum of angles on the same side of the transversal line is ${180}^{\circ }$ .

And hence, lines KL and MN are parallel to each other.

Quadrilateral KLMN has a pair of parallel lines, so it is a trapezium.

Q11 Find $m\mathrm{\angle }C$ in Fig 3.33 if $\overline{AB}||\overline{DC}$ .

Answer:

Given , $\overline{AB}||\overline{DC}$

$\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$ (Angles on same side of transversal)

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

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

$\mathrm{\angle }C={60}^{\circ }$

Hence, $m\mathrm{\angle }C={60}^{\circ }$.

Q12 Find the measure of $\mathrm{\angle }P$ and $\mathrm{\angle }S$ if $\overline{SP}||\overline{RQ}$ in Fig 3.34.
(If you find $m\mathrm{\angle }R$, is there more than one method to find $m\mathrm{\angle }P$ ?)

Answer:

Given, $\overline{SP}||\overline{RQ}$

$\mathrm{\angle }P+\mathrm{\angle }Q={180}^{\circ }$ (angles on the same side of the transversal)

$\mathrm{\angle }P={180}^{\circ }-{130}^{\circ }$

$\mathrm{\angle }P={50}^{\circ }$

$\mathrm{\angle }R+\mathrm{\angle }S={180}^{\circ }$ (angles on the same side of the transversal)

$\mathrm{\angle }S={180}^{\circ }-{90}^{\circ }$

$\mathrm{\angle }S={90}^{\circ }$

Yes, to find $m\mathrm{\angle }P$, there is more than one method.

PQRS is a quadrilateral, so the sum of all angles is 360

$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R+\mathrm{\angle }S={360}^{\circ }$

and we know $\mathrm{\angle }Q,\mathrm{\angle }R,\mathrm{\angle }S$

so put values of $\mathrm{\angle }Q,\mathrm{\angle }R,\mathrm{\angle }S$ and we get a measurement of $\mathrm{\angle }P$

Q1 (a) State whether True or False. All rectangles are squares

Answer:

(a) False, all squares are rectangles, but all rectangles can be squares.

Q1 (b) State whether True or False. All rhombuses are parallelograms

Answer:

True. The opposite sides of the rhombus are parallel and equal.

Q1 (c) State whether True or False. All squares are rhombuses and also rectangles.

Answer:

True. All squares are rhombuses because rhombuses have opposite sides parallel and equal, and same square has.

Also, all squares are rectangles because they have all interior angles of ${90}^{\circ }$.

Q1 (d) State whether True or False. Not all squares are not parallelograms.

Answer:

False.

All squares have their opposite sides equal and parallel. Hence, they are parallelograms.

Q1 (e) State whether True or False. All kites are rhombuses.

Answer:

False,

Kites do not have all sides equal, so they are not rhombuses.

Q1 (f) State whether True or False. All rhombuses are kites.

Answer:

True, all rhombuses are kites because they have two adjacent sides equal.

Q1 (g) State whether True or False. All parallelograms are trapeziums.

Answer:

True, all parallelograms are trapeziums because they have a pair of parallel sides.

Q1 (h) State whether True or False. All squares are trapeziums.

Answer:

True, all squares are trapeziums because all squares have pairs of parallel sides.

Q2 (a) Identify all the quadrilaterals that have four sides of equal length.

Answer:

The quadrilateral having four sides of equal length is are square and a rhombus.

Q2 (b) Identify all the quadrilaterals that have four right angles

Answer:

All the quadrilaterals that have four right angles are rectangles and squares

Q3 (i) Explain how a square is a quadrilateral

Answer:

A square is a quadrilateral because a square has four sides.

Q3 (ii) Explain how a square is a parallelogram

Answer:

A square is a parallelogram because the square has opposite sides are parallel to each other.

Q3 (iii) Explain how a square is a rhombus

Answer:

A square is a rhombus because a square has four sides equal.

Q3 (iv) Explain how a square is a rectangle

Answer:

A square is a rectangle since it has all interior angles of ${90}^{\circ }$.

Q4 (i) Name the quadrilaterals whose diagonals bisect each other

Answer:

The quadrilaterals whose diagonals bisect each other are a square, a rectangle, a parallelogram, and a rhombus.

Q4 (ii) Name the quadrilaterals whose diagonals are perpendicular bisectors of each other.

Answer:

The quadrilaterals whose diagonals are perpendicular bisectors of each other are a rhombus and a square.

Q4 (iii) Name the quadrilaterals whose diagonals are equal

Answer:

The quadrilaterals whose diagonals are equal are squares and rectangles.

Q5 Explain why a rectangle is a convex quadrilateral.

Answer:

A rectangle is a convex quadrilateral because it has two diagonals and both lie in the interior of the rectangle.

Q6 ABC is a right-angled triangle, and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B, and C. (The dotted lines are drawn additionally to help you.)

Answer:

Draw line AD and DC such that $AB\parallel CD$ and $AD\parallel BC$.

$AD=BC$ and $AB=CD$

ABCD is a rectangle as it has opposite sides equal and parallel.

All angles of the rectangle are ${90}^{\circ }$, and a rectangle has two diagonals equal and bisect each other.

Hence, AO = BO = CO = DO

$\therefore$ O is equidistant from A, B, C, D.

Q1 A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?

Answer:

(1)All the properties of a parallelogram.

(2) Each of the angles is a right angle.

(3) Diagonals are equal

Q2 A square was defined as a rectangle with all sides equal. Can we define it as a rhombus with equal angles? Explore this idea.

Answer:

Properties of the rectangle are :

(1) All the properties of a parallelogram.

(2) Each of the angles is a right angle.

(3) Diagonals are equal.

A square satisfies all the properties of rectangles so a square can be defined as a rectangle with all sides equal.

Properties of a rhombus are :

(1) All the properties of a parallelogram.

(2) Diagonals are perpendicular to each other.

A square satisfies all the properties of the rhombus, so we can define it as a rhombus with equal angles.

Q3 Can a trapezium have all angles equal? Can it have all sides equal? Explain.

Answer:

A trapezium has two sides parallel and the other two sides are non-parallel. Parallel sides may be equal or unequal, but we cannot have a trapezium with all sides and angles equal.

Understanding Quadrilaterals Class 8 Solutions - Topics

• Polygons
• Sum of the Measures of the Exterior Angles of a Polygon
• Kinds of Quadrilaterals
• Some Special Parallelograms

NCERT Solutions for Class 8 Maths: Chapter Wise

Importance of Solving NCERT Questions of Class 8 Maths Chapter 3

• NCERT questions help build a strong understanding of different types of quadrilaterals, their properties, and classification (like parallelograms, trapeziums, rhombus, etc.).
• This chapter lays the groundwork for geometry in higher classes. Understanding these basics makes it easier to grasp topics in Class 9 and 10.
• Regular practice of NCERT questions develops logical thinking and enhances the ability to solve geometric problems with accuracy.
• Regular practice boosts speed and accuracy in solving numerical problems, which is useful for competitive exams later on.
• Most exam questions in school tests and olympiads are directly or indirectly based on NCERT content, making it highly relevant for scoring well.

NCERT Solutions for Class 8 - Subject Wise

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

NCERT Books and NCERT Syllabus

• NCERT Books Class 8 Maths
• NCERT Syllabus Class 8 Maths
• NCERT Books Class 8
• NCERT Syllabus Class 8