NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Excercise: Angle Sum Property

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 $180\degree + 180\degree = 360\degree$ .

screenshot-from-2019-06-23-12-31-22

Answer:

As shown in $\triangle$ ACD,

$\angle 1 + \angle 2+\angle 3 = 180\degree$

As shown in $\triangle$ ABC,

$\angle 4 + \angle 5+\angle 6 = 180\degree$

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

$\angle D = \angle 3$ )

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

$= 180\degree + 180\degree$

$= 360\degree$

Hence proved, the sum of the measures of $\angle A,\angle B,\angle C$ and
$\angle D$ amounts to $180\degree + 180\degree = 360\degree$ .

Q2 Take four congruent card-board 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)].

screenshot-from-2019-06-23-12-32-48

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 $360\degree$ .

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

Hence, the sum of these angles is $360\degree$ = $\angle 1+\angle 2+\angle 3+\angle 4$

Angles ∠1, ∠2, ∠3, ∠4 meet at a point and sum of these angles is $360\degree$ = $\angle 1+\angle 2+\angle 3+\angle 4$ .

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 $\Delta PAB$ . From this we see $x = 180\degree - m\angle 2 - m\angle 3$ ; similarly from $\Delta PBC$ , $y = 180\degree - m\angle 4 - m\angle 5$ , from $\Delta PCD$ , $z = 180\degree - m\angle 6 - m\angle 7$ and from $\Delta PDA$ , $w = 180\degree - m\angle 8 - m\angle 1$ . Use this to find the total measure $m\angle 1 + m\angle 2 + ... + m\angle 8$ , does it help you to arrive at the result? Remember
$\angle x + \angle y + \angle z + \angle w = 360\degree$ .

screenshot-from-2019-05-14-14-57-54

Answer:

As we know $\angle x + \angle y + \angle z + \angle w = 360\degree$

$(180\degree - m\angle 2 - m\angle 3)$ $(180\degree - m\angle 4 - m\anlge 5)+$ $(180\degree - m\angle 6 - m\angle 7) +$ $( 180\degree - m\angle 8 - m\angle 1)$ $= 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $\left ( 180\degree+180\degree+180\degree+180\degree \right ) - 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 720\degree - 360\degree$

$m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 360\degree$

Hence, sum of $m\angle 1 + m\angle 2 + ... + m\angle 8$ $= 360\degree$ .

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).

screenshot-from-2019-06-23-12-33-46

Answer:

screenshot-from-2019-05-14-15-07-38

Draw a line matching points B and D.Line BD divides ABCD in two triangles . $\triangle BCD$ and $\triangle ABD$ .

Sum of angles of $\triangle BCD$ = $\angle DBC + \angle BDC + \angle C = 180\degree$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 1 \right )$

Sum of angles of $\triangle ABD$ = $\angle ADB + \angle ABD + \angle A = 180\degree$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left ( 2\right )$

Here, $\angle ADB + \angle BDC = \angle D$

$\angle DBC + \angle ABD = \angle B$

Adding equation 1 and equation 2,

$(\angle DBC + \angle BDC + \angle C ) +$ $(\angle ADB + \angle ABD + \angle A )$ $360\degree$ { $\angle ADB + \angle BDC = \angle D$ and

i.e. $\angle A + \angle B +\angle C +\angle D$ = $360\degree$ . $\angle DBC + \angle ABD = \angle B$ }

The sum of the interior angles in a quadilateral, $\angle A + \angle B +\angle C +\angle D$ = $360\degree$ .

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Excercise: 3.1

Q1 (a) Given here are some figures.

screenshot-from-2019-06-23-12-35-52

Classify each of them on the basis of the following.

(a) Simple curve

Answer:

(a) Simple curve : The curve which does not cross itself and have only one curve.

Some simple curves are

Q1 (b) Given here are some figures.

screenshot-from-2019-06-23-12-35-52
Classify each of them on the basis of the following.

(b) Simple closed curve

Answer:

Simple closed curve : The simple curves which are closed by line segment or curved line.

Some simple closed curve are

Q1 (c) Given here are some figures.

screenshot-from-2019-06-23-12-35-52
Classify each of them on the basis of the following.

Answer:

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

Some polygons are shown in figures

Q1 (d) Given here are some figures.

screenshot-from-2019-06-23-12-35-52
Classify each of them on the basis of the following.

(d) Convex polygon

Answer:

(d) Convex polygon :Convex polygon are polygons having all interior angles less than $180\degree$ .

Convex polygon = 2.

Q1 (e) Given here are some figures.

screenshot-from-2019-06-23-12-35-52
Classify each of them on the basis of the following.

(e) Concave polygon

Answer:

(e) Concave polygon : Concave polygon have one or more interior angles greater than $180\degree$ .

Concave polygon = 1.

Q2 (a) How many diagonals does each of the following have?
A convex quadrilateral

Answer:

(a) A convex quadrilateral :

There are 2 diagonals in a convex quadrilateral.

Q2 (b) How many diagonals does each of the following have?

A regular hexagon

Answer:

(b) A regular hexagon :

Total number of diagonal a regular hexagon have are 9 .

Q2 (c) How many diagonals does each of the following have?

A triangle

Answer:

A triangle does not have any diagonal.

Q3 What is the sum of the measures of the angles of a convex quadrilateral? Will this property
hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Answer:

The sum of the measures of the angles of a convex quadrilateral is $360\degree$ . A convex quadrilateral is made of two triangles and sum of angles of a triangle is $180\degree$ . Hence,sum of angles of two triangles is $360\degree$ which is also sum of angles of a quadrilateral.

Here,as we can see this quadrilateral has two triangles and the sum of angles of these triangles is $180\degree+ 180\degree=360\degree$ = sum of all angles of a quadrilateral.

Q4 (a) Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

screenshot-from-2019-06-23-12-43-19

What can you say about the angle sum of a convex polygon with number of sides?

(a) 7

Answer:

(a) 7

the sum of angles of a convex polygon $= (number of sides - 2)\ast 180\degree$

the sum of angle of a convex polygon with 7 sides $= (7 - 2)\ast 180\degree$

the sum of angles of a convex polygon with 7 sides $= (5)\ast 180\degree$

the sum of angles of a convex polygon with 7 sides $= 900\degree$

Q4 (a) Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

screenshot-from-2019-06-23-12-43-19

What can you say about the angle sum of a convex polygon with the number of sides?

(b) 8

Answer:

(b) 8

sum of the angles of a convex polygon $= (number of sides - 2)\ast 180\degree$

sum of the angles of a convex polygon with 8 sides $= (8 - 2)\ast 180\degree$

sum of the angles of a convex polygon with 8 sides $= (6)\ast 180\degree$

sum of the angles of a convex polygon with 8 sides $= 1080\degree$

Q4 (c) Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

screenshot-from-2019-06-23-12-43-19

What can you say about the angle sum of a convex polygon with number of sides?

Answer:

angle sum of a convex polygon $= (number of sides - 2)\ast 180\degree$

angle sum of a convex polygon with 10 sides $= (10 - 2)\ast 180\degree$

angle sum of a convex polygon with 10 sides $= (8)\ast 180\degree$

angle sum of a convex polygon with 10 sides $= 1440\degree$

Q4 (d) Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.)

screenshot-from-2019-06-23-12-43-19

What can you say about the angle sum of a convex polygon with number of sides?

(d)

Answer:

(d)

All the angles of a convex polygon added up to $= (number of sides - 2)\ast 180\degree$

the sum of the angles of a convex polygon having n sides $= (n - 2)\ast 180\degree$

Q5 (i) What is a regular polygon?
State the name of a regular polygon of 3 sides

Answer:

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

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

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

Each angle = $60\degree$

Q5 (ii) What is a regular polygon?
State the name of a regular polygon of 4 sides

Answer:

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

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

Square has all angles of $90\degree$ and all sides are equal.

Q5 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.

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All angles of the hexagon are $120\degree$ each.

Q6 Find the angle measure in the following figures.

screenshot-from-2019-05-14-20-49-26

Answer:

The solution for the above-written question is mentioned below,

Sum of angles of a quadrilateral $= 360\degree$

$130\degree+120\degree+50\degree+x= 360\degree$

$300\degree+x= 360\degree$

$x = 360\degree-300\degree$

$x = 60\degree$

Q6 Find the angle measure in the following figures.

<img alt="screenshot-from-2019-06-23-12-47-05" height="196"

src="https://lh3.googleusercontent.com/4e9fF6PnucwkV9sUA6cb4wugGNR3yQ6llYMnjKrv7UvD1G7FtrHH2j6vGxcbi3hHYrHhWvi1sN2Wp9vR1YxJcLF5x9r_7U4ev75SJXWOrQldA3JQVZqd0TJG1Rn33PrYsL38aMw" style="margin-left: 0px; margin-top: 0px;" width="228" />

Answer:

The Solution for the above-written question is mentioned below,

Sum of angles of a quadrilateral $= 360\degree$

$90\degree+60\degree+70\degree + x=360\degree$

$220\degree + x=360\degree$

$x=360\degree-220\degree$

$x=140\degree$

Q6 Find the angle measure in the following figures.

screenshot-from-2019-06-23-12-48-32

Answer:

This pentagon has all sides and all angles equal.

Sum of all angles of pentagon is $540\degree$

$x+x+x+x+x=540\degree$

$5\ast x=540\degree$

$x=108\degree$

Q6 Find the angle measure in the following figures.

screenshot-from-2019-06-23-12-47-56

Answer:

screenshot-from-2019-05-14-20-52-18

$a+70\degree=180\degree$ (linear pair) $b+60\degree =180\degree$ (linear pair)

$a =110\degree$ $b =120\degree$

Sum of all angles of pentagon = $540\degree$

$110\degree+120\degree+x+x+30\degree=540\degree$

$260\degree+2x=540\degree$

$2x=540\degree-260\degree$

$2x=280\degree$

$x=140\degree$

Q7 (a) Find

screenshot-from-2019-06-23-12-49-31

Answer:

(a) Find

screenshot-from-2019-05-14-22-45-01

$z+30\degree=180\degree$

$z=150\degree$

$x+90\degree=180\degree$

$x=90\degree$

Sum of all angles of triangle is $180\degree$

the unmarked angle of triangle be A. $A+90\degree+30\degree=180\degree$

$A=180\degree-120\degree$

$A=60\degree$

$A+y=180\degree$

$60\degree+y=180\degree$

$y=180\degree - 60\degree$

$y=120\degree$

= $90\degree+120\degree+150\degree$

= $360\degree$

Q7 (b) Find

screenshot-from-2019-05-14-22-47-05

Answer:

Here you will find the detailed solution of the above-written question,

screenshot-from-2019-05-14-22-47-05

$x+120\degree = 180\degree$

$x=180\degree - 120\degree$

$x=60\degree$

$y+80\degree = 180\degree$

$y=180\degree - 80\degree$

$y=100\degree$

$z+60\degree = 180\degree$

$z=180\degree - 60\degree$

$z=120\degree$

Sum of all angles of quadrilateral = $360\degree$

Let the unmarked angle be A.

$A+120\degree+60\degree+80\degree=360\degree$

$A=100\degree$

$A+w=180\degree$

$w= 80\degree$

= $60\degree+100\degree+120\degree+80\degree$

= $360\degree$

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Excercise: Sum of the Measures of the Exterior Angles of a Polygon

Take a regular hexagon Fig 3.10.
Q1 What is the sum of the measures of its exterior angles

screenshot-from-2019-05-14-22-49-20

Answer:

screenshot-from-2019-05-14-22-49-20

the sum of all its exterior angles will be equal to $360\degree$ .

x,y,z,p,q,r are all exterior angles.

Hence, $x+y+z+p+q+r=360\degree$

Q2 Take a regular hexagon Fig 3.10.

Is . Why?

screenshot-from-2019-05-14-22-49-20

Answer:

screenshot-from-2019-05-14-22-49-20

It is a hexagon with all sides equal.

All interior angles are also equal.

$a+r=a+x=a+y=a+z=a+p=a+q=180\degree$ (linear pairs)

$r=x=y=z=p=q=180\degree - a$

Hence,

because it is a hexagon with all sides and angles equal.

Q3 (i) What is the measure of each? exterior angle

screenshot-from-2019-05-14-22-49-20

Answer:

It is a hexagon with all sides equal.

All interior angles are also equal.

Sum of all angles of hexagon is $720\degree$ .

$6a = 720\degree$

$a = 120\degree$

$a+r=a+x=a+y=a+z=a+p=a+q=180\degree$ (linear pairs)

$r=x=y=z=p=q=180\degree - a$

screenshot-from-2019-05-14-22-49-20

Each exterior angle = $r=x=y=z=p=q=180\degree - a$ = $180\degree-120\degree$ = $60\degree$

Q3 (ii) What is the measure of each? interior angle

screenshot-from-2019-05-14-22-49-20

Answer:

It is a hexagon with all sides equal.

All interior angles are also equal.

Sum of all angles of hexagon is $720\degree$ .

$6a = 720\degree$

$a = 120\degree$

screenshot-from-2019-05-14-22-49-20

Q4 (i) Repeat this activity for the cases of a regular octagon

Answer:

(i) a regular octagon: It has all 8 angles equal and sum of all eight angles is 1080. Interior angles are equal so exterior angles are also equal.

Let interior angle be A.

$8\ast A= 1080\degree$

$A= 135\degree$ .

All exterior angles be B.

$B=180\degree - 135\degree$

$B=45\degree$

Q4 (ii) Repeat this activity for the cases of a regular 20-gon

Answer:

(ii) a regular 20-gon: It has all 20 angles equal and sum of all eight angles is 3240.Interior angles are equal so exterior angles are also equal.

Let interior angle be A.

$20\ast A= 3240\degree$

$A= 162\degree$ .

All exterior angles be B.

$B=180\degree - 162\degree$

$B=18\degree$

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Excercise: 3.2

Q1 (a) Find in the following figures.

screenshot-from-2019-05-15-11-05-10

Answer:

screenshot-from-2019-05-15-11-05-10

Sum of all exterior angles of polygon is $360\degree.$

$x+125\degree+125\degree=360\degree.$

$x=360\degree-250\degree.$

$x=110\degree.$

Q1 (a) Find in the following figures.

screenshot-from-2019-05-15-11-08-21

Answer:

screenshot-from-2019-05-15-11-08-21

Sum of all exterior angles of polygon is $360\degree$

$x+90\degree+60\degree+90\degree+70\degree=360\degree$

$x=360\degree-310\degree$

$x=50\degree$

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

Answer:

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

Sum of exterior angles of a polygon = $360\degree$

Let interior angle be A.

Sum of exterior angles of 9 sided polygon = $9 \ast A =$ $360\degree$

Exterior angles of 9 sided polygon $= A= 360\degree \div 9$

$A= 40\degree$

Hence, measure of each exterior angle of a regular polygon of 9 sides is $40\degree$

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

Answer:

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

Sum of exterior angles of a polygon = $360\degree$

Let interior angle be A.

Sum of exterior angles of 15 sided polygon = $15 \ast A =$ $360\degree$

Exterior angles of 15 sided polygon $= A= 360\degree \div 15$

$A= 24\degree$

Hence, measure of each exterior angle of a regular polygon of 15 sides is $24\degree$

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°

Regular polygon has all exterior angles equal.

Sum of exterior angles of a polygon = $360\degree$

Let number of sides be X.

Sum of exterior angles of a polygon = $X \ast 24\degree =$ $360\degree$

Exterior angles of 15 sided polygon $= X= 360\degree \div 24\degree$

Hence, sided regular polygon have measure of an exterior angle 24°

Q4 How 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 polygon has all exterior angles equal.

Let number of sides of polygon = n

Sum of Exterior angles of a polygon = $360\degree$

$\left ( n \right )\ast 15\degree= 360\degree$

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

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

Answer:

The measure of an exterior angle is 22°

Regular polygon has all exterior angles equal.

Sum of exterior angles of a polygon = $360\degree$

Let number of sides be X.

Sum of exterior angles of a polygon = $X \ast 22\degree =$ $360\degree$

Exterior angles of 15 sided polygon $= X= 360\degree \div 22\degree$

Hence,side of a polygon should be an integer but as shown above side is not a integer.So,it is not possible to have a regular polygon with measure of each exterior angle as 22 $\degree$ .

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

Answer:

The measure of an interior angle is 22°

Regular polygon has all interior angles equal.

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

Sum of interior angles of a polygon = $\left ( n-2 \right )\ast 180\degree$

Sum of interior angles of a polygon = $\left ( n-2 \right )\ast 180\degree$ = $\left ( n \right )\ast 22\degree$

$\left ( n \right )\ast 180\degree-\left ( 2 \right )\ast 180\degree=\left ( n \right )\ast 22\degree$

$\left ( n \right )\ast 180\degree-\left ( n \right )\ast 22\degree= 360\degree$

$\left ( n \right )\ast 158\degree= 360\degree$

Number of sides of a polygon should be an integer but since it is not an integer.So, it cannot be a regular polygon with interior angle as 22 $\degree$

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 = $\left ( 3-2 \right )\ast 180\degree=180\degree$

Interior angles of regular polygon are equal = A .

$\therefore$ $A+A+A=180\degree$

$3\ast A=180\degree$

$A=60\degree$

Hence,the minimum interior angle possible for a regular polygon is $60\degree$ .

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

Answer:

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

Exterior angles a equilateral triangle has maximum measure.

Sum of exterior angles of polygon = $360\degree$

Let exterior angle be A.

$\therefore$ $A+A+A=360\degree$

$3\ast A=360\degree$

$A=120\degree$

Hence,the maximum exterior angle possible for a regular polygon is $120\degree$ .

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Topic: Kinds of Quadrilaterals

Q1 Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, 5 cm. Arrange them as shown (Fig 3.11).

screenshot-from-2019-05-15-11-17-25

You get a trapezium. (Check it!) Which are the parallel sides here? Should the
non-parallel sides be equal?
You can get two more trapeziums using the same set of triangles. Find them out and
discuss their shapes.

Answer:

screenshot-from-2019-05-15-11-17-25

AB and CD are parallel sides.BC and AD are nonparallel sides. Non parallel sides need not be equal.

We can get two more trapeziums using the same set of triangles-

1596039255438 and 1596039263553

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Topic: Kite

Q1 Take a thick white sheet.
Fold the paper once.
Draw two line segments of different lengths as shown in Fig 3.12.
Cut along the line segments and open up.
You have the shape of a kite (Fig 3.13).
Has the kite any line symmetry

Fold both the diagonals of the kite. Use the set-square to check if they cut at

screenshot-from-2019-05-15-11-38-38

right angles. Are the diagonals equal in length? Verify (by paper-folding or measurement) if the diagonals bisect each other. By folding an angle of the kite on its opposite, check for angles of equal measure.
Observe the diagonal folds; do they indicate any diagonal being an angle bisector? Share your findings with others and list them. A summary of these results are given elsewhere in the chapter for your reference. Show that
$\Delta ABC$ and $\Delta ADC$ are congruent. What do we infer from this?

Answer :

screenshot-from-2019-05-15-11-38-46

Kite has symmetry along AC diagonal. $\triangleABC$ $\triangle ABC$ and $\triangle ACD$ are congruent and equal triangles.

Diagonals AC and BD are of different lengths.

Diagonals bisect each other.

The two diagonals AC and BD bisect $\angle A,\angle B,\angle C,\angle D$ .

NCERT solutions for class 8 maths 3 Understanding Quadrilaterals Excercise: Elements of Parallelogram

Q1 Take cut-outs of two identical parallelograms, say ABCD and A′B′C′D′ (Fig 3.19).

screenshot-from-2019-06-23-12-52-52

Here $\overline{AB}$ is same as $\overline{A'B'}$ except for the name. Similarly the other corresponding
sides are equal too.
Place $\overline{A'B'}$ over <img alt="\overline{DC}" height="15"

src="https://lh6.googleusercontent.com/UrVaihu13ALsyYTHIKb0UEhqQwC9ms7Bwt0QKOrwD3EHEtV0SL-qmDV9KQqqcV16eCsJsKZaGRIvtw5H7tNAYzfjxxPdZDxo1mNfTywVpiqBRF-FwtxM6CrzXMNqVDfRnKxyf1c" style="margin-left: 0px; margin-top: 0px;" width="30" /> . Do they coincide? What can you now say about the lengths
$\overline{AB}$ and $\overline{DC}$ ?
Similarly examine the lengths $\overline{AD}$ and $\overline{BC}$ . What do you find?
You may also arrive at this result by measuring $\overline{AB}$ and $\overline{DC}$ .

Answer:

screenshot-from-2019-05-15-12-07-29

Take cut-outs of two identical parallelograms, say ABCD and A′B′C′D′ (Fig 3.19).

Here $\overline{AB}$ is same as $\overline{A'B'}$ .

Also, $\overline{CD}$ is same as $\overline{C'D'}$

Place $\overline{A'B'}$ over $\overline{DC}$ . They coincide with each other.

The lengths $\overline{AB}$ and $\overline{DC}$ are equal and parallel lines.

The lengths $\overline{AD}$ and $\overline{BC}$ are equal and parallel lines.

Q2 Take two identical set squares with angles $30\degree - 60\degree - 90\degree$ and place them adjacently to form a parallelogram as shown in Fig 3.20. Does this help you to verify the above property?

Property: The opposite sides of a parallelogram are of equal length.

screenshot-from-2019-05-15-12-13-34

Answer:

screenshot-from-2019-05-15-12-13-34

Property : The opposite sides of a parallelogram are of equal length.

As we can see in the figure above, the opposite sides of figure are equal.

The figure above is a rectangle which is part of parallelogram.

Hence, the opposite sides of a parallelogram are of equal length.

Q3 Take two identical 30° – 60° – 90° set-squares and form a parallelogram as before. Does the figure obtained help you to confirm the above property?

Property: The opposite angles of a parallelogram are of equal measure.

Answer:

screenshot-from-2019-05-15-12-13-34

As shown in the above figure opposites angles are equal and are equal to $90 \degree$ .

Hence,the figure obtained help you to confirm the property: The opposite angles of a parallelogram are of equal measure.

Q4 Take a cut-out of a parallelogram, say,

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ABCD (Fig 3.29). Let its diagonals $\overline{AC}$ and $\overline{DB}$ meet at O.
Find the mid point of $\overline{AC}$ by a fold, placing C on A. Is the
mid-point same as O?
Does this show that diagonal $\overline{DB}$ bisects the diagonal $\overline{AC}$ at the point O? Discuss it
with your friends. Repeat the activity to find where the mid point of $\overline{DB}$ could lie.

Answer:

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ABCD . Let its diagonals $\overline{AC}$ and $\overline{DB}$ meet at O.
The mid point of $\overline{AC}$ is at point O.Hence, the
mid-point is same as O.

This show that diagonal $\overline{DB}$ bisects the diagonal $\overline{AC}$ at the point O.

The mid point of $\overline{DB}$ is point O.

NCERT solutions for class 8 maths chapter 3 Understanding Quadrilaterals Excercise: 3.3

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