NCERT solutions for class 8 Maths Chapter 3 Understanding Quadrilaterals

Solution for Exercise 3.1

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


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Answer :

The classification of the given figures is as under :
(a) (1), (2), (5), (6), (7) and (8) are Simple curves.

(b) (1), (2), (5), (6) and (7) are Simple closed curves.

(c)(1) and (2) are Polygon.

(d)(2) is Convex polygon.

(e) (1) and (4) are Concave polygons.

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

(b) A regular hexagon

(c) A triangle.


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Answer :

(a)


A convex quadrilateral has two diagonals.

(b)


A regular hexagon has nine diagonals.

(c)


A triangle does not have any diagonal.

3.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 j try!)


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Answer :

The sum of measures of the angles of a convex quadrilateral is \(\360^{\circ}\) because it is made up of two triangles and the sum of internal angle of convex quadrilateral is equal to the sum of interior angles of two triangles i.e, \(\360^{\circ}\).Yes, this property holds in case if the quadrilateral is not convex.

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


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

(a) 7

(b) 8

(c) 10

(d) n


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Answer :

From the given table, dearly we observe that the sum of angles (interior angles) of a polygon with n sides = \((n – 2)× 180^{\circ}\).
(a) n = 7

Sum of angle =\(\left( 7-2 \right) \times { 180 }^{ \circ }\)

\(= 5\times { 180 }^{ \circ }={ 900 }^{ \circ }\)

∴The sum of angle for 7 sided polygon=\(900^o\)

(b) n = 8

Sum of angles =\( \left( 8-2 \right) \times { 180 }^{ \circ }\) \(= 6\times { 180 }^{ \circ }={ 1080 }^{ \circ }\)

∴ The sum of the angles of a polygon of 8 sides=\(1080^o\)

(c) n = 10

Angle sum = \(\left( 10-2 \right) \times { 180 }^{ \circ }\)

\(= 8\times { 180 }^{ \circ }={ 1440 }^{ \circ }\)

∴ The sum of the angles of a polygon of 10 sides=\(1440^o\)

(d) We can observe from the given table that the number of triangles is two less them the number of sides in the polygon.

∴ If the polygon has n sides, the number of triangles formed will be (n – 2).

Also we know that the sum of angles of a triangle = \(180^{ \circ }\)

∴ The sum of angles of a polygon of n sides =\( (n – 2) \times180^{ \circ }\)

5.What is a regular polygon? State the name of a regular polygon of
(i) 3 slides

(ii) 4 slides

(iii) 6 slides


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Answer :

A polygon is said to be a regular polygon, if all its sides and interior angles and exterior angles are equal.
The regular polygon of:

(i)3 sides= equilateral triangle.

(ii)4 sides= square.

(iii)6 sides = regular hexagon
.

6.Find the angle measure x in the following figures.


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Answer :

(a) As we know the sum of interior angles of a quadrilateral is \(360^{\circ}\).
∴ \(x+ 120^{\circ} +130^{\circ} +50^{\circ}= 360^{\circ}\)

\(\Rightarrow x+300^{\circ} = 360^{\circ} \)

\(\Rightarrow x=360^{\circ} - 300^{\circ}= 60^{\circ} \)

(b) As we know the sum of interior angles of a quadrilateral is \(360^{\circ}\).
∴ \(x+ 70^{\circ} +60^{\circ} +90^{\circ}= 360^{\circ}\)

\(\Rightarrow x+220^{\circ} = 360^{\circ} \)

\(\Rightarrow x=360^{\circ} - 220^{\circ}= 140^{\circ} \)

(c) As we see the given figure with 5 sides, so the sum of interior angles of this polygon is =\((n-2) \times 180^{\circ} = 3 \times 180^{\circ}=540^{\circ}\)



So in the figure we have ,

\(m∠1 + 60^{\circ} =180^{\circ} \quad\)[Angle of straight line is \(180^{\circ}\)]

\(\Rightarrow m∠1=180^{\circ}- 60^{\circ} =120^{\circ} \)

and also, \(m∠2 + 70^{\circ} =180^{\circ} \quad \)[Angle of straight line is \(180^{\circ}\)]

\(\Rightarrow m∠2=180^{\circ}- 70^{\circ} =110^{\circ} \)

Therefore, the sum of internal angles:

\(\qquad m∠1+m∠1+x+30^{\circ} +x= 540^{\circ}\)

\(\Rightarrow 120^{\circ}+110^{\circ}+2x+30^{\circ}=540^{\circ} \)

\(\Rightarrow 2x+260^{\circ}=540^{\circ} \)

\(\Rightarrow 2x=540^{\circ}-260^{\circ}=280^{\circ} \)

\(\Rightarrow x=\frac{280^{\circ}}{2}=140^{\circ} \)

(d) As we see the given figure with 5 sides, so the sum of interior angles of this polygon is =\((n-2) \times 180^{\circ} = 3 \times 180^{\circ}=540^{\circ}\)

Therefore, the sum of internal angles:

\(\qquad x+x+x+x+x= 540^{\circ}\)

\(\Rightarrow 5x= 540^{\circ}\)

\(\Rightarrow x=\frac{280^{\circ}}{5}=108^{\circ} \)

7. (a) Find x + y + z
(b) Find x + y + z + w


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Answer :

(a)We know that the sum of interior angles of a triangle is \(180^{\circ}\).
So, we have \(m∠1+30^{\circ}+90^{\circ}=180^{\circ}\)
\(\Rightarrow m∠1+120^{\circ}=180^{\circ}\)

\(\Rightarrow m∠1=180^{\circ}-120^{\circ}=60^{\circ}\)

So we have,\( x+90^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow x=180^{\circ}-90^{\circ}=90^{\circ} \)

Similarly, \( y+m∠1=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow y=180^{\circ}-60^{\circ}=120^{\circ} \)

And also, \( z+30^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-30^{\circ}=150^{\circ} \)

So, \(x+y+z= 90^{\circ}+120^{\circ}+150^{\circ} =360^{\circ} \)

(b)We know that the sum of interior angles of a quadrilateral is \(360^{\circ\}\).

Therefore \(m∠1+120^{\circ}+80^{\circ}+60^{\circ}=360^{\circ}\)
\(\Rightarrow m∠1+260^{\circ}=360^{\circ}\)

\(\Rightarrow m∠1=360^{\circ}-260^{\circ}=100^{\circ}\)

So we have,\( x+120^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow x=180^{\circ}-120^{\circ}=60^{\circ} \)

Similarly, \( w+m∠1=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow w=180^{\circ}-100^{\circ}=80^{\circ} \)

And also, \( z+60^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-60^{\circ}=120^{\circ} \)

And also, \( y+80^{\circ}=180^{\circ} \quad \)[Linear pair]

\(\Rightarrow z=180^{\circ}-80^{\circ}=100^{\circ} \)

So, \(x+y+z+w= 60^{\circ}+80^{\circ}+120^{\circ}+100^{\circ} =360^{\circ} \)

Solution for Exercise 3.2

1.Find x in the following figures.


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Answer :

As 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°. Therefore here we have,
(a)\( x + 125^{\circ} + 125^{\circ} = 360^{\circ}\)

\(\Rightarrow x + 250^{\circ} = 360^{\circ}\)

\(\Rightarrow x = 360^{\circ}– 250° = 110^{\circ}\)

\(\Rightarrow x = 110^{\circ}\)

(b) \(x + 90^{\circ} +60^{\circ} + 90^{\circ} + 70^{\circ} = 360^{\circ} \)

\(\Rightarrow x + 310^{\circ} = 360^{\circ} \)

\(\Rightarrow x = 360^{\circ} – 310^{\circ} = 50^{\circ} \)

\(\Rightarrow x = 50^{\circ} \)

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

(ii) 15 sides


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Answer :

(i) We know that the measure of each exterior angle of a regular polygon = \(\frac{360^{\circ}}{n}\)
So, the measure of each exterior angle of 9 sided regular polygon =\( \frac { 360 }{ 9 } = 40^{\circ}\)

(ii) Similarly,the measure of each exterior angle of a regular polygon = \(\frac{360^{\circ}}n\)

So, the measure of each angle of 15 sided regular polygon =\(\frac { 360 }{ 15 } = 24^{\circ}\)

3.How many sides does a regular polygon have if the measure of an exterior angle is \(24^{\circ}\)?


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Answer :

We know that sum of all exterior angles of a regular polygon = \(360^{\circ}\)
So we have,Number of sides=\(\frac{360^{\circ}}{measure\; of \;an\;angle}\)
Hence, the number of sides of the regular polygon is= 15

4.How many sides does a regular polygon have if each of its interior angles is \(165^{\circ}\)?


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Answer :

Let the number of sides of a regular polygon be n.
We have, Sum of all interior angles = \((n – 2) \times 180^{\circ}\)

and, measure of its each angle=\(\frac{(n-2) \times 180^{\circ}}{n}\)

here we have each angle as \(165^{\circ}\)

\(\Rightarrow \frac{(n-2) \times 180^{\circ}}{n}=165\)

\(\Rightarrow 180n - 360 =165 n \quad \)[Cross multiplication and opening the brackets]

\(\Rightarrow 180n -165 n=360\)

\(\Rightarrow 15n = 360\)

\(\Rightarrow n = \frac{360}{15}=24\)

So, we have the number of sides of regular polygon as 24.

5.(a) Is it possible to have a regular polygon with measure of each exterior angle a is \(22^{\circ}\)?
(b) Can it be an interior angle of a regular polygon? Why?


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Answer :

(a) We have, Number of sides of regular polygon=\(\frac{360^{\circ} }{Exterior \;angle}\)

As we know number of sides by the formula should be an integer so, it is not possible that a regular polygon have its exterior angle of \(22^{\circ}\) because \(22^{\circ}\) is not divisible by \(360^{\circ}\) hence, we can't get a whole number for calculating the number of side.

(b) If interior angle =\(22^{\circ}\)

And we have: measure of each interior angle=\(\frac{(n-2)\times180^o}n\)

\(\Rightarrow 180n -2 \times 180=22n\)

\(\Rightarrow 180n -22n=360\)

\(\Rightarrow 158n=360\)

\(\Rightarrow n=\frac{360}{158}\)

But 158 does not divide 360 exactly.So, the polygon is not possible.

6.(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?


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Answer :

(a) A regular polygon of 3 sides i.e, equilateral triangle has the least measure of an interior angle= \(60^{\circ}\).
(b) Since the minimum interior angle of a regular polygon is equal to 60°, therefore, the maximum exterior angle possible for a regular polygon = \(180^{\circ} – 60^{\circ} = 120^{\circ}\)

Solution for Exercise 3.3

1.Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …………

(ii) ∠DCB = ………

(iii) OC = ………

(iv) m?DAB + m∠CDA = ……..




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Answer :

(i) AD = BC as the opposite sides of a parallelogram are equal.
(ii)∠DCB = ∠DAB as the opposite angles of a parallelogram are equal.

(iii) OC = OA as the diagonals of a parallelogram bisect each other.

(iv) m∠DAB + m∠CDA = \(180^{\circ}\) as the adjacent angles of a parallelogram are supplementary.

2.Consider the following parallelograms. Find the values of the unknowns x, y, z.


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Answer :

(i)Here, ABCD is a parallelogram.


\(∠B = ∠D\quad \) [Opposite angles of a parallelogram are equal]

\(∠D = 100^{\circ}\)

\(\Rightarrow y = 100^{\circ}\)

\(∠A + ∠B = 180^{\circ}\quad\) [Adjacent angles of a parallelogram are supplementary]

\(\Rightarrow z + 100^{\circ} = 180^{\circ}\)

\(\Rightarrow z = 180^{\circ} – 100^{\circ} = 80^{\circ}\)

\(∠A = ∠C\quad\) [Opposite angles of a parallelogram are equal]

\(x = 80^{\circ}\)

\(Hence, x = 80^{\circ}, y = 100^{\circ} and\; z = 80^{\circ}\)

(ii)We have PQRS as a parallelogram.



\(∠P + ∠S = 180^{\circ}\quad \) [Adjacent angles of parallelogram]

\(\Rightarrow x + 50^{\circ} = 180^{\circ}\)

\(\Rightarrow x = 180^{\circ} – 50^{\circ} = 130^{\circ}\)

\(Now, ∠P = ∠R\quad\)[Opposite angles are equal]

\(\Rightarrow x = y\)

\(\Rightarrow y = 130^{\circ}\)

\(Also, y = z\quad\) [Alternate angles]

\(\Rightarrow z = 130^{\circ}\)

\(Hence, x = 130^{\circ}, y = 130^{\circ} and z = 130^{\circ}\)

(iii) Here we have, ABCD as a rhombus because diagonals intersect at \(90^{\circ} \)



\(So, x = 90^{\circ}\)

So,now in ∆OCB we have,

\(x + y + 30^{\circ} = 180^{\circ}\quad \) [Angle sum property]

\(\Rightarrow 90^{\circ} + y + 30^{\circ} = 180^{\circ}\)

\(\Rightarrow y + 120^{\circ} = 180^{\circ}\)

\(\Rightarrow y = 180^{\circ} – 120^{\circ} = 60^{\circ}\)

\(y = z \quad \)[Alternate angles]

\(\Rightarrow z = 60^{\circ}\)

\(Hence, x = 90^{\circ}, y = 60^{\circ} and z = 60^{\circ}\)

(iv) ABCD is a parallelogram



∠A + ∠B = 180^{\circ} (Adjacent angles of a parallelogram are supplementary)

\(\Rightarrow x + 80^{\circ} = 180^{\circ}\)

\(\Rightarrow x = 180^{\circ} – 80^{\circ} = 100^{\circ}\)

\(Now, ∠D = ∠B\quad \) [Opposite angles of a parallelogram are equal]

\(\Rightarrow y = 80^{\circ}\)

\(Also, z = ∠B = 80^{\circ}\quad\) (Alternate angles)

\(So, x = 100^{\circ}, y = 80^{\circ} and z = 80^{\circ}\)

(v) In this case we have ABCD as a parallelogram.



\(∠D = ∠B \quad\)[Opposite angles of a parallelogram are equal]

\(y = 112^{\circ}\)

\(x + y + 40^{\circ} = 180^{\circ}\quad\) [Angle sum property]

\(\Rightarrow x + 112^{\circ} + 40^{\circ} = 180^{\circ}\)

\(\Rightarrow x + 152^{\circ} = 180^{\circ}\)

\(\Rightarrow x = 180^{\circ} – 152^{\circ} = 28^{\circ}\)

\(z = x = 28^{\circ} \quad \)[Alternate angles]

\(So\; we \;have, x = 28^{\circ}, y = 112^{\circ}, z = 28^{\circ}\).

3.Can a quadrilateral ABCD be a parallelogram if
(i) \(∠D + ∠B = 180^{\circ}?\)

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?

(iii)\(∠A = 70^{\circ} and ∠C = 65^{\circ}?\)


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Answer :

(i) For ∠D + ∠B = \(180^{\circ}\), quadrilateral ABCD may be a parallelogram if the sum of measures of adjacent angles is \(180^{\circ}\) and also the opposite angles should be of same measures. So here, ABCD can be but need not be a parallelogram.

(ii) Given: AB = DC = 8 cm, AD = 4 cm, BC = 4.4 cm

In general the opposite sides od a parallelogram is equal.

But here, \(AD \neq BC\)

Thus, ABCD cannot be a parallelogram.

(iii) \(∠A = 70^{\circ} and ∠C = 65^{\circ}\)

Since \(∠A \neq ∠C\quad \)[Opposite angles of quadrilateral are not equal.]

Hence, ABCD is not a parallelogram.

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


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Answer :

In taking a kite as reference we have exactly two opposite angles of equal measure but not a parallelogram.

5.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.


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Answer :

As per the given conditions, let ABCD be parallelogram such that
m∠B : m∠C = 3 : 2



So,let m∠B = \(3x^{\circ}\) and m∠C = \(2x^{\circ}\)

m∠B + m∠C = \(180^{\circ}\quad\) [Sum of adjacent angles = \(180^{\circ}\)]

\(\Rightarrow 3x + 2x = 180^{\circ}\)

\(\Rightarrow 5x = 180^{\circ}\)

\(\Rightarrow x = 36^{\circ}\)

Thus,\(∠B = 3 \times 36 = 108^{\circ}\)

\(∠C = 2 \times 36^{\circ}\) = \(72^{\circ}\)

\(∠B = ∠D = 108^{\circ}\)

and \(∠A = ∠C = 72^{\circ}\)

Hence, the angles of the parallelogram are \(108^{\circ}\), \(72^{\circ}\), \(108^{\circ}\) and \(72^{\circ}\).

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


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Answer :

Let ABCD be a parallelogram in which we have, ∠A = ∠B

We know \(∠A + ∠B = 180^{\circ}\quad\) [Sum of adjacent angles = \(180^{\circ}\)]

\(∠A + ∠A = 180^{\circ}\)

\(\Rightarrow 2∠A = 180^{\circ}\)

\(\Rightarrow ∠A = 90^{\circ}\)

Thus, \(∠A =∠C = 90^{\circ} and ∠B = ∠D = 90^{\circ}\quad \) [Opposite angles of a parallelogram are equal]

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


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Answer :

We have, \(∠y = 40^{\circ}\) (Alternate angles)

\(∠z + 40°^{\circ} = 70^{\circ}\) (Exterior angle property)

\(\Rightarrow ∠z = 70^{\circ} – 40^{\circ} = 30^{\circ}\)

\(z = ∠EPH \quad\)[Alternate angle]

In ∆EPH, \(∠x + 40^{\circ} + ∠z = 180^{\circ}\quad\) [Adjacent angles]

\(\Rightarrow ∠x + 40^{\circ} + 30^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠x + 70^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠x = 180^{\circ} – 70^{\circ} = 110^{\circ}\)

Hence we have,\( x = 110^{\circ}, y = 40^{\circ} and\; z = 30^{\circ}.\)

8.The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)


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Answer :

(i)Opposite sides of a parallelogram is equal so we have,
GU=SN

\(\Rightarrow 3y-1=26\)

\(\Rightarrow 3y=26+1\)

\(\Rightarrow 3y=27\)

\(\Rightarrow y=\frac{27}3=9\)

Similarly, we have GS=UN

\(\Rightarrow 3x=18\)

\(\Rightarrow x=\frac{18}3=6\)

Hence, we have \(x=6cm\;and \;y=9cm\)

(ii)As we know that the diagonals of parallelogram bisect each other so we have

OU=OS

\(\Rightarrow y + 7 = 20\)

\(\Rightarrow y = 20 – 7 = 13\)

Similarly, ON = OR

\(\Rightarrow x + y = 16\)

\(\Rightarrow x + 13 = 16\)

\(\Rightarrow x = 16 – 13 = 3\)

Hence,\( x = 3 cm\; and \;y = 13 cm.\)

9.In the above figure both RISK and CLUE are parallelograms. Find the value of x.


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Answer :

Here we have RISK and CLUE as two parallelograms.
\(∠1 = ∠L = 70^{\circ}\quad\) [Opposite angles of a parallelogram]

\(∠K + ∠2 = 180^{\circ}\)

Sum of adjacent angles is \(180^{\circ}\)

\(120^{\circ} + ∠2 = 180^{\circ}\)

\(∠2 = 180^{\circ} – 120^{\circ} = 60^{\circ}\)

In ∆OES we have,

\(∠x + ∠1 + ∠2 = 180^{\circ}\quad\) [Angle sum property]

\(\Rightarrow ∠x + 70^{\circ} + 60^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠x + 130^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠x = 180^{\circ} – 130^{\circ} = 50^{\circ}\)

Hence \(∠x = 50^{\circ}\)

10.Explain how this figure is a trapezium. Which of its two sides are parallel?


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Answer :

We have, \(∠M + ∠L = 100^{\circ} + 80^{\circ} = 180^{\circ}\)
∠M and ∠L are the adjacent angles, and sum of adjacent interior angles is \(180^{\circ}\) and also KL is parallel to NM.Hence KLMN is a trapezium.

11.Find m∠C in below figure if \(\bar { AB } ∥ \bar { DC }\)


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Answer :

Since AB ∥ DC and transversal BC intersects them.So we have,
\(∠B + ∠C= 180^{\circ}\quad \) [∵ Sum of interior angles is 180^{\circ}]

\(\Rightarrow 120^{\circ}+ ∠C =180^{\circ} \)

\(\Rightarrow ∠C = 180^{\circ}- 120^{\circ} = 60^{\circ}\)

Hence, \(m∠C =60^{\circ}\)

12.Find the measure of ∠P and ∠S if \(\bar { SP } ∥ \bar { RQ }\) in figure, is there any other method to find m∠P?)


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Answer :

We have, \(∠Q = 130° and ∠R = 90^{\circ}\) and \(\bar { SP } || \bar { RQ }\)
\(∠P + ∠Q = 180^{\circ}\quad\) [Adjacent angles]

\(\Rightarrow ∠P + 130^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠P = 180^{\circ} – 130^{\circ} = 50^{\circ}\)

and also we have, \(∠S + ∠R = 180^{\circ}\quad\) [Adjacent angles]

\(\Rightarrow ∠S + 90^{\circ} = 180^{\circ}\)

\(\Rightarrow ∠S = 180^{\circ} – 90^{\circ} = 90^{\circ}\)
Hence, \(m∠P = 50^{\circ} \;and \; m∠S = 90^{\circ}\)
Alternate Method:
\(∠Q = 130^{\circ}, ∠R = 90^{\circ} and ∠S = 90^{\circ}\)

We know that

\(∠P + ∠Q + ∠R + ∠Q = 360^{\circ}\quad\) [Angle sum property of quadrilateral]

\(\Rightarrow ∠P + 130^{\circ}+ 90^{\circ} + 90^{\circ} = 360^{\circ}\)

\(\Rightarrow ∠P + 310^{\circ} = 360^{\circ}\)

\(\Rightarrow ∠P = 360^{\circ}– 310° = 50^{\circ}\)

Hence \(m∠P = 50^{\circ}\)

Solution for Exercise 3.4

1.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.


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Answer :

(a) False

(b) True

(c) True

(d) False

(e) False

(f) True

(g) True

(h) True

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

(b) four right angles
.


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Answer :

(a) Rhombus and square.
(b) Rectangle and square.

3.Explain how a square is
(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle


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Answer :

(i) Square is a quadrilateral because it is a closed polygon with four line segments.
(ii) Square is a parallelogram because the opposite sides are equal and parallel and also opposite angles are equal.

(iii) Square is a rhombus because its opposite sides are parallel and all the sides are equal.

(iv) Square is a rectangle because its opposite sides are equal and have equal diagonal.

4.Name the quadrilaterals whose diagonals
(i) bisect each other

(ii) are perpendicular bisectors of each other

(iii) are equal


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Answer :

(i) Parallelogram, rectangle, square and rhombus
(ii) Square and rhombus

(iii) Rectangle and square

5.Explain why a rectangle is a convex quadrilateral.


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Answer :

It is a convex quadrilateral because in a rectangle, both of its diagonal lie in its interior.

6.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).


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Answer :

Since the right-angled triangle ABC makes a rectangle ABCD by the dotted lines. Therefore OA = OB = OC = OD as the diagonals of a rectangle are equal and bisect each other.Hence, O is equidistant from A, B and C.



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