Quadrilaterals Class 8 Solutions Maths Ganita Prakash Chapter 4

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Class 8 Maths Chapter 4 Quadrilaterals Solutions

Ganita Prakash Class 8 Chapter 4 Solutions

Class 8 Maths Ganita Prakash Chapter 4 Solutions Quadrilaterals

Intext (NCERT Page 82)

Question 1.
Observe the following figures.

Figures (i), (ii), and (iii) are quadrilaterals, and the others are not. Why?
Solution:
A quadrilateral is a polygon with four sides and four angles; They must be a closed shape and their sides must be line segments (no curves).
Here, figures (i), (ii), and (iii) are quadrilaterals because each has exactly four straight sides and is a closed shape. Figures, (iv) and (v) are not quadrilaterals because their all sides are not line segments.

Intext (NCERT Page 89)

Question 1.
In the earlier definition, we stated that a rectangle has (a) opposite sides of equal length and (b) all angles equal to 90°. Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are 90°?
Solution:
No, we would not be wrong. A quadrilateral with all angles equal to 90° must have equal opposite sides, making the simplified definition correct and equivalent to the original.

Question 2.
If you think that this definition is incomplete, try constructing a quadrilateral in which the angles are all 90° but the opposite sides are not equal.
Solution:
It is impossible to construct such a quadrilateral. If all angles are 90° then quadrilateral becomes a special type of parallelogram in which the opposite sides are parallel and equal, making the figure a rectangle by definition.

Intext (NCERT Page 90)

Question 1.
If ΔBAD = ΔDCB then is it wrong to write ΔBAD ≅ ΔCDB? Why?
Solution:
Yes, it is wrong to write ΔBAD ≅ ΔCDB.
The vertex order does not correctly represent the correspondence of angles and sides between the two triangles.
A valid congruence statement would require matching corresponding vertices, i.e. ΔBAD ≅ ΔCDB.

Intext (NCERT Page 93)

Question 1.
Using this fact, construct a square with a diagonal of length 8 cm.
Solution:
The steps of construction are given below.
(i) Draw a line segment AC of length 8 cm.
(ii) Draw the perpendicular bisector of AC and denote the line joining the points M and Nby Z, which intersects AC at O.
(iii) Now, mark two points B and D on line l, above and below the line segment AC, respectively, such that OB – OD = 4 cm, because the diagonals of a square are perpendicular bisector to each other and both are equal.
(iv) Join AB, AD, CB and CD to get the required square.

Intext (NCERT Page 93)

Question 1.
Using this fact, construct a square with a diagonal of length 8 cm.
Solution:
The steps of construction are given below.
(i) Draw a line segment AC of length 8 cm.
(ii) Draw the perpendicular bisector of AC and denote the line joining the points M and N by Z, which intersects AC at O.
(iii) Now, mark two points B and D on line l, above and below the line segment AC, respectively, such that OB – OD = 4 cm, because the diagonals of a square are perpendicular bisector to each other and both are equal.
(iv) Join AB, AD, CB and CD to get the required square.

Figure it Out (NCERT Page 94)

Question 1.
Find all the other angles inside the following rectangles.

Solution:
(i) Let the diagonals intersect at O.
Given, ∠BAO = 30°

In rectangle ABCD,
∠ABC = ∠BCD = ∠DAB = ∠CDA = 90°
∴ ∠BAD = ∠BAO + ∠OAD
⇒ 90° = 30° + ∠OAD
⇒ ∠OAD = 90° – 30° = 60°
Since, AB || CD
∠BAC = ∠ACD = 30°
and ∠BCA = ∠DAC – 60° [alternate interior angles]
Since, OA = OB
⇒ ∠OAB = ∠OBA = 30°
Similarly, ∠BDC = 30°, ∠DBC = 60° and ∠ADB = 60°

In ΔAOB, by angle sum property ∠AOB + ∠OAB + ∠ABO = 180°
⇒ ∠AOB + 30° + 30° = 180°
⇒ ∠AOB = 180° – 60° = 120°
Here, ∠AOB = ∠COD [vertically opposite angles]
and ∠AOD = 180° – 120° = 60° [linear pair]
∠BOC = ∠AOD = 60°
[vertically opposite angles]

(ii) Let the diagonals intersect at O.
Given, ∠QOR = 110°

In rectangle PQRS
∠PSR = ∠SRQ = ∠RQP = ∠QPS = 90°
Let ∠OQR = x
⇒ ∠OQR = ∠ORQ = x [OQ = OR]
In AQOR, by angle sum property
∠OQR + ∠ORQ + ∠QOR = 180°
⇒ x + x + 110° = 180°
⇒ 2x = 180° – 110°
⇒ x = \(\frac{70}{2}\) = 35°
Since, PS || QR
∴ ZSPR = ∠PRQ = 35° [alternate interior angles]
Also, ∠SRP = ∠RPQ = 90° – 35° = 55°
and ∠QSR = ∠PQS = 55°
∠POS = ∠QOR = 110°
[vertically opposite angles]
By linear pair,
∠POS + ∠POQ = 180°
⇒ ∠POQ = 180° – 110° = 70°
Also, ∠SOR = 70°. [vertically opposite angles]

Question 2.
Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other and intersect at an angle of
(i) 30° (ii) 40° (iii) 90° (iv) 140°
Solution:
The steps of construction are as follows.
I. DrawalinesegmentACoflength8cm.
II. Mark the mid-point O of AC.
III. Draw another line segment BD of length 8 cm, passing through O, such that O is also the mid-point of BD and ensure that angle ZCOD (or ZAOB or ZBOC or ZDOA) is the specified angle.
IV. Connect the end points A B,C and D to form the quadrilateral AB CD.

(i) Angle of 30° Construct the quadrilateral as described above, ensuring the diagonals intersect at 30° angle.
This will result in a rectangle.

(ii) Angle of 40° Construct the quadrilateral as described above, ensuring the diagonals intersect at 40° angle. This will result in a rectangle.

(iii) Angle of 90° Construct the quadrilateral as described above, ensuring the diagonal intersect at 90° angle. This will result in a square.

(iv) Angle of 140° Construct the quadrilateral as described above, ensuring the diagonal intersect at 140° angle. This will result into a rectangle.

Question 3.
Consider a circle with centre 0. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.
Solution:
Let the circle with centre O. The line segments PL and AM are two perpendicular diameters of the circle. Join the points A, P, M and L

So, the figure APML is a square because its diagonals are equal and bisect at right angle.

Question 4.
We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length and a thread. How do we make an exact 90° using these?
Solution:
Do yourself.

Question 5.
We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal, a rectangle?
Solution:
A quadrilateral with opposite sides parallel and equal is a parallelogram.
While all rectangles are parallelogram, not all parallelograms are rectangles.
A rectangle specifically requires all interior angles to be 90 degrees.
Therefore, a quadrilateral with only opposite sides parallel and equal is not necessarily a rectangle. It could be a rhombus or a general parallelogram with non-right angles.

Intext (NCERT Page 96)

Question 1.
What are the remaining angles of the parallelograrri? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out.

Solution:
Given, ∠BAD = 30°
Since, in a parallelogram, opposite angles and opposite sides are equal.
∴∠BCD = 30°, BC = AD = 5 cm
and CD = AB 4 cm
Also, ∠ABC = ∠ADC = x (say)
By angle sum property of quadrilateral
∠ABC + ∠BCD + ∠CDA + ∠DAB = 360°
⇒ x + 30° + x + 30° = 360°
⇒ 2x = 360° – 60°
⇒ x = \(\frac{300}{2}\) = 300°
Therefore, ∠ABC = ∠ADC = 150°

Intext (NCERT Page 98)

Question 1.
If ΔABD ≅ ΔCDB, then is it wrong to write ΔABD ≅ ΔCBD? Why?
Solution:
Yes, it is wrong to write ΔABD ≅ ΔCBD.
The vertex order does not guarantee congruency unless the quadrilateral has symmetric properties (e.g. rhombus).

Intext (NCERT Page 99)

Question 1.
If ΔAOE ≅ ΔYOS, then is it wrong to write ΔAOE ≅ ΔSOY? Why?
Solution:
Yes, it is wrong to write ΔAOE ≅ ΔSOY in most cases. The vertex order does not ensure congruency unless the quadrilateral has symmetric properties (e.g. rhombus or kite).

Question 2.
Do the diagonals of a parallelogram intersect at a particular angle?
Solution:
No, the diagonals of a general parallelogram do not intersect at a fixed angle. The angle depends on the side lengths and angles between them.

Figure it Out (NCERT Page 102)

Question 1.
Find the remaining angles in the following quadrilaterals.

Solution:
We know that in a parallelogram, opposite angles are equal.
∴ ∠RPE = ∠EAR = 40°
and ∠PEA = ∠PRA = x (let)
By angle sum property of a quadrilateral,
∠RPE + ∠PEA + ∠EAR + ∠ARP = 360°
⇒ 40° + x + 40° + x = 360°
⇒ 2x = 360° – 80° = 280°
⇒ x = \(\frac{280^{\circ}}{2}\) = 140°
Therefore, ∠PEA = ∠PRA = 140°

Solution:
We know that in a parallelogram, opposite angles are equal.
∴ ∠SPQ = ∠QRS = 110°
and ∠PQR = ∠PSR = x (let)
By angle sum property of a quadrilateral,
∠PQR + ∠QRS + ∠RSP + ∠SPQ = 360°
⇒ x + 110° + x + 110° = 360°
⇒ 2x + 220° = 360°
⇒ 2x = 360° – 220° = 140°
⇒ x = \(\frac{140}{2}\) = 70°

Solution:
Given, quadrilateral UVWX is a rhombus because all sides are equal and opposite angles are equal but not 90°.
∴ ∠UXV= ∠UVX= 30° [∵ UV = UX]

In ΔXUV, by angle sum property
∠XUV + ∠UVX + ∠VXU= 180°
⇒ ∠XUV + 30° + 30° = 180°
⇒ ∠XUV= 180° – 60° = 120°
and ∠XUV = ∠VWX= 120°

Since, rhombus is a type of a parallelogram.
∴ ∠UVX = ∠VXW = 30° [∵ UV ∥ XW]
and ∠UXV= ∠XVW = 30°
So, ∠UXW = ∠UVW = 60°

Solution:
Given, quadrilateral AEIO is a rhombus because all sides are equal and opposite angles are equal but not 90°.
∴ ∠AEO = ∠AOE = 20° [∵ AE = AO]
In ΔAOE, by angle sum property
∠AEO + ∠AOE + ∠OAE = 180°
⇒ 20° + 20° + ∠OAE = 180°
⇒ ∠OAE = 180° – 40° = 140°
So, ∠OAE = ∠OIE = 140°
Since, rhombus is a type of parallelogram.
∴ ∠AEO = ∠EOI = 20° [∵ AE ∥ OI]
and ∠AOE = ∠OEI = 20°
So, ∠AOI = ∠AEI = 40°

Question 2.
Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm and intersect at an angle of 140°.
Solution:
The steps of construction are as follows.
(i) Draw a line segment AC to be one of its diagonals of length 7 cm.

(ii) Let O be the mid-point of AC
i.e. OA = OC = \(\frac{7}{2}\) = 35 cm.

(iii) Draw another line segment BD of length 5 cm intersecting AC at O making an angle of 140°
(i.e. ∠BOC = 140°), such that OB = OD = \(\frac{5}{2}\) = 25 cm.
This is because diagonals of a parallelogram bisect each other.

(iv) Join AB, AD, BC and CD to get the required parallelogram.

Question 3.
Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Solution:
The steps of construction are as follows.
(i) Draw a line segment AC of length 5 cm as one of diagonals.
(ii) Locate the mid-point O of AC.
(iii) Since, the diagonals of a rhombus bisect each other at right angles. So, draw another line segment BD of length 4 cm which makes an angle of 90° with line segment AC such that BO = DO = 2 cm.
(iv) Join AB, AD, BC and CD to get the rhombus ABCD.

Geoboard Activity (NCERT Page 103)

Question 1.
Take a geoboard and some rubber bands. If you do not have these, you could just use the dot grid papers given at the end of the book for this activity.

(i) Place two rubber bands perpendicular to each other, forming diagonals of equal length. Join the ends.

What is the quadriferfferal that you get? Justify your answer.
Answer:
Here, after joining the ends of the perpendiculars, we get a quadrilateral which has all the sides equal, all the angles equals 90° and its diagonals bisect perpendicularly each other.

So, the quadrilateral is a square.

(ii) Extend one of the diagonals on both sides by 2 cm.

What quadrilateral will you get now? Justify your answer.
Solution:
Here, after extending the one of the diagonals on both sides by 2 cm and then joining the ends of the perpendiculars, we get a quadrilateral which has all the sides equal, all the angles necessarily need not to be 90° and its diagonals bisect perpendicularly each other.

So, the quadrilateral is a square.

Joining Triangles (NCERT Page 104-105)

Question 1.
(i) Take two cardboard cutouts of an equilateral triangle or side length 8 cm.

Can you join them to get a quadrilateral?
Solution:
Yes, we can join them to get a quadrilateral.

(ii) What type of a quadrilateral is this? justify your answer.

Solution:
Clearly, the formed quadrilateral has all the four sides equal, opposite angles are equal (60° and 120°) and if we draw a line as a diagonal to connect two unjoined vertices, we see that diagonals bisects each other at 90°.

Hence, this is a rhombus.

Question 2.
(i) Take two cardboard cutouts of an isosceles triangle with side lengths 8 cm, 8 cm and 6 cm.

What are the different ways they can be joined to get a quadrilateral?
Solution:
Here, figure (a) is a quadrilateral has all the properties of a rhombus. So, it is a rhombus.

(ii) Joining them in this way you get

What quadrilaterals are these? Justify your answers.
Solution:
Here, in figure (b), opposite sides are equal and parallel and we can observe that diagonals bisect each other. So, it is a parallelogram.

Question 3.
(i) Take two cardboard cutouts of a scalene triangle with sides 6 cm, 9 cm and 12 cm.

What are the different ways they can be joined to get a quadrilateral?
Solution:
We can join these scalene triangles along their common length sides as follows.

(ii) Are you able to identify the different quadrilaterals that are obtained by joining the triangles? Justify your answer whenever you identify a quadrilateral.
Solution:
Here, figures (a), (b) and (c) satisfy the properties of a parallelogram. So, all these are parallelograms.

Figure it Out (NCERT Page 107-109)

Question 1.
Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Solution:
Let the two equilateral triangles be ΔABC and ΔADC, with the common side being AC.
Since, all sides of an equilateral triangle are equal,
AB = BC = AC = 4 cm and AD = DC = AC = 4 cm.
The quadrilateral formed is ABCD.
Therefore, all sides of the quadrilateral are 4 cm.

So, two angles are 60° each and the other two angles are 120° each.

Question 2.
Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Solution:
The steps of construction are as follows
(i) Draw a line segment AC to be one of diagonal of length 6 cm.
(ii) Since, we know the longer diagonal of a kite bisects the shorter diagonal at 90°. So, draw a perpendicular bisector of AC and denote the line joining the points M andNby/, which intersects AC at O.
(iii) Now, mark points B and D on line l, above and below the line AC, respectively, such that OB ≠ OD ≠ 4 cm. Let’s say, OB = 3 cm and OD = 5 cm because
OB + OD = 3 cm + 5 cm = 8 cm, which is the length of another longer diagonal.
(iv) Join AB,BC, AD and DC to get the required kite ABCD.

Question 3.
Find the remaining angles in the following trapeziums.

Solution:
(i) In trapezium ABCD, ∠A + ∠D = 180°
[Y adjacent angles are supplementary]
⇒ 135° + ∠D = 180°
⇒ ∠D = 180°-135° = 45°
Also, ∠B + ∠C= 180°
⇒ 105°+ ∠C = 180°
⇒ ∠C = 180° – 105° = 75°

(ii) In the trapezium PQRS,
∠P + ∠S = 180°
⇒ ∠P + 100° = 180°
⇒ ∠P = 180° – 100 =80°
Since, it’s an isosceles trapezium, the other base angles are equal.
∴∠Q = ∠P = 80° and ∠R = 100°

Question 4.
Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles and squares. Then, answer the following questions.
(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?
Solution:
The Venn diagram is shown below.

(i) A rhombus is the only quadrilateral that is both a kite and a parallelogram.
This is because a rhombus has all the properties of a parallelogram (opposite sides parallel, opposite angles equal, diagonals bisect each other) and also the properties of a kite (two pairs of adjacent sides of equal length, one pair of opposite angles equal, diagonals are perpendicular.

(ii) Yes, a square is a quadrilateral that is both a kite and a rectangle. A square possesses the properties of a rectangle (four right angles, opposite sides parallel and equal) and also the properties of a kite (two pairs of adjacent sides of equal length, diagonals are perpendicular).

(iii) No, not every kite is a rhombus.
A rhombus is a special type of kite where all four sides are equal in length. A kite only requires two distinct pair of adjacent sides having equal length.

Question 5.
If PAIR and RODS are two rectangles, find ∠IOD.

Solution:
In a rectangle PAIR, ∠I = 90° and ∠IRO = 30°.
In ΔRIO, by angles sum property of a triangle
∠IRO + ∠RIO + ∠IOR = 180°
⇒ 30° + 90° + ∠IOR = 180°
⇒ ∠IOR = 180° -120° = 60°
Since, ∠DOR = 90° [v RODS is a rectangle]
∠IOD = ∠DOR – ∠IOR
= 90° – 60°
= 30°

Question 6.
Construct a square with diagonal 6 cm without using a protractor.
Solution:
The steps of construction are as follows
(i) Draw a line segment AC of length 6 cm.
(ii) Since, the diagonals of square bisects at right angle then draw the perpendicular bisector of AC.
(iii) Let O be the mid-point of line segment AC.
(iv) Mark two points B and D on the perpendicular bisector ofAC; above and below the diagonal AC, respectively, such that OB = OD = 3 cm. This is because both the diagonals of a square are equal.
(v) Join AB, AD, CB and CD. So, ABCD is the required square.

Question 7.
CASE is a square. The points U, V, IV and X are the mid-points of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).

Solution:

Since, U, V, W andXare mid-points of the sides of the square CASE then UVWX is a square. By measuring the sides and angles of UVWX we shall confirm it is a square.
Figure (b) shows a square whose vertices lie on the sides of outer square ABCD, but not necessarily at the mid-points.
Here are other ways to construct such a square PQRS.
(i) Rotation Rotate the inner square around the centre of the outer square by an arbitrary angle. The vertices of the rotated square will still lie on the sides of the outer square, forming a new inner square.

(ii) Using Diagonals Draw the diagonals of the outer square. Then, draw lines parallel to these diagonals, intersecting the sides of the outer square.
By carefully choosing the distance of these parallel lines from the centre, we can form an inner square.

Question 8.
If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Solution:
Yes, if a quadrilateral has four equal sides and one angle of 90°, it will be a square.
Construction
Draw a line segment AB of a specific length.
At point A construct a perpendicular line segment AD of the same length as AB, forming a 90° angle at A. From point D, draw a line segment DC of the same length parallel to AB.
Connect point C to point B.

(v) Measure the lengths of all sides AB,BC, CD and DA and all angles ∠A, ∠B, ∠C and ∠D.
We observe that all sides are equal and all angles are 90°, confirming it is a square.

Question 9.
What type of a quadrilateral is one in which the opposite sides are equal? Justify your answer.
[Hint Draw a diagonal and check for congruent triangles.]
Solution:
A quadrilateral in which the opposite sides are equal is a parallelogram.
Consider a quadrilateral ABCD such that
AB = CD and
BC = DA.
Draw a diagonal AC,

Now, in ΔABC and ΔCD A, we have
AD = BC [opposite side of a parallelogram]
AB = CD [opposite side of a parallelogram]
AC = AC [common side]
∴ ΔABC ≅ ΔCD A [by SSS congruence rule]
Since, triangles are congruent, so their corresponding angles are equal
i.e. ∠BAC = ∠DCA
and ∠BCA = ∠D AC.
But these equal angles are alternate interior angles.
Then, AB is parallel to CD and BC is parallel to DA.
Hence, the quadrilateral ABCD is a parallelogram.

Question 10.
Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.

Solution:
The sum of the angles in any quadrilateral including the shown one is indeed 360°.
Draw a diagonal line within the quadrilateral from vertex B to vertex D in the given figure.
This divides the quadrilateral into two triangles as shown below.

We know that, the sum of the interior angles of any triangle is 180°.
So, the sum of the angles of quadrilateral is the sum of the angles of these two triangles.
Therefore, the sum of the angles in the quadrilateral = 180°+ 180° = 360°
By measuring the angles in given figures, we get
∠A = 40°, ∠B = 50°, ∠C = 40° and ∠D = 230°.
Since, ∠A + ∠B + ∠C + ∠D
= 40°+ 50° + 40° + 230°
= 360°.
Hence, the sum of the angles in the given figure of quadrilateral will be 360°.

Question 11.
State whether the following statements are true or false. Justify your answers.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
(ii) A quadrilateral having three right angles must be a rectangle.
(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
(vi) A quadrilateral in which all the angles are equal is a rectangle.
(vii) Isosceles trapeziums are parallelograms.
Solution:
(i) False, a quadrilateral whose diagonals are equal and bisect each other is a rectangle, while a square is a special type of rectangle whose diagonals bisect at 90°.
(ii) True, the sum of angles in a quadrilateral is 360°, the fourth angle must also be 360 – (90 + 90 + 90) i.e. 90°.
(iii) True, this is a property of a parallelogram.
(iv) False, while a rhombus does have perpendicular diagonals, but other quadrilaterals can also have perpendicular diagonals.
(v) True, if the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram.
(vi) True, if all angles in a quadrilateral are equal, then the quadrilateral is a rectangle.
(vii) False, an isosceles trapezium has only one pair of parallel sides and another pair of non-parallel sides of equal length.