Quadrilaterals Class 8
Maths Ganita Prakash Chapter 4 Solutions
Page 82
Observe the following figures.
Page 83
Are there other ways to define a rectangle?
Answer Yes.
There are other ways to define a rectangle, even though they
must mean the same thing.
► Definition (using parallelogram
idea): A rectangle is a parallelogram whose all angles are right angles.
► Definition (using one right angle):
A parallelogram with even one right angle is a rectangle (because if one angle
is 90°, all others automatically become 90°).
► Definition (using diagonals): A
quadrilateral whose diagonals are equal and bisect each other is a rectangle.
► Definition (opposite sides equal +
right angle): A quadrilateral with opposite sides equal and one right angle is
a rectangle.
A carpenter needs to put together two thin strips of wood, as
shown in Fig. 1, so that when a thread is passed through their endpoints, it
forms a rectangle.
She already has one 8 cm long strip. What should be the
length of the other strip? Where should they both be joined?
Answer The length of
the other strip should be 8 cm because diagonals of rectangle are equal in
length.
The both be joined at the mid-point of the strip that is, at
4 cm from the end point.
1. What is the length of the other diagonal?
Answer The length of
other diagonal is 8 cm because diagonals of rectangle are equal in length.
2. What is the point of intersection of the two diagonals?
Answer We know that
diagonals of rectangle bisect each other. Therefore, the point of intersection
of the two diagonals is mid-point of two diagonals.
3. What should the angle be between the diagonals?
Answer One angle
between the diagonals should be acute but the other is obtuse.
Page 85
Can the following equalities be used to establish that ΔAOD ≅ ΔCOB?
AO = CO (proved above)
∠AOB = ∠COD
(vertically opposite angles)
AD = CB
Answer No.
As per to establish that ΔAOD ≅ ΔCOB,
we need angle between two sides to fulfil the condition of SAS. Here SSA is
being applied which is not the rule for coungruency.
Can you find all the remaining angles?
Answer ∠AOB = ∠COD = 60° [Vertically opposite
angles]
∠AOB + ∠BOC =
180° [Linear Pair]
⇒ ∠BOC = 180° – ∠AOB
⇒ ∠BOC = 180° – 60° = 120°
∠AOD = ∠COB =
120° [Vertically opposite angles]
Can you find the value of a?
Answer In ΔAOB, OA =
OB
Therefore, ∠OAB = ∠OBA = a
[The angles opposite them are equal]
So, ∠OAB + ∠OBA + ∠AOB = 180° [Angle sum property of triangle]
⇒ a + a + 60° = 180°
⇒ 2a = 180° – 60° = 120°
⇒ a = 120°/2 = 60°
Page 86
Can we now identify what type of quadrilateral ABCD is?
Answer All angles of
the quadrilateral are of 90° and diagonals are not intersecting at right angle,
therefore it is a rectangle.
What can we say about its sides?
Answer The opposite
sides of rectangle are equal, so it opposite sides should be equal.
Will ABCD remain a rectangle if the angles between the
diagonals are changed? Can we generalise this?
Answer:
It will remain rectangle until the angles become right angle.
At right angle it would be a square.
Page 87
Can you find the other angles?
Answer Using angle sum
property in ΔCOD, ΔAOB, ΔBOC and ΔAOD, we can find each angle.
What is the value of a (in degrees) in terms of x?
Answer In ΔAOB, OA =
OB
Therefore, ∠OAB = ∠OBA = a
[The angles opposite them are equal]
So, ∠OAB + ∠OBA + ∠AOB = 180° [Angle sum property of triangle]
⇒ a + a + x = 180°
⇒ 2a = 180° – x
⇒ a = (180° – x)/2.
What can we say about AB and CD, and AD and BC?
Answer Opposite sides
of a rectangle are equal.
Therefore, AB = CD and AD = BC.
Page 89
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°?
Answer No, we would
not be wrong — but the definition would be incomplete.
If we say: “A rectangle is a quadrilateral in which all
angles are 90°”, then it automatically becomes true that the opposite sides are
equal.
Because any quadrilateral with all angles 90° must be a
parallelogram and in a parallelogram, opposite sides are always equal.
So even if we don’t mention “opposite sides are equal”, the
condition all angles = 90° already forces that to happen.
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.
Are you able to construct such a quadrilateral?
Answer If I try to
draw a quadrilateral with all angles = 90°, this is what happens:
► First, I draw one side AB.
► At A and B, I make right angles and
draw the next two sides AD and BC.
► These two sides become parallel to
each other.
► Now, to close the figure, I join D
and C.
► That last side DC automatically
becomes parallel to AB.
So I always end up with both pairs of opposite sides
parallel, which means it becomes a parallelogram with all right angles → that
is exactly a rectangle.
And in a parallelogram, opposite sides are equal.
So, it is not possible to construct a quadrilateral with all
angles 90°, but opposite sides not equal.
Therefore, the definition “a quadrilateral in which all
angles are 90°” is actually complete enough for a rectangle.
Consider a quadrilateral ABCD with all angles measuring 90°.
What can we say about the opposite sides of such a quadrilateral?
Answer If a
quadrilateral ABCD has all angles measuring 90°, then:
► Opposite sides must be parallel:
AB ∥ CD
BC ∥ AD
This happens because when all angles are right angles, the
sides automatically become parallel in pairs.
► Opposite sides must also be equal:
Since the opposite sides are parallel and the angles are 90°,
the quadrilateral becomes a parallelogram with all right angles — that is, a
rectangle.
In every rectangle ABCD, AB = CD and BC = AD
Thus, in a quadrilateral with all angles = 90°, the opposite
sides are equal and parallel.
Page 90
Is it wrong to write ΔBAD ≅ ΔCDB? Why?
Answer Yes, it is
wrong to write ΔBAD ≅ ΔCDB in a rectangle ABCD.
In rectangle ABCD, triangles BAD and CDB do not match in
size.
In a rectangle, AB = CD and BC = AD (opposite sides are
equal).
But triangle BAD uses the sides BA, AD and diagonal BD.
Triangle CDB uses the sides CD, CB and the same diagonal BD.
Even though the diagonal BD is common, the other sides are:
BA vs CD → equal
AD vs CB → equal
But the included angles are different:
In ΔBAD, the angle between BA and AD is ∠BAD.
In ΔCDB, the angle between CD and CB is ∠DCB.
These two angles are not equal.
Therefore, even if two sides are equal, the included angles
are not the same.
So the SAS (Side–Angle–Side) condition fails and the
triangles cannot be congruent.
Are the opposite sides of a rectangle parallel?
Answer Yes.
The opposite sides of a rectangle parallel as well as equal
in length.
Page 92
Let us consider the Carpenter’s Problem again. If the wooden
strips have to be placed such that the thread passing through their endpoints
forms a square, what must be done?
Answer The two equal
threads should bisect at 90°.
What more needs to be done to get equal sidelengths as well?
Can this be achieved by properly choosing the angle between the diagonals?
Answer If the two
equal threads bisect at 90°, the sidelenghts automatically become equal.
Yes, it can be achieved by properly choosing 90° angle
between the diagonals.
By the SSS condition for congruence, ΔBOA ≅ ΔBOC.
Can this be used to find the angles ∠BOA and ∠BOC formed by the diagonals?
Answer In congrueant
triangles, the corresponding parst are equal. So, this can be used to find the
angles ∠BOA and ∠BOC formed by the diagonals.
Page 93
Using this fact, construct a square with a diagonal of length
8 cm.
Answer ► Use a ruler to draw a line segment
AC of length 8 cm.
► Find the midpoint O of AC.
Take compass more than half of AC.
Draw arcs above and below AC from A.
With the same radius, draw arcs from C.
The arcs intersect at two points.
Join these points to get a perpendicular line.
This perpendicular line meets AC at O, which is the midpoint.
► The line through O perpendicular to
AC is the other diagonal BD of the square.
► In a square, diagonals are equal. So
BD must also be 8 cm.
With compass at O, radius = half of AC = 4 cm.
Mark two points on the perpendicular bisector:
These are points B and D, each 4 cm from O. So, BO = OD = 4
cm → BD = 8 cm.
► Join A to B, B to C, C to D, and D
to A.
This is the required square ABCD.
Verify if this is true by going through geometric reasoning
in Deduction
1 and Deduction 2, and see if they apply to a square as well.
Answer In a square,
both deductions are fully true based on geometric reasoning. First, the
diagonals of a square are always equal in length. This is because a square is a
special parallelogram in which all angles are 90° and all sides are equal.
Since the diagonals meet at right angles and split the square into four
congruent right triangles, each half of both diagonals is equal. This makes the
full lengths of AC and BD equal, so if one diagonal is 8 cm, the other must
also be 8 cm. Second, the diagonals of a square always bisect each other. A
square has opposite sides parallel, so it behaves like a parallelogram and in
every parallelogram, diagonals cut each other into two equal parts. Therefore,
the intersection point of the diagonals is the midpoint of both AC and BD.
Thus, both Deduction 1 and Deduction 2 hold true for every square.
What are the measures of ∠1, ∠2, ∠3, and ∠4? See if you can reason and/or
experiment to figure this out!
Answer In ΔADC, we
have,
AD = DC, therefore ∠1 = ∠3.
Using angle sum property in ΔADC:
∠1 + ∠3 + 90°
= 180°
⇒ ∠1 + ∠1 + 90°
= 180° [Since ∠1 = ∠3]
⇒ 2∠1 + 90° = 180°
⇒ 2∠1 = 90°
⇒ ∠1 = 90°/2 = 45°
Thus, ∠1 = ∠3 = 45°.[Since
∠1 = ∠3]
Similarly, In ΔABC, we have,
AB = BC, therefore ∠4 = ∠2.
Using angle sum property in ΔABC:
∠2 + ∠4 + 90°
= 180°
⇒ ∠2 + ∠2 + 90°
= 180° [Since ∠2 = ∠4]
⇒ 2∠2 + 90° = 180°
⇒ 2∠2 = 90°
⇒ ∠2 = 90°/2 = 45°
Thus, ∠2 = ∠4 = 45°.[Since
∠2 = ∠4]
Page 94
1. Find all the other angles inside the following rectangles.
AO = BO [Diagonals of rectangle bisect each other]
∠OBA = ∠OAB =
30° [Angles opposite to equal sides are equal]
Similarly, ∠OCD = ∠ODC =
30°
Now, in ΔOAB
∠OBA + ∠OAB + ∠AOB = 180° [Angle sum property of triangle]
⇒ 30° + 30° + ∠AOB = 180°
⇒ ∠AOB = 180° – 60°
⇒ ∠AOB = 120°
∠AOB = ∠DOC =
120° [Vertically opposite angles]
∠BOC + ∠AOB =
180° [Linear Pair]
⇒ ∠BOC + 120° = 180°
⇒ ∠BOC = 180° – 120° = 60°
∠BOC = ∠AOD =
60° [Vertically opposite angles]
In ΔOBC
BO = CO [Diagonals of rectangle bisect each other]
∠OBC = ∠OCB
[Angles opposite to equal sides are equal]
∠OBC + ∠OCB + ∠COB = 180° [Angle sum property of triangle]
⇒ ∠OBC + ∠OBC +
60° = 180° [∠OBC = ∠OCB]
⇒ 2∠OBC = 180° – 60°
⇒ ∠OBC = 120°/2 = 60°
So, ∠OBC = ∠OCB =
60°
Similarly, ∠OAD = ∠ODA =
60°.
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°
Answer:
3. Consider a circle with centre O. 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.
See Explanation Figure APML is a square.
PL and AM are diameters of the circle and are perpendicular
to each other. So they cut the circle into four equal parts (four 90° arcs).
The points A, P, M, L lie on the circle at the ends of these
diameters. So the arcs AP, PM, ML, LA are all equal.
In a circle, equal arcs have equal chords, so AP = PM = ML =
LA
So, all four sides of quadrilateral APML are equal.
Now look at angle at P, ∠APM. It is an angle in a semicircle
(made by diameter AM) and we know that angle in a semicircle is 90°.
So ∠APM = 90°.
Similarly, all other angles (at A, M, L) are also 90° because
each is formed using a diameter.
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?
Answer You can make an
exact 90° angle using two sticks of equal length and a thread by using the idea
of a right-angled triangle.
► Take the thread and tie it so that
its length is exactly the same as the diagonal of the square you want to form.
► Fix the two sticks so that they meet
at one end (like two sides of a square).
► Now stretch the thread between the
free ends of the two sticks.
► Adjust the sticks until the
stretched thread becomes tight and forms the longest side between the ends.
If the two sticks represent two equal sides and the thread
represents the diagonal, then in a square, the diagonal opposite a right angle
is the longest side.
When the thread becomes tight, it forces the angle between
the sticks to become exactly 90°, because only a right angle fits the diagonal
length correctly.
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?
Answer No, this cannot
be used as a definition of a rectangle.
If a quadrilateral has opposite sides parallel and equal, it
becomes a parallelogram.
But not every parallelogram is a rectangle.
A rectangle needs one more special condition: All angles must
be 90°.
In many parallelograms:
► opposite sides are equal
► opposite sides are parallel
► but angles are not 90° (they are
slanted)
So a parallelogram may look like a tilted shape, not a
rectangle.
Page 94
Is it possible to construct a quadrilateral with three angles
equal to 90° and the fourth angle not equal to 90°? But why not?
Answer No, it is not
possible to construct a quadrilateral with three angles equal to 90° and the
fourth angle not equal to 90°.
The sum of all angles in any quadrilateral is 360°.
If three angles are 90°, then 90° + 90° + 90° = 270°
The fourth angle must be 360° − 270° = 90°
So the fourth angle is forced to be 90°.
Page 95
Are there quadrilaterals that have parallel opposite sides
that are not rectangles?
Answer:
Yes, there are quadrilaterals with parallel opposite sides
that are not rectangles.
Any quadrilateral whose opposite sides are parallel is called
a parallelogram.
But not every parallelogram is a rectangle.
Examples of such quadrilaterals:
► Parallelogram
• Opposite sides are parallel.
• Opposite sides are equal.
• Angles are not necessarily 90°.
► Rhombus
• All sides are equal.
• Opposite sides are parallel.
• Angles are not all 90° (only in a square they are).
► Square is a special case of a
parallelogram, but you asked for shapes that are not rectangles, so rhombus and
general parallelograms qualify.
Construct such a figure by recalling how parallel lines can
be constructed using a ruler and a set-square or a compass and a ruler.
Answer:
Construction of a Parallelogram (Not a Rectangle) (using
ruler and set-square OR compass)
Step 1: Draw a line segment AB.
Step 2: Construct a line parallel to AB through a point C.
• Choose a point C anywhere above AB (not vertically above).
• Using a set-square, slide it along the ruler to draw a line
through C parallel to AB.
• Mark a point D on this new line.
Now CD ∥ AB.
Step 3: Construct a line through B parallel to CD.
• Again use the set-square to draw a line through B parallel
to CD.
• This line will meet the line through C at point D
automatically.
So BC ∥ AD.
Step 4: Join A to D and B to C to get quadrilateral ABCD
where AB ∥ CD and BC ∥ AD but angles are not 90°
This is a parallelogram, not a rectangle.
Page 96
Is a rectangle a parallelogram?
Answer Yes, a
rectangle is a parallelogram.
A parallelogram is a quadrilateral in which opposite sides
are parallel.
In a rectangle:
• Opposite sides are parallel
• Opposite sides are equal
• All angles are 90°
So a rectangle satisfies all the conditions of a
parallelogram and has an extra special property (right angles).
Draw a parallelogram with adjacent sides of lengths 4 cm and
5 cm and an angle of 30° between them.
Answer Step 1: Draw
line segments AB = 4 cm and AD = 5 cm with an angle of 30° between them.
Step 2: Draw a line parallel to AB through the point D and a
line parallel to AD through B. Mark the point at which these lines intersect as
C.
ABCD is the required parallelogram.
What are the remaining angles of the parallelogram? What are
the lengths of the remaining sides? See if you can reason out and/or experiment
to figure these out.
Answer In ABCD, ∠A + ∠B = 180° [Adjacent angles of
parallelogram]
⇒ 30° + ∠B = 180°
⇒ ∠B = 180° – 30° = 150°
We know that the opposite angles of parallelogram are equal.
∠B = ∠D = 150°
and
∠A = ∠C = 30°
Given that AD = 5 cm and AB = 4 cm.
BC = AD = 5 cm [Opposite sides of parallelogram are equal]
DC = AB = 4 cm [Opposite sides of parallelogram are equal]
Page 97
What can we say about the angles of a parallelogram?
Answer In a
parallelogram, the adjacent pairs of angles add up to 180° and opposite pairs
of angles are equal.
What about the opposite angles? Will they be equal in all
parallelograms? If yes, how can we be sure?
Answer Yes, the
opposite angles of a parallelogram are always equal.
We can prove it in the following manner:
Let ∠P = x
Since ∠P + ∠R = 180°
[Adjacent angles of parallelogram]
⇒ ∠R = 180° – ∠P = 180° – x.
Similarly, since ∠A + ∠R = 180°,
∠A = 180° – ∠R = 180° – (180 – x) = 180° – 180° + x = x.
Thus, ∠P = ∠A = x.
Similarly, we can deduce that ∠ R = ∠E = 180° – x.
Therefore, it proves that the opposite angles of a
parallelogram are always equal.
What can we say about the sides of a parallelogram?
Answer The opposite
sides of a parallelogram are equal.
Page 98
Is it wrong to write ΔABD ≅ ΔCBD? Why?
Answer Yes, it is
wrong to write ΔABD ≅ ΔCBD according to the figure.
In the figure:
► The two triangles ABD and CBD share
the common side BD.
► They also have some equal angles
shown in the figure.
► But for congruence, we need one of
these conditions:
• SSS (all three sides equal)
• SAS (two sides and the included angle equal)
• ASA or AAS
• RHS (for right triangles)
In the picture:
1. Only one side (BD) is common.
There is no information showing that AB = BC or AD = CD
So the three-side equality (SSS) is missing.
2. The equal angle markings are not the included angle
between known equal sides.
Even if some angles look equal:
• We do not know that the sides around those angles are
equal.
• So SAS congruence does not apply.
3. The shape looks like a parallelogram or general
quadrilateral.
In such shapes the diagonals (BD) divide the quadrilateral
into two triangles,
but these triangles are usually not congruent because the
sides differ.
Are the diagonals of a parallelogram always equal? Check with
the parallelogram that you have constructed.
Answer No, the
diagonals of a parallelogram are not always equal.
In any parallelogram:
► Opposite sides are parallel
► Opposite sides are equal
► Diagonals bisect each other
► But they do not have to be equal in
length.
In the parallelogram, that I have constructed, one diagonal
is 8 cm where as the other is 4.6 cm.
Do they bisect each other (do they intersect at their
midpoints)? Reason and/or experiment to figure this out.
Answer Yes. The
diagonals of a parallelogram always bisect each other.
In a parallelogram:
► Opposite sides are parallel.
► When the diagonals cross, they form
pairs of alternate interior angles that are equal.
Because of these equal angles and parallel sides, the
triangles formed by the diagonals are congruent, which shows AO = OC and BO =
OD
So the diagonals meet at point O, which is the midpoint of
both diagonals.
Page 99
Is it wrong to write ΔAOE ≅ ΔSOY? Why?
Answer Yes, it is
wrong to write ΔAOE ≅ ΔSOY.
To prove two triangles congruent, we need one of these rules:
► SSS (all three sides equal)
► SAS (two sides and included angle
equal)
► ASA (two angles and included side
equal)
► RHS (right angle, hypotenuse, side)
In the figure:
► Triangle AOE is on the left.
► Triangle SOY is on the right.
Even though the diagram shows some equal sides and angles,
the positions of angles and sides do not match in the same order, so the
triangles cannot be paired correctly under any congruency rule.
Do the diagonals of a parallelogram intersect at a particular
angle?
Answer No, the
diagonals of a parallelogram do not intersect at any fixed or particular angle.
In a parallelogram:
► The diagonals always bisect each
other (they cut each other into two equal halves).
► But the angle at which they
intersect can vary.
This angle depends on the shape of the parallelogram:
► Square → diagonals intersect at 90°
► Rectangle → diagonals intersect at
not 90° (usually acute/obtuse)
► Rhombus → diagonals intersect at 90°
► General parallelogram → diagonals
may intersect at any acute or obtuse angle.
Are squares the only quadrilaterals that have equal
sidelengths?
Answer No, squares are
not the only quadrilaterals that have all sides equal.
A square has all four sides equal and all angles equal (90°).
But there is another quadrilateral that also has all sides
equal:
► Rhombus: A rhombus has:
All four sides equal
Opposite sides are parallel
So, a rhombus is a quadrilateral with all sides equal, but it
is not a square unless its angles become 90°.
Can we complete this quadrilateral so that all its sides are
of the same length?
Answer Measure AB
using a compass, mark a point C whose distance from B and D is equal to AB (or
AD). Keeping this length as the radius, cut arcs from B and D.
Now we have a quadrilateral with equal side lengths and one
of its angles 50°.
Page 100
What are the other angles of the rhombus ABCD that we have
constructed?
Answer ∠A + ∠B = 180° [Adjacent angles]
⇒ 50° + ∠B = 180°
[∠A = 50°]
⇒ ∠B = 180° – 50° = 130°
Now,
∠A = ∠C = 50°
[Opposite angles of parallelogram]
∠B = ∠D = 130°
[Opposite angles of parallelogram]
It can be seen that ΔGAE ≅ ΔMAE (How?)
Answer In ΔGAE and
ΔMAE
GE = ME [Sides of rhombus]
GA = MA [Sides of rhombus]
AE = EA [Common]
So, by SSS rule, ΔGAE ≅ ΔMAE.
Page 101
So a rhombus is a parallelogram and a rectangle is also a parallelogram.
How can this be represented using a Venn diagram?
Where will the set of squares occur in this diagram?
Answer:
Are the diagonals of a rhombus equal?
Answer No, the
diagonals of a rhombus are not equal.
Page 102
Do the diagonals of a rhombus intersect at any particular
angle?
Answer Yes, the
diagonals of a rhombus always intersect at a particular angle: they meet at a
right angle (90°).
In a rhombus:
► The diagonals are not equal, but
► They always cut each other exactly
in half (bisect each other) and
► They always intersect at 90°,
forming four right angles at their intersection point.
This property is true for every rhombus.
In the rhombus GAME, we have ΔGEO ≅ ΔMEO. why?
Answer In ΔGEO and
ΔMEO
EG = EM [Sides of rhombus]
GO = MO [Digonals of rhombus bisect each other]
EO = EO [Common]
So, by SSS rule ΔGEO ≅ ΔMEO.
Page 102
1. Find the remaining angles in the following quadrilaterals.
See Explanation(i) ∠P + ∠E = 180°
[Adjacent angles]
⇒ 40° + ∠E = 180°
[∠P = 40°]
⇒ ∠E = 180° – 40° = 140°
∠R = ∠E = 140°
[Opposite angles of parallelogram]
∠A = ∠P = 40°
[Opposite angles of parallelogram]
(ii) ∠P + ∠Q = 180°
[Adjacent angles]
⇒ 110° + ∠Q = 180° [∠P = 110°]
⇒ ∠Q = 180° – 110° = 70°
∠S = ∠Q = 70°
[Opposite angles of parallelogram]
∠R = ∠P = 110°
[Opposite angles of parallelogram]
(iii) In ΔXUV
UX = UV [Sides of rhombus]
So, ∠UXV = ∠UVX =
30° [Angles opposite to equal sides]
Now, ∠UXV + ∠UVX + ∠U = 180° [Angle sum property of triangle]
⇒ 30° + 30°+ ∠U = 180°
⇒ ∠U = 180° – 60° = 120°
∠W = ∠U = 120°
[Opposite angles of rhombus]
In ΔXWV
WX = WV [Sides of rhombus]
So, ∠WXV = ∠WVX
[Angles opposite to equal sides]
Now, ∠WXV + ∠WVX + ∠W = 180° [Angle sum property of triangle]
⇒ ∠WXV + ∠WXV +
120° = 180° [∠WXV = ∠WVX]
⇒ 2∠WXV + 120° = 180°
⇒ 2∠WXV = 180° – 120° = 60°
⇒ ∠WXV = 60°/2 = 30°
(iv) In ΔEAO
AE = AO [Sides of rhombus]
So, ∠AOE = ∠AEO =
20° [Angles opposite to equal sides]
Now, ∠AOE + ∠AEO + ∠A = 180° [Angle sum property of triangle]
⇒ 20° + 20°+ ∠A = 180°
⇒ ∠A = 180° – 40° = 140°
∠I = ∠A = 140°
[Opposite angles of rhombus]
In ΔOIE
OI = IE [Sides of rhombus]
So, ∠IEO = ∠IOE
[Angles opposite to equal sides]
Now, ∠IEO + ∠IOE + ∠I = 180° [Angle sum property of triangle]
⇒ ∠IEO + ∠IEO +
140° = 180° [∠IEO = ∠IOE]
⇒ 2∠IEO + 140° = 180°
⇒ 2∠IEO = 180° – 140° = 40°
⇒ ∠IEO = 40°/2 = 20°
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°.
Steps of ConstructionSteps of Construction:
► Draw a straight line segment AC = 7
cm.
► To find the midpoint of AC, mark the
midpoint O of AC by measuring 3.5 cm from A or C.
► At point O, use a protractor to
construct an angle of 140°.
► Draw a line through O along this
angle. This will be the direction of the second diagonal.
► Mark half of the shorter diagonal
(2.5 cm) on both sides of O. Since the diagonals of a parallelogram bisect each
other, the half-length of the shorter diagonal = 5 ÷ 2 = 2.5 cm.
► From O, measure 2.5 cm on each side
along the 140° line. Mark these points as B and D.
► Join the vertices, to get ABCD, the
required parallelogram.
3. Using the diagonal properties, construct a rhombus whose
diagonals are of lengths 4 cm and 5 cm.
Steps of ConstructionSteps of Construction:
► Draw a point O. This will be the
point where the diagonals intersect.
► Draw a line through O and mark 2.5
cm on both sides of O.
► Mark point A 2.5 cm to the left of
O. Mark point C 2.5 cm to the right of O. Thus, AC = 5 cm (full diagonal).
► Construct a perpendicular line to AC
at point O. Use a set-square or compass to draw a line at 90° to AC.
► On the perpendicular line, mark 2 cm
on both sides of O. Mark point B 2 cm above O.
► Mark point D 2 cm below O. Thus, BD
= 4 cm (full diagonal).
► Join the points A–B, B–C, C–D, and
D–A.
The figure formed is the required rhombus ABCD.
Page 103
Join the ends.
What is the quadrilateral that you get? Justify your answer.
Answer We will get a
square because the diagonals are equal and bisecting each other at right angle.
Class 8 Maths Ganita Prakash Chapter 4 Image 29
Extend one of the diagonals on both sides by 2 cm.
What quadrilateral will you get now? Justify your answer.
Answer We will get a
rhombus because the diagonals are unequal and bisecting each other at right
angle.
Page 104
Can you join them to get a quadrilateral?
What type of a quadrilateral is this? Justify your answer.
Answer Yes we can join
them taking one side as common to get a quadrilateral. Since all sides of
quadrilateral are equal and its diagonals are not equal so it is a rhombus.
Take two cardboard cutouts of an isosceles triangle with
sidelengths 8 cm, 8 cm and 6 cm.
What are the different ways they can be joined to get a
quadrilateral?
What quadrilaterals are these? Justify your answers.
Explanation:
We can join them taking one 6 cm side as common to get a
quadrilateral. In this condition all sides of quadrilateral are equal and its
diagonals are not equal so it is a rhombus.
We can join them taking one 8 cm side as common to get a
quadrilateral. In this condition opposite sides of quadrilateral are equal so
it is a parallelogram.
Page 105
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?
Are you able to identify the different quadrilaterals that
are obtained by joining the triangles? Justify your answer whenever you
identify a quadrilateral.
We can join them taking one 12 cm side as common to get a
quadrilateral. In this condition its two pair of adjacent sides of
quadrilateral are equal so it is a kite.
We can join them taking one 6 cm side as common to get a
quadrilateral. In this condition opposite sides of quadrilateral are equal so
it is a parallelogram.
We can join them taking one 9 cm side as common to get a
quadrilateral. In this condition opposite sides of quadrilateral are equal so
it is a parallelogram.
Property 1: In the kite, show that the diagonal BD
(i) bisects ∠ABC and ∠ADC,
(ii) bisects the diagonal AC, that is, AO = OC, and is
perpendicular to it.
Answer Let the
diagonals of kite ABCD intersect at O.
Given: AB = BC and AD=DC (pairs of equal sides of the kite).
(i) To prove BD bisects ∠ABC and ∠ADC
Consider △ABD and △CBD:
AB = CB (sides of the kite)
AD = CD (sides of the kite)
BD = BD (common side)
So, △ABD ≅△ CBD
by SSS.
Hence, by CPCTC: ∠ABD = ∠DBC and
∠ADB = ∠CDB
So, BD bisects ∠ABC and ∠ADC.
(ii) To prove BD bisects AC (i.e. AO = OC) and is ⟂ to AC
Now consider △AOB and △COB:
AB = CB (given, sides of kite)
∠ABO = ∠CBO
(since BD bisects ∠ABC, from part (i))
OB = OB (common side)
So, △AOB ≅ △COB by SAS.
Therefore, by CPCTC: AO = OC
So BD bisects AC.
Also, ∠AOB = ∠COB.
But ∠AOB + ∠COB =
180 (they form a straight line along AC).
Since they are equal and add to 180, each is 90. So BD ⊥ AC.
Page 106
Construct a trapezium. Measure the base angles (marked in the
figure).
Can you find the remaining angles without measuring them?
Answer By measurment: ∠S = 75° and ∠R = 65°
Since PQ||SR, we have
∠S + ∠P = 180°
⇒ 75° + ∠P = 180°
⇒ ∠P = 180° – 75° = 105°
and ∠R + ∠Q = 180°
⇒ ∠R + 65° = 180°
⇒ ∠R = 180° – 65° = 115°
How do we construct an isosceles trapezium?
Construct an isosceles trapezium UVWX, with UV||XW. Measure ∠U.
Answer Steps to
Construct an Isosceles Trapezium UVWX (with UV ∥ XW)
• Draw a straight line segment UV of any convenient length.
• Draw a parallel line to UV.
• At point U, draw an arc which intersect the parallel line
at X.
• At point V, with the same distance, draw an arc which
intersect parallel line at W.
• Join X to U and and W to V.
Using a protractor we get ∠U = 80°.
Quadrilateral UVWX is now an isosceles trapezium because UV ∥ XW and UX = VW.
What type of quadrilateral is XWZY?
Answer Since XW||UV,
a = 180° – ∠XYZ = 90° and
b = 180° – ∠WZY = 90° (since the internal angles
on the same side of a transversal
add up to 180°)
Hence, XWZY is a rectangle.
Page 107
Now, it can be shown that ΔUXY ≅ ΔVWZ.
(How?)
Answer In ΔUXY and
ΔVWZ
∠XYU = ∠WZV
[Each 90°]
UX = VW [Sides of isosceles trapezium]
XY = WZ [Opposite sides of rectangle]
Therefore, by RHS rule
ΔUXY ≅ ΔVWZ
Page 107
1. Find all the sides and the angles of the quadrilateral
obtained by joining two equilateral triangles with sides 4 cm.
Answer The lengths of
all sides are 4 cm and angles are of 60°, 120°, 60° and 120°.
2. Construct a kite whose diagonals are of lengths 6 cm and 8
cm.
Answer:
3. Find the remaining angles in the following trapeziums —
⇒ ∠A + 135° = 180° [Since ∠D = 135°]
⇒ ∠A = 180° – 135° = 45°
Similarly ∠B + ∠C = 180°
⇒ ∠B + 105° = 180° [Since ∠C = 105°]
⇒ ∠B = 180° – 105° = 75°
∠S + ∠P = 180°
[Adjacent angles]
⇒ ∠S + 100° = 180° [Since ∠P = 100°]
⇒ ∠S = 180° – 100° = 80°
∠P = ∠Q = 100°
[Adjacent angles of issoceles trapezium]
∠R + ∠Q = 180°
[Adjacent angles]
⇒ ∠R + 100° = 180° [Since ∠Q = 100°]
⇒ ∠R = 180° – 100° = 80°
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?
Answer (i) A rhombus
is the quadrilateral that is both:
► a kite (because it has two pairs of
equal adjacent sides) and
► a parallelogram (because opposite
sides are parallel).
So, rhombus = kite + parallelogram.
(ii) Yes, but only in one special case.
A square is the quadrilateral that is both:
► a kite (all sides equal → two pairs
of adjacent equal sides) and
► a rectangle (all angles 90°).
So, a square is both a kite and a rectangle.
(iii) No, every kite is NOT a rhombus.
Correct relationship:
► Every rhombus is a kite (because it
has two pairs of equal adjacent sides).
► But every kite is not a rhombus,
because in a rhombus all four sides are equal, but in a kite only two pairs of
adjacent sides are equal.
5. If PAIR and RODS are two rectangles, find ∠IOD.
∠ORI + ∠IOR + ∠I = 180° [Angles sum property of triangle]
⇒ ∠ORI + ∠IOR +
90° = 180° [Since ∠I = 90°, angle of rectangle PAIR]
⇒ 30° + ∠IOR +
90° = 180° [Since ∠ORI = 30°]
⇒ ∠IOR + 120° = 180°
⇒ ∠IOR = 180° – 120° = 60°
Now, ∠IOD + ∠IOR =
90° [Since ∠DOR = 90°, angle of rectangle RODS]
⇒ ∠IOD + 60° = 90° [Since ∠IOR = 60°]
⇒ ∠IOD = 90° – 60° = 30°
6. Construct a square with diagonal 6 cm without using a
protractor.
Answer:
7. CASE is a square. The points U, V, W and X are the midpoints
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).
Geometric Reasoning:
Let CASE be a square.
Points U, V, W, X are midpoints of CA, AS, SE, EC
respectively.
(a) UVWX is a parallelogram
Using the midpoint theorem:
In △CAS, U and V are midpoints ⇒ UV ∥ CS
In △ASE, V and W are midpoints ⇒ VW ∥ AE
In △SEC, W and X are midpoints ⇒ WX ∥ CS
In △CEA, X and U are midpoints ⇒ XU ∥ AE
Hence, UVWX is a parallelogram.
UV ∥ CS and VW ∥ AE.
In a square, diagonals CS ⟂ AE. A parallelogram with all sides
equal and one right angle is a square.
The 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 are
given below:
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.
See Explanation Geometric Reasoning: Let the quadrilateral be
ABCD with the following properties:
AB = BC = CD = DA (all four sides equal)
∠D = 90°
A quadrilateral with all four sides equal is a rhombus.
In any rhombus, opposite sides are parallel because it is
also a parallelogram.
In a parallelogram, if one angle is 90°, then all four angles
become 90° (opposite and adjacent angles adjust automatically).
Therefore, a rhombus with even one right angle becomes a
rectangle.
So we get a figure that is: a rhombus (all sides equal) and a
rectangle (all angles 90°).
A quadrilateral that is both a rhombus and a rectangle is a
square.
9. What type of a quadrilateral is one in which the opposite
sides are equal? Justify your answer.
Answer A quadrilateral
in which the opposite sides are equal is a parallelogram.
Justification:
If in a quadrilateral ABCD:
► AB = CD (one pair of opposite sides
equal), and
► BC = AD (other pair of opposite
sides equal),
then this matches a known property of parallelograms:
In a parallelogram, opposite sides are equal and parallel.
Also, the converse is true:
If both pairs of opposite sides in a quadrilateral are equal,
then the quadrilateral must be a parallelogram.
• The equal sides ensure that the triangles formed by drawing
one diagonal become congruent (by SSS).
• From congruency, alternate interior angles become equal.
• Equal alternate interior angles imply the opposite sides
are parallel.
• A quadrilateral with both pairs of opposite sides parallel
is a parallelogram.
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.
Answer The sum of the
angles in a quadrilateral such as the following one also will be 360°.
After measurement, we get:
∠A = 40°
∠B = 50°
∠C = 40°
∠D = 230°
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.
Answer (i) False.
If a quadrilateral has diagonals that are equal and bisect
each other, it is a rectangle. A square is a special rectangle with all sides
equal. Here, nothing is said about all four sides being equal, so it need not
be a square.
(ii) A quadrilateral having three right angles must be a
rectangle.
Answer (ii) True.
Sum of angles of a quadrilateral = 360°.
If three angles are 90° each, then
90° + 90° + 90° = 270°
Fourth angle = 360° − 270° = 90°.
So all four angles are right angles → it is a rectangle (or
possibly a square, which is also a rectangle).
(iii) A quadrilateral whose diagonals bisect each other must
be a parallelogram.
Answer (iii) True.
If the diagonals bisect each other, each diagonal is cut into
two equal parts. This is the converse property of a parallelogram. So any
quadrilateral whose diagonals bisect each other must be a parallelogram.
(iv) A quadrilateral whose diagonals are perpendicular to
each other must be a rhombus.
Answer (iv) False.
If diagonals are perpendicular, the quadrilateral could be a
kite or even some other shape, not necessarily a rhombus. In a rhombus,
diagonals are perpendicular and bisect each other, but perpendicularity alone
is not enough to guarantee it is a rhombus.
(v) A quadrilateral in which the opposite angles are equal
must be a parallelogram.
Answer (v) True.
If in a quadrilateral, opposite angles are equal, then the
quadrilateral is a parallelogram (converse of a property of parallelograms: in
a parallelogram, opposite angles are equal).
(vi) A quadrilateral in which all the angles are equal is a
rectangle.
Answer (vi) True.
If all angles in a quadrilateral are equal, each must be 360°
÷ 4 = 90°.
A quadrilateral with all angles 90° is a rectangle (a square
is a special case of a rectangle).
(vii) Isosceles trapeziums are parallelograms.
Answer (vii) False.
An isosceles trapezium has only one pair of opposite sides
parallel and the other pair equal but not parallel. A parallelogram must have
both pairs of opposite sides parallel. So, in general, isosceles trapeziums are
not parallelograms (only a very special case would become a rectangle, but that
would no longer be just a trapezium).
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