 Exercise 3.1
Question 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
(a) 1, 2, 5, 6, 7
(b) 1, 2, 5, 6, 7
(c) 1, 2 (d) 2
(e) 1
Solution:
(a) Simple curve: (1), (2), (5), (6), (7).
(b) Simple closed curve: (1), (2), (5), (6), (7)
(c) Polygon: (1), (2)
(d) Convex polygon: (2) (e) Concave polygon: (1)
Question 2
How many diagonals does each of the following have?
(b) A regular hexagon
(c) A triangle
(a) 2 (b) 9 (c) 0
Solution:
(a) 2 (b) 9 (c) 0
Question 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 try)
360°; yes.
Solution:
360°, Yes
Question 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 a number of sides? (a) 7 (b) 8 (c) 10 (d) n
(a) 900° (b) 1080° (c) 1440° (d) (n 2)180°
Solution:
(a) Angle sum of a convex polygon
= (n 2) × 180°
= (7 2) × 180° ( n = 7)
= 5 × 180° = 900°
(b) Angle sum of a convex polygon
= (n 2) × 180° = (8 2) × 180° ( n = 8)
= 6 × 180° = 1080°
(c) Angle sum of a convex polygon
= (n 2) × 180° = (10 2) × 180° = 8 × 180° = 1440°
(d) Angle sum of a convex polygon = (n 2) × 180°
Question 5
What is a regular polygon? State the name of a regular polygon of.
(i) 3 Sides (ii) 4 Sides (iii) 6 Sides
A polygon with equal sides and equal angles is called a regular polygon.
(i) Equilateral triangle
(ii) Square
(iii) Regular hexagon a polygon with equal sides and equal angles is called a regular
polygon.
Triangle:
(i) Equilateral
(ii) Square
(iii) Regular hexagon Solution:
Regular polygon:
A regular polygon is both equiangular and equilateral. In other words, a polygon
with equal sides and equal angles is called a regular polygon.
(i) Equilateral triangle (ii) Square (iii) Regular hexagon
Question 6
Find the angle measure x in the following figures:
(a) 60° (b) 140° (c) 55° (d) 108°
Solution:
(a) From the given figure,
50° + 130° + 120° + x = 360°
Or, x = 360° 300° = 60°
(b) From the given figure,
x + 70° + 60° + 90° = 360° or,
0 0 0
360 220 140x
(c) From the given figure,
a + 70° = 180° (Straight angle) or, a = 180° 70° = 110°
Also, b + 60° = 180° or, b = 180° 60° = 120°
Again, from the figure, 30° + x + x + 110° + 120° = 360°
or 2x = 360° 250° = 110° or
0
110
2
x
= 55°
(d) From the figure, sides of the given polygon are equal.
Here, it is a regular pentagon.
Measure of each angle of regular polygon =
540
n
Where n is the number of sides i.e., x =
540
n
(n = 5) or, x = 108°
Question 7
Find x + y + z Find x + y + z + w
(a) x + y + z = 360° (b) x + y + z + w = 360°
Solution:
(a) We known that,
Sum of the angles of a triangle = 180° or
A B C
= 180°
Or, 30° + B + 90° = 180°
( A = 30°, C = 90°)
Or B = 18 120° = 60°
Again from the figure, we have
z + ABC = 180° (Linear pair)
Or, z + 30° = 180° or, z = 180° 30° = 150°
Also from given figure, y + ABC = 180° (Linear pair)
Or, y + 60° = 180° or,y = 180° 60° = 120°
Again, x + BCA = 180° (Linear pair) Or, x + 90° = 180° or, x = 180° 90° = 90°
Thus, the measures are x = 90°, y = 120° and z = 150°
Therefore, x + y + z = 360°
(b) In quadrilateral ABCD, we have, A + B + C + D = 360°
Or, 60° + B + 120° + 80° = 360°
Or, 260° + B = 360° or, B = 360° 260° = 100°
Exercise 3.2
Question 1
Find x in the following figures:
(a) 360° 250° = 110° (b) 360° 310° = 50° Solution:
(a) From the figure, x° + 125° + 125° = 360° or, x° = 360° 250° = 110°
(b) From the figure, x° + 60° + 90° + 90° + 70° = 360°
Or, x + 310° = 360° or, x = 360° 310° = 50°
Question 2
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides (ii) 15 sides.
(i)
o
o
360
40
9
(b)
o
o
360
24
15
Solution:
(i) Measure of each exterior of a regular polygon of n sides
=
360
n
=
360
9
(n = 9) = 40°
Thus, the required measure of each angle is 40°.
(ii) n = 15
Measure of each exterior of a regular polygon of 15 sides =
360
15
= 24°
Question 3
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
360
15
24
(Sides)
Solution:
Given, measure of each exterior angle of a regular polygon = 24°
Measure of each exterior angle of a regular polygon of n sides
=
360
n
or, 24° =
360
n
or, n =
360
24
= 15
Thus, the required number of sides of the regular polygon is 15. Question 4
How many sides does a regular polygon have if each of its interior angles is 165°?
Number of sides = 24
Solution:
Each interior angle of a polygon = 165°
Each exterior angle = 180° 165° = 15°
Sum of exterior angles = 360°
Number of sides =
360
15
= 24
Thus, the required number of sides of the regular polygon is 24.
Question 5
(a) Is it possible to have regular polygon with measure of each exterior angle as 22°?
(b) Can it be an interior angle of a regular polygon? Why?
(i) No (ii) No
Solution:
(a) No, since 22 is not a divisor of 360
(b) No, because each exterior angle is 180° 22° = 158°, which is not a divisor of 360°
Question 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?
(a) 60° (b) 120°.
Solution:
(a) The equilateral triangle being a regular polygon of three sides has the least
measure of an interior angle = 60°
(b) From (a) above, we can see that the greatest exterior angle = 180° 60° = 120° Exercise 3.3
Question 1
Given a parallelogram ABCD.
Complete each statement along with the definition or property used.
(ii) DCB =.........
(iii) OC =........
(iv) m DAB + m CDA …….
(i) BC (Opposite sides are equal)
(ii) DAB (Opposite angles are equal)
(iii) OA (Diagonals bisect each other)
(iv) 180° (Adjacent angles are supplementary)
Solution:
(i) AD = BC (Opposite sides are equal)
(ii) DCB = DAB (Opposite angles are equal)
(iii) OA (Diagonals bisect each other)
(iv) m DAB + m CDA = 180° (Adjacent angles are supplementary.) Question 2
Consider the following parallelograms.
Find the values of the unknowns x, y, z.
(i) x = 80°; y = 100°; z = 80°
(ii) x = 130°; y = 130°; z = 130°
(iii) x = 90°; y = 60°; z = 60°
(iv) x = 100°; y = 80°; z = 80°
(v) y = 112°; x = 28°; z = 28°
Solution:
(i)  D = B (Opposite angles are equal)
y = 100° ( B = 100°)
In parallelogram ABCD,
 A +  B +  C + D = 360° or, z + 100° + x + 100° = 360°
Or, z + x + 200° = 360° or, z + x = 360° 200° = 160°
But, z = x [Opposite angles are equal]
z + z = 160° or, 2 z = 160° or, z = 80° = x
Hence, A = z = 80°, C = x = 80°, D = y = 100°
(ii) ABCD is a parallelogram
D = ABC [Opposite angles are equal]
ABC = 50°
ABC +  CBP = 180° [Linear pair]
50° + CBP = 180°
CBP = 180° 50° = 130°
CBP = z = 130° in || gm ABCD,
A +  B + C + D = 360°
Or, x + 50° + y + 50° = 360°
Or, x + y = 100° = 360°
Or, x + y = 360° 100° = 260°
But, x = y [Opposite angles are equal]
2 x = 260°
x = 130° = y
Hence, x = 130°, z = 130°, y = 130°.
(iii) AOD =  BOC
Here, BOC = 90°
AOD = 90°
In AOD,
Or, 90° + 30° + OAD = 180° or, 120° + OAD = 180°
OAD = 180° 120° = 60°
y = z [AD || CD (alternate angle)]
z = 60°
(iv) ABCD is a parallelogram,
B =  D
y = 80° [D = y]
 B = z [Corresponding angles since AB || DC]
z = 80°
z + BCD = 180°
[Linear pair]
80° + BCD = 180°
BCD = 180° 80° = 100°
BCD = x [Opposite angles are equal]
x = 100° Hence, x = 100°, z = 80°, y = 80°.
(v) B = D [Opposite angles are equal]
D = 112°
In parallelogram ABCD, A +  B + C + D = 360°
Or, A + 112° + C + 112° = 360°
or A + C + 224° = 360°
or A + C = 360° 224°
or A + C = 136°
But A = C
2 A = 136°
A = 68°
A = C = 68°
Also, A = 40° + z
68° = 40° + z
z = 68° 40° = 28°
z = x = 28° [Alternate angles, AB || DC]
H ence, x = 28°, y =112°, z = 28°. Question 3
Can a quadrilateral ABCD be a parallelogram if
D + B = 180°?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
(iii) A = 70° and C = 65°
(i) Can be, but need not be (ii) No (iii) No
Solution:
(i) Yes, it can be but not always.
(ii) No, because in a parallelogram, opposite sides are equal, but here,
.
(iii) No, because in a parallelogram, opposites sides are equal, but here,
AC
.
Question 4
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly
two opposite angles of equal measure.
A kite, for example
Solution:
A kite, for example Question 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.
A = C = 72° and D = B = 108°
Solution:
Suppose ABCD is the parallelogram, two of whose adjacent angles, B and C, say,
are in the ratio of 3:2
B + C = 180° (Interior angles on the same sides of the transversal)
but it is given that,
B: C = 3: 2
Let B = 3x and C = 2x
Hence, 3x + 2x = 180°
Or, 3x + 2x = 180° or, 5x = 180°
x =
180
5
= 36°
So, B = 3x = 3 × 36° = 108°
And C = 2x = 2 × 36° = 72°
Thus, A = C = 72° and D = B = 108°
(Opposite angle of a parallelogram are equal.) Question 6
Two adjacent angles of a parallelogram have equal measure.
Find the measure of each of the angles of the parallelogram.
The measure of each angle is 90°.
Solution:
Let the measure of each of the two adjacent angles be x°.
Since adjacent angles of a parallelogram are supplementary.
x° + x° = 180°
Or, 2x° = 180°
x =
180
2
= 90.
In a parallelogram, opposite angles are equal.
Hence, the measure of each angle is 90°.
Question 7
The adjacent figure HOPE is a parallelogram.
Find the angle measures x, y and z.
State the properties used to find them.
x = 110°; y = 40°; z = 30° Solution:
From the given figure,
70° + POH = 180° or, POH = 180° 70° = 110° (Linear pair)
Opposite angles of a parallelogram are equal.
110° = x x = 110°
Further, EHP =  HPO [Alternate interior angles, EH || PO] or, 40° = y
y = 40°
Again from the figure, we have
y + z + HOP = 180° (Angle sum property of a triangle) or, 40° + z + 110° = 180° or,
z = 180° 150° = 30°
Question 8
The following figures GUNS and RUNS are parallelograms.
Find x and y.
(Lengths are in cm)
(i) x = 6; y = 9 (ii) x = 3; y = 13;
Solution:
(i) GUNS is a parallelogram,
GU = SN [Opposite sides of a || gm] or, 3y 1 = 26 or, 3y = 26 + 1 or, 3y = 27
y =
27
3
= 9
Also, GS = UN [Opposite sides of a || gm] or, 3x = 18
x = 6 Hence, x = 6, y = 9.
(ii) Here, y + 7 = 20 [Diagonals of a || gm bisect each other] or, y = 20 7 y = 13
x + y = 16 [Diagonals of a || gm. bisect each other] or x, + 3 = 16
 x = 16 13 = 3
Hence, x = 3, y = 13.
Question 9
In the given figure both RISK and CLUE are parallelograms.
Find the value of x.
x = 50°
Solution:
RISK is a parallelogram,
K = RIS = 120°
In parallelogram RISK, KRI = ISK = x (say)
SKR +  KRI +  RIS +  ISK = 360°
Or, 120° + x + 120° + x = 360° or, 240° + 2x = 360°
Or, 2x = 360° 240° = 120° or, x = (
120
2
)° = 60° ISK = 60°
CLUE is a parallelogram.
L = CEU = 70°
In triangular shape,
x + CEU + ISK = x + S + E = 180° = x + 70° + 60° = 180° x = 180° 70° 60° = 50°.
Question 10
Explain how figure is a trapezium.
Which of its two sides are parallel?
NM || KL (sum of interior angles on the same side of a transversal is 180°).
So, KLMN is a trapezium.
Solution:
NM || KL (sum of interior angles on the same side of a transversal is 180°).
So, KLMN is a trapezium. Question 11
Find m C in fig if
AB|| DC
.
60°
Solution:
B + C = 180° (
AB|| DC
, interior angles on the same side of a transversal)
Or, 120° + C = 180° or, C = 180° 120° = 60°
C = 60°. Question 12
Find the measure of P and S if
SP || RQ
in fig.
(If you find m R, is there more than one method to find mP)?
P = 50°; S = 90°
Solution:
QRS + RSP = 180° [Sum of interior angles on the same side of a transversal is
180° because
SP || RQ
]
90° + RSP = 180°
RSP = 90°
P + Q + R + S = 360°
P + 130° + 90° + 90° = 360° or, P + 310° = 360°
P = 360° 310° = 50°
P = 50°; S = 90° Exercise 3.4
Question 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 trapezium.
(b), (c), (f), (g), (h) are true; others are false.
Solution:
(a) False (b) True (c) True (d) False (e) False
(f) True (g) True (h) True
Question 2
Identify all the quadrilaterals that have:
(a) Four sides of equal length (b) Four right angles
(a) Rhombus; Square (b) Squares : Rectangle
Solution:
(a) Rhombus; Square
(b) Squares : Rectangle
Question 3
Explain how a square is
(i) A quadrilateral (ii) A parallelogram (iii) A rhombus (iv) A rectangle
(i) A square has four sides, so it is a quadrilateral.
(ii) A square is a rectangle (with adjacent sides equal).
A rectangle is a parallelogram (with an angle a right angle).
Hence, a square is a parallelogram.
(iii) A square is a rectangle with adjacent sides equal.
A rectangle is a parallelogram (with an angle a right angle).
Therefore, a square is a parallelogram with adjacent sides equal.
In a parallelogram, opposite sides are equal. Therefore, a square is a parallelogram with all four sides equal; so it is a rhombus.
(iv) By definition, a square is a rectangle (with adjacent sides equal).
Solution:
(i) A square has four sides, so it is a quadrilateral.
(ii) A square is a rectangle (with adjacent sides equal).
A rectangle is a parallelogram (with an angle a right angle).
Hence, a square is a parallelogram.
(iii) A square is a rectangle with adjacent sides equal.
A rectangle is a parallelogram (with an angle a right angle).
Therefore, a square is a parallelogram with adjacent sides equal.
In a parallelogram, opposite sides are equal.
Therefore, a square is a parallelogram with all four sides equal; so it is a rhombus.
(iv) By definition, a square is a rectangle (with adjacent sides equal).
Question 4
(i) Bisect each other
(ii) Are perpendicular bisectors of each other?
(iii) Are equal.
(i) Parallelogram; rhombus; square; rectangle
(ii) Rhombus; square
(iii) Square; rectangle
Solution:
(i) Parallelogram; rhombus; square; rectangle
(ii) Rhombus; square
(iii) Square; rectangle
Question 5
Explain why a rectangle is a convex quadrilateral.
Both of its diagonals lie in its interior.
Solution:
Both of its diagonals lie in its interior. Question 6
ABC is a right-angled triangle and O is the mid-point of the side opposite to the
right angle.
Explain why O is equidistant from A, B and C.
AD || BC; AB||DC. So, in parallelogram ABCD, the mid-point of diagonal AC is O.
Solution:
Construction: Join BO. Produce it to the point D such that BO = OD. Join DA and DC.
Proof: In quad ABCD, BO = OD (By construction)
AO = OC (O is the mid-point of AC)
i.e., diagonals of quad ABCD bisect each other.
Hence, quad ABCD is a parallelogram.
Also, C = a right angle.
Therefore, quad ABCD is a rectangle.
AC = BD (Diagonals of a rectangle are equal.)
2AO = 2BO
AO = BO
AO = OC
Therefore, AO = BO = CO