Class VIII Math
Exercise 8.1
Q1.  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
Ans:
(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
Q2.  How many diagonals does each of the following have?
Ans. Two
(b)   A regular hexagon
Ans. 9
(c)   A triangle
Ans. 0 (zero)
Q3.  What is the sum of the measures of the angles of a convex quadrilateral? Why this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)
Ans. Angle sum of a convex quadrilateral = (4 – 2) × 180° = 2 × 180° = 360°
Since, quadrilateral, which is not convex, i.e. concave has same number of sides i.e. 4 as a convex quadrilateral have, thus, a quadrilateral which is not convex also hold this property. i.e. angle sum of a concave quadrilateral is also equal to 360°
Q4.  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
Ans. Given number of sides = 7
Angle sum of a polygon with 7 sides
= (7 – 2) × 180° = 5 × 180° = 900°
(b) 8
Ans. Given number of sides = 8
Angle sum of a polygon with 8 sides
= (8 – 2) × 180° = 6 × 180° = 1080°
(c) 10
Ans. Given number of sides = 10
Angle sum of a polygon with 10 sides
= (10 – 2) × 180° = 8 × 180° = 1440°
(d) n
Ans. Given number of sides = 7
Angle sum of a polygon with n sides = (n – 2) × 180° = (n – 2)180°
Q5.  What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides (ii) 4 sides
(iii) 6 sides
Ans. A polygon with equal sides and equal angles is called reagular polygon.
(i) Equilateral triangle
(ii) Square
(iii) Regular hexagon
Q6.  (a) Find the angle measures x in the following figures.

Ans. We know that, angle sum of a quadrilateral = 360°
∴ 50° + 130° + 120° + x = 360°
⇒ 300° + x = 360°
⇒ x = 360° – 300°
Q6.           (b)
Ans. We know that, angle sum of a quadrilateral = 360°
∴ 90° + 60° + 70° + x = 360°
⇒ 220° + x = 360°
⇒ x = 360° – 220°
⇒ x = 140° Amswer
Q6.           (c)

∴ 110° + 120° + 30° + x + x = 540°
⇒ 260° + 2x = 540°
⇒ 2x = 540° – 260°
⇒ 2x = 280°

Q6.           (d)
Ans. Angle sum of a pentagon = (5 – 2) x 180°
= 3 × 180° = 540°
Since, it is a regular pentagon, thus, its angles are equal
∴ x + x + x + x + x = 540°
⇒ 5x = 540°

Q7.           (a) Find x + y + z

Ans.
We know that angle sum of a triangle = 180°
Thus, 30° + 90° + C = 180°
Or, 120° + C = 180°
Or, C = 180° – 120°
Or, C = 60°
Now,
y = 180° – C
⇒ y = 180° – 60° = 120°
z = 180° – 30° = 150°
x = 180° – 90° = 90°
∴ x + y + z = 90° + 120° + 150°
⇒ x + y + z = 360° Answer
Alternate method
We know that sum of external angles of a polygon = 360°
(b) Find x + y + z + w
Ans.
We know that angle sum of a quadrilateral = 360°
∴ A + 60° + 80° + 120° = 360°
⇒ A + 260° = 360°
⇒ A = 360° – 260° = 100°
∴ w = 180° – 100° = 80°
x = 180° – 120° = 60°
y = 180° – 80° = 100°
z = 180° – 60° = 120°
∴ x + y + z + w = 60° + 100° + 120° + 80°
⇒ x + y + z + w = 360°
Alternate method
We know that sum of external angles of a polygon = 360°
∴ x + y + z + w = 360° Answer
Exercise 8.2
Q1.  Find x in the following figures.
Ans. We know that sum of exterior angles of a polygon = 360?
∴ 125° + 125° + x° = 360°
⇒ 250° + x° = 360°
⇒ x° = 360° – 250°
⇒ x° = 110°
We know that sum of exterior angles of a polygon = 360°
∴ 70° + x + 90° + 60° + 90° = 360°
⇒ 310° + x = 360°
⇒ x = 360° – 310°
⇒ x = 50°
Q2.  Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
Ans. Since, 9 sides of a polygon has nine angles
And we know that sum of exterior angles of a polygon = 360°
∴ 9 exterior angles = 360°
⇒ 1 exterior angle
∴ each exterior angle = 40°
(ii)    15 sides
Ans. Since, 15 sides of a polygon has 15 angles
And we know that sum of exterior angles of a polygon = 360°
∴ 15 exterior angles = 360°
⇒ 1 exterior angle
∴ each exterior angle = 24°
Q3.  How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Ans. We know that number of angles of a polygon = number of sides
And we know that sum of exterior angles of a polygon = 360°
∵ measure of each angle = 24°
∴ number of exterior angles

∴ Number of sides = 15
Q4.  How many sides does a regular polygon have if each of its interior angles is 165°?
Ans.
∵ Each interior angle = 165°
∴ each exterior angle
= 180° – 165° = 15°
Now ∵ measure of each angle = 15°
∴ number of exterior angles

∴ Number of sides = 24
Q5.  (a) Is it possible to have regular polygon with measure of each exterior angle as 22°?
Ans.
∵Number of sides of a polygon

∴ No. of sides for given polygon

Since, answer is not a whole number, thus, a regular polygon with measure of each exterior angle
as 22° is not possible.
(b) Can it be an interior angle of a regular polygon? Why?
Ans.
Here each interior angle = 20°
∴ each exterior angle = 180° – 22° = 158°
Now ∵measure of each angle = 158°
∴ number of exterior angles

Since, answer is not a whole number, thus, a regular polygon with measure of each interior angle
as 22° is not possible.
Q6.  (a)  What is the minimum interior angle possible for a regular polygon? Why?
Ans. Triangle is the polygon with minimum number of sides and an equilateral triangle is a regular polygon because all sides are equal in this. We know that each angle of an equilateral triangle measures 60 degree. Hence, 60 degree is the minimum possible value for internal angle of a regular polygon.
(b)      What is the maximum exterior angle possible for a regular polygon?
Ans. Each exterior angle of an equilateral triangle is 120 degree and hence this the maximum possible value of exterior angle of a regular polygon. This can also be proved by another principle; which states that each exterior angle of a regular polygon is equal to 360 divided by number of sides in the polygon. If 360 is divided by 3, we get 120.
Exercise 8.3
Q1.  Given a parallelogram ABCD. Complete each statement along with the definition or property used.
Ans. (i)    AD = Opposite Sides Are Equal
(ii)   ∠DCB = Opposite Angles are equal
(iii)  OC = Diagonals Bisect Each Other
(iv)  m ∠DAB + m ∠CDA = 180°
Q2.  Consider the following parallelograms. Find the values of the unknowns x, y, z.
Ans. x = 180° – 100° = 80°
As Opposite angles are equal in a parallelogram
Ans. x, y and z will be complementary to 50°.
So, Required angle = 180° – 50° = 130°
Ans. z being opposite angle= 80°
x and y are complementary, x and y
= 180° – 80° = 100°
Ans. As angles on one side of a line are always complementary
So, x = 90°
So, y = 180° – (90° + 30°) = 60°
The top vertex angle of the above figure
= 60° × 2 = 120°
Hence,
bottom vertex Angle = 120° and z = 60°
Ans. y= 112°, as opposite angles are equal in a parallelogram
x= 180° – (112° – 40°) = 28°
As adjacent angles are complementary so angle of the bottom left vertex
=180° – 112° = 68°
So, z = 68° – 40° = 28°
Another way of solving this is as follows:
As angles x and z are alternate angles of a transversal so they are equal in measurement.
Q3.  Can a quadrilateral ABCD be a parallelogram if
(i)    ∠D = ∠B = 180°?
(ii)   ∠AB = DC = 8 cm,
AD = 4cm and BC = 4.4cm?
(iii)  ∠A = 70° and ∠C = 65°?
Ans. (i)    It can be , but not always as you need to look for other criteria as well.
(ii)   In a parallelogram opposite sides are always equal, here AD BC, so its not a parallelogram.
(iii)  Here opposite angles are not equal, so it is not a parallelogram.
Q5.  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.
Ans. Opposite angles of a parallelogram are always addupto 180°.
So, 180° = 3x + 2x
⇒ 5x = 180°
⇒ x = 36°
So angles are;
36° × 3 = 108° and 36° × 2 = 72°
Q6.  Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Ans. 90°, as they add up to 180°
Q7.  The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Ans. Angle opposite to y = 180° – 70° = 110°
Hence, y = 110°
x = 180° – (110° + 40°) = 30°,
(triangle’s angle sum)
z = 30° (Alternate angle of a transversal)
Q8.  The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
Ans. As opposite sides are equal in a parallelogram
So, 3y – 1 = 26
⇒ 3y = 27
⇒ y = 9
Similarly, 3x = 18
⇒ x = 6
Ans. As you know diagonals bisect each other in a parallelogram.
So, y + 7 = 20
⇒ y = 20 – 7 = 13
Now, x + y = 16
⇒ x + 13 = 16
⇒ x = 16 – 13 = 3
Q9.  In the given figure both RISK and CLUE are parallelograms. Find the value of x.
Ans. In parallelogram RISK
∠ISK = 180° – 120° = 60°
Similarly, in parallelogram CLUE
∠CEU = 180° – 70° = 110°
Now, in the triangle
x = 180° – (110° – 60°) = 10°
Exercise 8.4
Q1.  State whether True or False.
(a) All rectangles are squares
Ans. All squares are rectangles but all rectangles can’t be squares, so this statement is false.
(b) All kites are rhombuses.
Ans. All rhombuses are kites but all kites can’t be rhombus.
(c) All rhombuses are parallelograms
Ans. True
(d) All rhombuses are kites.
Ans. True
(e) All squares are rhombuses and also rectangles
Ans. True; squares fulfill all criteria of being a rectangle because all angles are right angle and opposite sides are equal. Similarly, they fulfill all criteria of a rhombus, as all sides are equal and their diagonals bisect each other.
(f) All parallelograms are trapeziums.
Ans. False; All trapeziums are parallelograms, but all parallelograms can’t be trapezoid.
(g) All squares are not parallelograms.
Ans. False; all squares are parallelograms
(h) All squares are trapeziums.
Ans. True
Q2.  Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
Ans.
(a) If all four sides are equal then it can be either a square or a rhombus.
(b) All four right angles make it either a rectangle or a square.
Q3.  Explain how a square is.
(b) a parallelogram
(c) a rhombus
(d) a rectangle
Ans. (a) Having four sides makes it a quadrilateral
(b) Opposite sides are parallel so it is a parallelogram
(c) Diagonals bisect each other so it is a rhombus
(d) Opposite sides are equal and angles are right angles so it is a rectangle.
Q4.  Name the quadrilaterals whose diagonals.
(a) bisect each other
(b) are perpendicular bisectors of each other
(c) are equal
Ans. Rhombus; because, in a square or rectangle diagonals don’t intersect at right angles.
Q5.  Explain why a rectangle is a convex quadrilateral.
Ans. Both diagonals lie in its interior, so it is a convex quadrilateral.
Q6.  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.
Ans. If we extend BO to D, we get a rectangle ABCD. Now AC and BD are diagonals of the rectangle.
In a rectangle diagonals are equal and bisect each other.
So, AC = BD
AO = OC
BO = OD
And AO = OC = BO = OD
So, it is clear that O is equidistant from A, B and C.