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Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.3

1. Draw the graphs for the following tables of values, with suitable scales on the axes.

Question (a)
Cost of apples

Number of apples	1	2	3	4	5
Cost (in ₹)	5	10	15	20	25

Solution:

1. Draw 2 lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale :
On X-axis : 1 cm = 1 apple
On Y-axis 1 cm = ₹ 5
3. Plot the points (1, 5), (2, 10), (3, 15), (4, 20) and (5, 25) on graph paper.
4. Join these points and extend line.

(b) Distance travelled by a car

Time (in hours)	6 a.m.	7 a.m.	8 a.m.	9 a.m.
Distance (in km)	40	80	120	160

Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = 1 unit (Time in hours.)
On Y-axis : 1 cm = 10 km
3. Plot the points (6, 40), (7, 80), (8, 120) and (9, 160) on graph paper.
4. Join these points and extend line.

Question (i)
How much distance did the car cover during the period 7:30 a.m. to 8:00 a.m.?
Solution:
In the graph, draw a perpendicular at the point indicating 7:30 a.m. on the X-axis such that it meets the graph at P.
From P draw a line parallel to X-axis to meet Y-axis at 100 km.
∴ Distance travelled between 7:30 am and 8:00 am.
= (120 – 100) km
= 20 km

Question (ii)
What was the time when the car had covered a distance of 100 km since it’s start?
Solution:
When the car had covered a distance of 100 km, the time was 7 : 30 am.

(c) Interest on deposits for a year.

Deposit (in ₹)	1000	2000	3000	4000	5000
Simple Interest (in ₹)	80	160	240	320	400

Solution :
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 2 cm = ₹ 1000 (deposit)
On Y-axis : 1 cm = ₹ 40 (simple interest)
3. Plot the points (1000, 80), (2000, 160), (3000, 240), (4000, 320) and (5000, 400) on graph paper.
4. Join these points and extend line.

Question (i)
Does the graph pass through the origin?
Solution:
Yes, it passes through the origin.

Question (ii)
Use the graph to find the interest on ₹ 2500 for a year.
Solution:
From the graph, the interest on ₹ 2500 for a year is ₹ 200.

Question (iii)
To get an interest of ₹ 280 per year, how much money should be deposited?
Solution:
From the graph an interest of ₹ 280 can be got by depositing ₹ 3500.

2. Draw a graph for the following:

Question (i)

Side of square (in cm)	2	3	3.5	5	6
Perimeter (in cm)	8	12	14	20	24

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm = 4 cm (Perimeter of a square) )
3. Plot the points (2, 8), (3, 12), (3.5, 14), (5, 20) and (6, 24) on graph paper.
4. Join these points and extend line.

Yes, it is a linear graph.

Question (ii)

Side of square (in cm)	2	3	4	5	6
Area (in cm2)	4	9	16	25	36

Is it a linear graph?
Solution:
1. Draw two lines perpendicular to each other as X-axis and Y-axis on graph paper.
2. Take scale:
On X-axis : 1 cm = 1 cm (Side of a square)
On Y-axis : 1 cm 5 cm (Area of a square)
3. Plot the points (2, 4), (3, 9), (4, 16), (5, 25) and (6, 36) on graph paper.
4. Join these points.

No, this graph is not a straight line. So it is not a linear graph.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral LIFT.
LI = 4 cm,
IF = 3 cm,
TL = 2.5 cm,
LF = 4.5 cm,
IT = 4 cm.
Solution:

Steps of construction:

• Draw a line segment LI = 4 cm.
• With L as centre and radius = 2.5 cm, draw an arc.
• With I as centre and radius = 4 cm, draw an arc to intersect the previous arc at T.
• With L as centre and radius = 4.5 cm draw an arc.
• With I as centre and radius 3 cm, draw an arc to intersect the previous, arc at F.
• Draw \(\overline{\mathrm{LT}}, \overline{\mathrm{IF}}, \overline{\mathrm{FT}}, \overline{\mathrm{LF}}\) and \(\overline{\mathrm{IT}}\).

Thus, LIFT is the required quadrilateral.

Question (ii).
Quadrilateral GOLD
OL = 7.5 cm,
GL = 6 cm,
GD = 6 cm,
LD = 5 cm,
OD = 10 cm.
Solution:

Steps of construction:

• Draw a line segment LD = 5 cm.
• With L as centre and radius = 7.5 cm, draw an arc.
• With D as centre and radius = 10 cm, draw another arc to intersect the previous arc at O.
• With L as centre and radius = 6 cm, draw an arc.
• With D as centre and radius = 6 cm, draw another arc to intersect previous arc at G.
• Draw \(\overline{\mathrm{LO}}, \overline{\mathrm{GO}}, \overline{\mathrm{DG}}, \overline{\mathrm{LG}}\) and \(\overline{\mathrm{DO}}\).

Thus, GOLD is the required quadrilateral.

Question (iii).
Rhombus BEND
BN = 5.6 cm,
DE = 6.5 cm.
Solution:

[Note: Diagonals of a rhombus are perpendicular bisectors of each another. Here, diagonals of □ BEND \(\overline{\mathrm{DE}}\) and \(\overline{\mathrm{BN}}\) intersect at A. So AN = 2.8 cm and AB = 2.8 cm (BN = 5.6 cm, \(\frac {1}{2}\) BN = AN and AB)]
Steps of construction:

• Draw a line segment DE = 6.5 cm.
• Draw perpendicular bisector \(\overleftrightarrow{X Y}\) of \(\overline{\mathrm{DE}}\), which intersects \(\overline{\mathrm{DE}}\) at A.
• With centre at A and radius = 5.6 × \(\frac {1}{2}\) = 2.8 cm, draw two arcs intersecting \(\overleftrightarrow{X Y}\) in points B and N.
• Draw \(\overline{\mathrm{DN}}, \overline{\mathrm{EN}}, \overline{\mathrm{EB}}\) and \(\overline{\mathrm{DB}}\).

Thus, BEND is the required quadrilateral.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.2

1. Plot the following points on a graph sheet. Verify if they lie on a line:

Question (a)
A(4, 0), B (4, 2), C(4, 6), D(4, 2.5)
Solution:

Plotting the given points and then l joining them we find that they all S lie on the same line.

Question (b)
P(1, 1), Q(2, 2), R(3, 3), S(4, 4)
Solution:

Plotting the given points and then joining them we find that they all lie on the same line.

Question (c)
K(2, 3), L(5, 3), M(5, 5), N(2, 5)
Solution:

Plotting the given points and then joining them we find that all of them do not lie on the same line.

2. Draw the line passing through (2, 3) and (3, 2). Find the coordinates of the points at which this line meets the x-axis and y-axis.
Solution:

Plot the given points and join them to make a line. When you extend this line, it meets the X-axis at C (5, 0) and the Y-axis at D (0, 5).

3. Write the coordinates of the vertices of each of these adjoining figures:

Solution:
Figure:
(i) The coordinates of the vertices of quadrilateral OABC:
O are (0, 0)
A are (2, 0)
B are (2, 3)
C are (0, 3)

(ii) The coordinates of the vertices of quadrilateral PQRS:
P are (4, 3)
Q are (6, 1)
R are (6, 5)
S are (4, 7)

(iii) The coordinates of the vertices of triangle KLM:
K are (10, 5)
L are (7, 7)
M are (10, 8)

4. State whether True or False. Correct that are false:

Question (i)
A point whose x-coordinate is zero and y- coordinate is non-zero will lie on y-axis.
Solution:
True

Question (ii)
A point whose y-coordinate is zero and x-coordinate is 5 will lie on y- axis.
Solution:
False
Correct statement: A point whose y-coordinate is 0 and x-coordinate is 5 will lie on X-axis at a distance 5 units from origin.

Question (iii)
The coordinates of the origin are (0, 0).
Solution:
True

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 4 Practical Geometry Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.1

1. Construct the following quadrilaterals:

Question (i).
Quadrilateral ABCD.
AB = 4.5 cm,
BC = 5.5 cm,
CD = 4 cm,
AD = 6 cm,
AC = 7 cm.
Solution :

Steps of construction:

• Draw a line segment AB = 4.5 cm.
• With A as centre and radius = 7 cm, draw an arc.
• With B as centre and radius = 5.5 cm, draw another arc to intersect previous arc at C.
• With A as centre and radius 6 cm draw an arc.
• With centre at C and radius 4 cm, draw another arc to intersect previous arc at D.
• Draw \(\overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{AD}}\) and \(\overline{\mathrm{AC}}\).

Thus, ABCD is the required quadrilateral.

Question (ii).
Quadrilateral JUMP
JU = 3.5 cm,
UM = 4 cm,
MP = 5 cm,
PJ = 4.5 cm,
PU = 6.5 cm.
Solution:

Steps of construction :

• Draw a line segment JU = 3.5 cm.
• With J as centre and radius = 4.5 cm, draw an arc.
• With U as centre and radius = 6.5 cm, draw another arc to intersect previous arc at P
• With U as centre and radius = 4 cm draw an arc.
• With P as centre and radius 5 cm, draw an arc which intersects previous arc at M.
• Draw \(\overline{\mathrm{JP}}, \overline{\mathrm{UM}}, \overline{\mathrm{MP}}\) and \(\overline{\mathrm{UP}}\).

Thus, JUMP is the required quadrilateral.

Question (iii).
Parallelogram MORE ?
OR = 6 cm,
RE = 4.5 cm,
EO = 7.5 cm.
Solution:

[Note: □ MORE is a parallelogram. So length of opposite sides are equal. ]
∴ RE = MO = 4.5 cm; OR = ME = 6 cm
Steps of construction:

• Draw a line segment MO = 4.5 cm.
• With M as centre and radius = 6 cm, draw an arc.
• With O as centre and radius = 7.5 cm, draw another arc to intersect the previous arc at E.
• With O as centre and radius = 6 cm, draw an arc.
• With E as centre and radius = 4.5 cm, draw another arc to intersect the previous arc at R.
• Draw \(\overline{\mathrm{ME}}, \overline{\mathrm{OR}}, \overline{\mathrm{RE}}\) and \(\overline{\mathrm{OE}}\).

Thus, MORE is the required parallelogram.

Question (iv).
Rhombus BEST
BE = 4.5 cm,
ET = 6 cm.
Solution:

[Note : BEST is a rhombus. So length of all four sides are equal.]
BE = 4.5 cm
∴ ES = ST = TB = 4.5 cm
ET = 6 cm (Given)
Steps of construction:

• Draw a line segment BE = 4.5 cm.
• With B as centre and radius = 4.5 cm, draw an arc.
• With E as centre and radius = 6 cm, draw another arc to intersect the previous arc at T.
• With E as centre and radius = 4.5 cm, draw an arc.
• With T as centre and radius = 4.5 cm, draw another arc to intersect previous arc at S.
• Draw \(\overline{\mathrm{BT}}, \overline{\mathrm{ES}}, \overline{\mathrm{ST}}\) and \(\overline{\mathrm{ET}}\).

Thus, BEST is the required quadrilateral.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 15 Introduction to Graphs Ex 15.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 15 Introduction to Graphs Ex 15.1

1. The following graph shows the temperature of a patient in a hospital, recorded every hour:

Question (a)
What was the patient’s temperature at 1 p.m.?
Solution:
The patient’s temperature at 1 p.m. was 36.5 °C.

Question (b)
When was the patient’s temperature 38.5 °C?
Solution:
The patient’s temperature was 38.5 °C at 12 noon.

Question (c)
The patient’s temperature was the same two times during the period given. What were these two times?
Solution:
The patient’s temperature was same (36.5 °C) at 1 p.m. and 2 p.m.

Question (d)
What was the temperature at 1:30 p.m.? How did you arrive at your answer?
Solution:
The patient’s temperature at 1:30 p.m. was 36.5 °C.
(The temperature did not change during interval of 1 p.m. and 2 p.m. So the temperature did not show any change and it was 36.5 °C at 1:30 p.m.)

Question (e)
During which periods did the patients’ temperature showed an upward trend?
Solution:
The patient’s temperature showed an upward trend during the periods 9 a.m. to 10 a.m., 10 a.m. to 11a.m. and 2 p.m. to 3 p.m., because the temperature increased during these intervals.

2. The following line graph shows the yearly sales figures for a manufacturing company:

Question (a)
What were the sales in (i) 2002 (ii) 2006?
Solution:
1. The sales in the year 2002 was ₹ 4 crores.
2. The sales in the year 2006 was ₹ 8 crores.

Question (b)
What were the sales in (i) 2003 (ii) 2005?
Solution:
1. The sales in the year 2003 was ₹ 7 crores.
2. The sales in the year 2005 was ₹ 10 crores.

Question (c)
Compute the difference between the sales in 2002 and 2006.
Solution:
The difference between the sales in 2002 and 2006 = ₹ (8 – 4) crore
= ₹ 4 crores

Question (d)
In which year was there the greatest difference between the sales as compared to its previous year?
Solution:
In year 2005, there was the greatest difference between the sales as compared to its previous year.

3. For an experiment in Botany, two different plants, plant A and plant B were grown under similar laboratory conditions. Their heights were measured at the end of each week for 3 weeks. The results are shown by the following graph:

Question (a)
How high was Plant A after
1. 2 weeks
2. 3 weeks?
Solution:
1. The plant A was 7 cm high after 2 weeks.
2. The plant A was 9 cm high after 3 weeks.

Question (b)
How high was Plant B after
1. 2 weeks
2. 3 weeks?
Solution:
1. The plant B was 7 cm high after 2 weeks.
2. The plant B was 10 cm high after 3 weeks.

Question (c)
How much did Plant A grow during the 3rd week?
Solution:
Plant A grew (9 cm – 7 cm) = 2 cm during 3rd week.

Question (d)
How much did Plant B grow from the end of the 2nd week to the end of the 3rd week?
Solution:
The plant B grew (10cm-7cm) = 3 cm from the end of 2nd week to the end of 3rd week.

Question (e)
During which week did Plant A grow most?
Solution:
The growth of the plant A During the 1st week = 2 cm – 0 cm = 2 cm
During the 2nd week = 7 cm – 2 cm = 5 cm
During the 3rd week = 9 cm – 7 cm = 2 cm
Thus, during the 2nd week, the plant A grew the most.

Question (f)
During which week did Plant B grow least?
Solution:
The growth of the plant B.
During the 1st week = 1cm – 0 cm
= 1 cm
During the 2nd week = 7 cm – 1 cm
= 6 cm
During the 3rd week = 10 cm-7 cm
= 3 cm
Thus, the plant B grew the least in the first week.

Question (g)
Were the two plants of the same height during any week shown here? Specify.
Solution:
At the end of 2nd week, both the plants were of the same height, that is 7 cm.

4. The following graph shows the temperature forecast and the actual temperature for each day of a week.

Question (a)
On which days was the forecast temperature the same as the actual temperature?
Solution:
The forecast temperature was the same as the actual temperature on Tuesday, Friday and Sunday.

Question (b)
What was the maximum forecast temperature during the week?
Solution:
The maximum forecast temperature during the week was 35 °C.

Question (c)
What was the minimum actual temperature during the week?
Solution:
The minimum actual temperature during the week was 15 °C.

Question (d)
On which day did the actual temperature differ the most from the forecast temperature?
Solution:
On Thursday, the actual temperature differed the most from the forecast temperature (7.5 °C).

Difference of temperature:

• Monday : 17.5 °C – 15 °C = 2.5 °C
• Tuesday : 20 °C – 20 °C = o°c
• Wednesday : 30 °C – 25 °C = 5°C
• Thursday : 22.5 °C – 15 °C = 7.5 °C
• Friday : 15 °C – 15 °C = o°c
• Saturday : 30 °C – 25 °C = 5°C
• Sunday : 35 °C – 35 °C = o°c

5. Use the tables below to draw linear graphs:

Question (a)
The number of days a hillside city received snow in different years:

Year	2003	2004	2005	2006
Days	8	10	5	12

Solution:
Linear graph to show snowfall in different years:

Question (b)
Population (in thousands) of men and women in a village in different years:

Year	2003	2004	2005	2006	2007
Number of Men	12	12.5	13	13.2	13.5
Number of Women	11.3	11.9	13	13.6	12.8

Solution:
Draw two perpendicular lines on the graph paper. Take year along X-axis (horizontal line) and population (in thousand) along Y-axis (vertical line).
For men: Mark the points (2003, 12), (2004, 12.5); (2005, 13); (2006, 13.2) and (2007, 13.5) and join them.
For women: Mark the points (2003, 11.3); (2004, 11.9); (2005, 13); (2006, 13.6) and (2007, 12.8) and join them.

6. A courier cycles from a town to a neighboring suburban area to deliver a parcel to a merchant. His distance from the town at different times is shown by the following graph:

Question (a)
What is the scale taken for the time axis?
Solution:
The time is taken along the X-axis. The scale along X-axis is 4 units = 1 hour.

Question (b)
How much time did the person take for the travel?
Solution:
Total travel time taken by a courier : = 8:00 am to 11:30 am = 3\(\frac {1}{2}\) hours

Question (c)
How far is the place of the merchant from the town?
Solution:
Distance of the merchant from the town is 22 km.

Question (d)
Did the person stop on his way? Explain.
Solution:
Yes, the stopage time = 10:00 am to 10:30 am. This is indicated by the horizontal part of the graph.

Question (e)
During which period did he ride fastest?
Solution:
He rode fastest between 8:00 am and 9:00 am.

7. Can there be a time-temperature graph as follows? Justify your answer.

Solution:
In case of (iii), the graph shows different number of temperatures at the same time which is not possible.
∴ Case (iii) does not represent a time-temperature graph.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 3 Understanding Quadrilaterals InText Questions

Try These (Textbook Page No. 43)

Take a regular hexagon Fig 3.10.

Question 1.
What is the sum of the measures of its exterior angles x, y, z, p, q r?
Solution:
∠x + ∠y + ∠z + ∠p + ∠q + ∠r = 360°
(∵ Sum of exterior angles of a polygon = 360°)

Question 2.
Is x = y = z = p = q = r ? Why?
Solution:
Since, all the sides of the polygon are equal, it is a regular hexagon. So its interior angles are equal.
∴ x = (180° – a), y = (180° – a),
z = (180° – a), p = (180° – a),
q = (180° – a), r = (180° – a)
∴ x = y = z = p = q = r

Question 3.
What is the measure of each ?
(i) exterior angle
(ii) interior angle
Solution:
(i) x + y + z + p + q + r = 360°
(∵ Sum of exterior angles = 360°)
All angles are equal.
∴ Measure of each exterior angle = \(\frac{360^{\circ}}{6}\) = 60°

(ii) Exterior angle = 60°
∴ Interior angle = 180° – 60° = 120°.

Question 4.
Repeat this activity for the cases of:
(i) a regular octagon
(ii) a regular 20-gon
Solution:
(i) In a regular octagon, number of sides (n) = 8.
∴ Each exterior angle = \(\frac{360^{\circ}}{8}\) = 45°
∴ Each interior angle = 180° – 45° = 135°

(ii) For a regular 20-gon, the number of sides (n) = 20.
∴ Each exterior angle = \(\frac{360^{\circ}}{20}\) = 18°
∴ Each interior angle = 180° – 18° = 162°

Try These (Textbook Page No. 47)

Question 1.
Take two identical set squares with angles 30°-60°-90° and place them adjacently to form a parallelogram as shown in figure. Does this help you to verify the above property ?

Solution:
Yes, the given figure helps us to verify that opposite sides of a parallelogram are equal.

Try These (Textbook Page No. 48)

Question 1.

Solution:
Yes, this figure also helps us to confirm that opposite angles of a parallelogram are equal.

Think, Discuss and Write (Textbook Page No. 50)

Question 1.
After showing m∠R = m∠N = 70°, can you find m∠I and m∠G by any other method ?
Solution:

Yes, without using the property of parallelogram, we can find m∠I and m∠G.
m∠R = m∠N = 70° (Given)
RG || IN, the transversal RI intersecting them,
∴ m∠R + m∠I = 180° (Sum of interior angles is 180°)
∴ 70° + m∠I = 180° (∵ m∠R = m∠N = 70°)
∴ m∠I = 180° – 70°
∴ m∠I = 110°
Similarly, m∠G = 110°

Think, Discuss and Write (Textbook Page No. 56)

Question 1.
A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular?
Solution :
He can make sure that it is rectangular using the following different ways :

• By making opposite sides of equal length.
• By keeping each angle at the corners as 90°.
• By keeping the diagonals of equal length.
• By making opposite sides parallel.

Question 2.
A square was defined as a rectangle with all sides equal. Can we define it as rhombus with equal angles? Explore this idea.
Solution:
Yes, because a rhombus becomes a square if its all angles are equal.

Question 3.
Can a trapezium have all angles equal ?
Can it have all sides equal ? Explain.
Solution:
Yes, a trapezium can have all angles equal. In this case, it becomes a square or rectangle.
Yes, it can have all sides equal. In this case, it becomes a rhombus or square.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 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 trapeziums.
Solution :
(a) False
(b) True
(c) True
(d) False
(e) False
(f) False
(g) False
(h) False

Question 2.
Identify all the quadrilaterals that have:
(a) four sides of equal length
(b) four right angles
Solution:
(a) Squares as well as rhombuses have four sides of equal length.
(b) Squares as well as rectangles have four right angles.

Question 3.
Explain how a square is
(i) a quadrilateral
(ii) a parallelogram
(iii) a rhombus
(iv) a rectangle
Solution:
(i) A square is a four sided closed figure, so it is a quadrilateral.
(ii) The opposite sides of a square are equal and parallel, so it is a parallelogram.
(iii) All the sides of a square are equal, so it is a rhombus.
(iv) Each angle of a square is a right angle, so it is a rectangle.

Question 4.
Name the quadrilaterals whose diagonals:
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal
Solution:
(i) The diagonals of the following quadrilaterals bisect each other :

• Parallelogram
• Rectangle
• Square
• Rhombus

(ii) The diagonals of the following quadrilaterals are perpendicular bisectors of each other :

• Square
• Rhombus

(iii) The diagonals are equal in following quadrilaterals :

• Square
• Rectangle

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution:

• All the angles have measure less than 180°.
• Both diagonals lie wholly in the interior of the rectangle.

∴ The rectangle is a convex quadrilateral.

Question 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).

Solution:
Produce \(\frac {1}{2}\) to D such that BO = OD.
Joining \(\frac {1}{2}\) and \(\frac {1}{2}\), we get a □ ABCD.
AO = OC (Given)
BO = OD (Constration)
∴ The diagonals AC and BD bisect each other.
∴ □ ABCD is a parallelogram.
∠B is a right angle. (Given)
∴ □ ABCD is a rectangle.
∴ BO = OD = AO = OC
∴ Point O is equidistant from points A, B and C.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 16 Playing with Numbers InText Questions and Answers.

PSEB 8th Class Maths Solutions Chapter 16 Playing with Numbers InText Questions

Try These : [Textbook Page No. 250]

1. Write the following numbers in generalised form:

Question (i)

1. 25
2. 73
3. 129
4. 302

Solution:

1. 25 = 10 × 2 + 5 [ ∵ ab = 10a + b]
2. 73 = 10 × 7 + 3 [ ∵ ab = 10a + b]
3. 129 = 100 × 1 + 10 × 2 + 9 [ ∵ abc = 100a + 10b + c]
4. 302 = 100 × 3 + 10 × 0 + 2 [ ∵ abc = 100a + 10b + c]

Question (ii)
Write the following in the usual form:

1. 10 × 5 + 6
2. 100 × 7 + 10 × 1 + 8
3. 100 × a + 10 × c + b

Solution:

1. 10 × 5 + 6 = 50 + 6 = 56
2. 100 × 7 + 10 × 1 + 8
   = 700 + 10 + 8 = 718
3. 100 × a + 10 × c + b
   = 100a + 10c + b = acb

Try These : [Textbook Page No. 251 ]

1. Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 27
Solution:
Let Sundaram choose the number = 27
Then reversed number = 72
∴ Sum of these two numbers
= 27 + 72 = 99
Now, 99 = 11 (9) = 11 (2 + 7)
= 11 (Sum of the digits of the chosen number)

Question (ii)
2. 39
Solution:
Let Sundaram choose the number = 39
Then reversed number = 93
∴ Sum of these two numbers
= 39 + 93 = 132
Now, 132 = 11 (12) = 11 (3 + 9)
= 11 (Sum of the digits of the chosen number)

Question (iii)
3. 64
Solution:
Let Sundaram choose the number = 64
Then reversed number = 46
∴ Sum of these two numbers
= 64 + 46 = 110
Now, 110= 11 (10) = 11 (6+ 4)
= 11 (Sum of the digits of the chosen number)

Question (iv)
4. 17
Solution:
Let Sundaram choose the number =17
Then reversed number = 71
∴ Sum of these two numbers
= 17 + 71 = 88
Now, 88 = 11 (8) = 11 (1 + 7)
= 11 (Sum of the digits of the chosen number)
[Note: From above results, it is clear that sum of 2-digit number and number obtained by interchanging its digit is multiple of 11, i.e., is divisible by 11, leaving remainder 0.]

Try These [Textbook Page No. 251]

Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 17
Solution:
Let Sundaram choose the number =17
Then reversed number = 71
∴ Difference of these numbers
= 71 – 17 = 54
Now, 54 = 9(6) = 9(7- 1)
= 9 (Difference of the digits of the chosen number)

Question (ii)
2. 21
Solution:
Let Sundaram choose the number = 21
Then reversed number =12
∴ Difference of these numbers
= 21 – 12 = 9
Now, 9 = 9(1) = 9(2 – 1)
= 9 (Difference of the digits of the chosen number)

Question (iii)
3. 96
Solution:
Let Sundaram choose the number = 96
Then reversed number = 69
∴ Difference of these numbers = 96 – 69 = 27
Now, 27 = 9 (3) = 9 (9-6)
= 9 (Difference of the digits of the chosen number)

Question (iv)
4. 37
Solution:
Let Sundaram choose the number = 37
Then reversed number = 73
∴ Difference of these numbers = 73 – 37 = 36
Now, 36 = 9(4) = 9(7 – 3)
= 9 (Difference of the digits of the chosen number)
[Note: From above results, it is clear that difference of 2-digit number and number obtained by interchanging its digit is a multiple of 9, i.e., is divisible by 9, leaving remainder 0.]

Try These: [Textbook Page No. 252]

Check what the result would have been if Minakshi had chosen the numbers shown below. In each case keep a record of the quotient obtained at the end:

Question (i)
1. 132
Solution:
Let Minakshi choose the number =132
Then reversed number = 231
∴ Difference of these numbers = 231 – 132 = 99
Now, 99 ÷ 99 = 1, so remainder = 0

Question (ii)
2. 469
Solution:
Let Minakshi choose the number = 469
Then reversed number = 964
∴ Difference of these numbers = 964 – 469 = 495
Now, 495 ÷ 99 = 5, so remainder = 0

Question (iii)
3. 737
Solution:
Let Minakshi choose the number = 737
Then reversed number = 737
∴ Difference of these numbers = 737 – 737 = 0
Now, 0 ÷ 99 = 0, so remainder = 0

Question (iv)
4. 901
Solution:
Let Minakshi choose the number = 901
Then reversed number =109
∴ Difference of these numbers = 901 – 109 = 792
Now, 792 ÷ 99 = 8, so remainder = 0

[Note: From above results, it is clear s that difference of 3-digit number and number obtained by reversing | its digits (Interchanging unit and ; hundred’s place) is multiple of 99, i.e., is divisible by 99, leaving ’ remainder 0.]

Try These : [Textbook Page No. 253]

Check what the result would have been if Sundaram had chosen the numbers shown below:

Question (i)
1. 417
Solution:
Let Sundaram choose the number = 417
By interchanging the digits, we get two ) numbers, which are 741 and 147.
∴ Sum of these three numbers = 417 + 741 + 174 = 1332
Now, dividing the sum by 37,
1332 ÷ 37 = 36, so remainder = 0

Question (ii)
2. 632
Solution:
Let Sundaram choose the number = 632
By interchanging digits, we get two numbers, which are 263 and 326.
∴ Sum of these three numbers = 632 + 263 + 326 = 1221
Now, dividing the sum by 37,
1221 ÷ 37 = 33, so remainder = 0

Question (iii)
3. 117
Solution:
Let Sundaram choose the number =117 By interchanging digits, we get two numbers, which are 711 and 171.
∴ Sum of these three numbers = 117 + 171 + 711 = 999
Now, dividing the sum by 37,
999 ÷ 37 = 27, so remainder = 0

Question (iv)
4. 937
Solution:
Let Sundaram choose the number = 937
By interchanging digits, we get two numbers, which are 379 and 793.
∴ Sum of these three numbers = 937 + 379 + 793 = 2109
Now, dividing the sum by 37,
2109 ÷ 37 = 57, so remainder = 0
[Note: Sum of 3-digit number and numbers formed by interchanging their digit is divisible by 37, leaving no remainder. ]

Try These : [Textbook Page No. 257]

Question (i)
If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N? (The ones digit, when divided by 5, must leave a remainder of 3. So the ones digit must be either 3 or 8.)
Solution:
The ones digit, when divided by 5 leaves a remainder of 3. So the ones digit must be either 3 or 8.
(Given in textbook, but let us make it simple by solving numerically.)

Question (ii)
If the division N ÷ 5 leaves a remainder of 1, what might be the ones digit of N?
Solution:
The ones digit, when divided by 5 leaves a remainder of 1. So the ones digit must be either 1 or 6.

Question (iii)
If the division N ÷ 5 leaves a remainder of 4, what might be the ones digit of N?
Solution:
The ones digit, when divided by 5 leaves a remainder of 4. So the ones digit must be either 4 or 9.

Try These : [Textbook Page No. 257 – 258]

Question (i)
If the division N ÷ 2 leaves a remainder of 1, what might be the ones digit of N? (N is odd; so its ones digit is odd. Therefore, the ones digit must be 1, 3, 5, 7 or 9.)
Solution:
N is odd, so its ones digit is odd. Therefore, the ones digit must be 1, 3, 5, 7 or 9.
(Given in textbook, but let us make it simple by solving numerically.)

Question (ii)
If the division N ÷ 2 leaves no remainder (i.e., zero remainder), what might be the one’s digit of N?
Solution:
Here, the remainder = 0. So N is an even number, i.e., its ones digit is even. Therefore, the one’s digit must be 0, 2, 4, 6 or 8.

Question (iii)
Suppose that the division N ÷ 5 leaves a remainder of 4 and the division N ÷ 2 leaves a remainder of 1. What must be the ones digit of N?
Solution:
Here, the remainder is 4. So ones digit of N should be 4 or 9. Again, N ÷ 2 leaves a remainder 1.
So ones digit of N is odd, i.e., ones digit is one of these 1, 3, 5, 7 or 9.
∴ 9 is a common ones digit in both the cases.
Therefore, ones digit of N must be 9.

Try These : [Textbook Page No. 259]

Check the divisibility of the following numbers by 9:

Question (i)
1. 108
Solution:
Sum of digits of 108 = 1 + 0 + 8 = 9
Now, 9 ÷ 9 = 1 and remainder = 0
Thus, 108 is divisible by 9.

Question (ii)
2.616
Solution:
Sum of digits of 616 = 6 +1 + 6 = 13
Now, 13 ÷ 9 = 1 and remainder = 4
Thus, 616 is not divisible by 9.

Question (iii)
3. 294
Solution:
Sum of digits of 294 = 2 + 9 + 4 = 15
Now, 15 ÷ 9 = 1 and remainder = 6
Thus, 294 is not divisible by 9.

Question (iv)
4. 432
Solution:
Sum of digits of 432 = 4 + 3 + 2 = 9
Now, 9 ÷ 9 = 1 and remainder = 0
Thus, 432 is divisible by 9.

Question (v)
5. 927
Solution:
Sum of digits of 927 = 9 + 2 + 7 = 18
Now, 18 ÷ 9 = 2 and remainder = 0
Thus, 927 is divisible by 9.

Think, Discuss and Write: [Textbook Page No. 259]

1. You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible by 3 which is a factor of 9. Can you say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m?
Solution:
Yes, if a number is divisible by any number m, then it will also be divisible by each of the factors of m. e.g. 12 is divisible by 6.
Factors of 6 are 2 and 3.
∴ 12 is also divisible by 2 and 3.

2.

Question (i)
Write a 3-digit number abc as 100a + 10b + c
= 99a + 11b + (a – b + c)
= 11 (9a + b) + (a – b + c)
If the number abc is divisible by 11, then what can you say about (a – b + c) ? Is it necessary that (a + c – b) should be divisible by 11?
Solution:
If the number abc is divisible by 11, then (a-b + c) is either 0 or a multiple of 11. Yes, it is necessary that (a + c-b) should be divisible by 11.

Question (ii)
Write a 4-digit number abed as 1000a + 100b + 10c + d = (1001a + 99b + 11c) – (a – b + c – d) = 11 (91a + 9b + c) + [(b + d) – (a + c)] If the number abed is divisible by 11, then what can you say about ((b + d) – (a + c)]?
Solution:
If the number abcd is divisible by 11, then [(b + d) – (a + c)] must be divisible by 11, i.e., [(b + d) – (a + c)] must be 0 or a multiple of 11.

Question (iii)
From (i) and (ii) above, can you say that a number will be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11?
Solution:
Yes, we can say that a number will be divisible by 11, if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11.

Try These : [Textbook Page No. 260]

Check the divisibility of the following numbers by 3:

Question (i)
1. 108
Solution:
Sum of digits of 108 = 1 + 0 + 8 = 9
Now, 9 is divisible by 3.
(∵ 9 ÷ 3 = 3, remainder = 0)
∴ 108 is divisible by 3.

Question (ii)
2. 616
Solution:
Sum of digits of 616 = 6 + 1 + 6 = 13
Now, 13 is not divisible by 3 leaving remainder 0.
(∵ 13 ÷ 3 = 4, remainder = 1)
∴ 616 is not divisible by 3.

Question (iii)
3. 294
Solution:
Sum of digits of 294 = 2 + 9 + 4 = 15
Now, 15 is divisible by 3.
(∵ 15 ÷ 3 = 5, remainder = 0)
∴ 294 is divisible by 3.

Question (iv)
4. 432
Solution:
Sum of digits of 432 = 4 + 3 + 2 = 9
Now, 9 is divisible by 3.
(∵ 9 ÷ 3 = 3, remainder = 0)
∴ 432 is divisible by 3.

Question (v)
5. 927
Solution:
Sum of digits of 927 = 9 + 2 + 7= 18
Now, 18 is divisible by 3.
(∵ 18 ÷ 3 = 6, remainder = 0)
∴ 927 is divisible by 3.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 10 Circles Ex 10.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

In Question 1 to 3, choose the corred option and give justification.

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Solution:
A circle with centre O from a point, Q the length of the tangent to a circle is 24 cm and distance of Q from the centre is 25 cm.

∴ ∠QPO = 90°
Now, in right angled ∠OPQ,
OQ² = PQ² + OP²
(25)² = (24)² + OP²
Or 625 = 576 + OP²
Or OP² = 625 – 576
Or OP² = 49 = (7)²
Or OP = 7 cm
∴ Option (A) is correct.

Question 2.
In Fig., if TP and TQ and tangents to a circle with centre O so ∠POQ = 110°, then ∠PTQ is equal to
(A) 60°
(B) 70°
(C) 80°
(D) 90°

Solution:
In figure, OP is radius and PT is tangent to circle.
∠OPT = 90°
Similarly, ∠OQT = 90° and ∠POQ = 110° (Given)
Now, POQT is a Quadrilateral.
∴ ∠POQ + ∠OQT + ∠PTQ + ∠TPO = 360°
110° + 90° + ∠PTQ + 90° = 360°
Or ∠PTQ + 290° = 360°
Or ∠PTQ = 360° – 290°
Or ∠PTQ = 70°
∴ Option (B) is correct.

Question 3.
In tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 800, then LPOA is equal to
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
In given firgure, OA is radius and AP is a tangent to the circle.

∴ ∠OAP = 90°
Similarly, ∠OBP = 90°
Now, in right angled ∆PAO and ∆PBO
∠PAO = ∠PBO = 90°
OP = OP (Common side)
OA = OB (radii of same Circle)
∴ ∆PAO ≅ ∆PBO [RHS congruence]
∴ ∠AOP = ∠BOP [CPCT]
Or ∠AOP =∠BOP = \(\frac{1}{2}\) ∠AOB
Also, In Quad. OAPB,
∠OBP + ∠BPA + ∠PAO + ∠AOP = 360°
90° +80° +90° + ∠AOB = 360°
∠AOB = 360° – 260°
∠AOB = 100°
Form (1) and (2), we get
∠AOP = ∠BOP = \(\frac{1}{2}\) × 100° = 50°

∴ Option (A) is correct.

Question 4.
Prove that, the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
Given: A circle with center O and AB as its diameter l and m are tangents at points A and B.
To Prove: l || m
Proof: OA is the radius and l is the tangent to the circle.

∴ ∠1 = 90°
Similarly, ∠2 = 90°
Or ∠1 = ∠2 = 90°
But these are alternate angles between two lines, when one transversal cuts them.
∴ l || m
Hence, tangents drawn at the ends of a diameter of a circle are parallel.

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
Given. A circle with centre O. AB its tangent meet circle at P.
i.e., P is the point of contact.

To Prove: Perpendicular at the point of contact to the tangent to a circle passes through the centre.
Construction: Join OP.
Proof: The perpendicular to a tangent line AB through the point of contact passes through the centre of the circle because only one perpendicular, OP can be drawn to the line AB through the point P.
Hence, perpendicular at the point of contact to the tangent to a circle passes through the centre.

Question 6.
The length of a tangent from a point A at a distance 5 cm. from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
A circle with centre ‘O’ A is any point outside the circle at a distance of 5 cm from the centre.

Length of tangent = PA = 4 cm
Since, OP is the radius and PA is the tangent to the circle.
∠OPA = 90°
Now, in right angled ∠OPA.
Using Pythagoras Theorem.
OA² = OP² + PA²
(5)² = OP² + (4)²
Or OP² = 25 – 16
Or OP² = 9 = (3)²
Or OP = 3 cm.
Hence, radius of the circle is 3 cm.

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of Ihe chord of the larger circle which touches the smaller circle.
Solution:
Two concentric circles having same centre O, and radii 5 cm and 3 cm respectively. Let PQ be the chord of larger circle
but tangent to the smaller circle.

Since, OM be the radius of smaller circle and PMQ be the tangent.

∴ ∠OMP = ∠OMQ = 90°
Consider, right angled triangles OMP and OMQ,
∠OMP = ∠OMQ = 90°
OP = OQ [radii of same circle]
OM = OM [common side]
∴ ∆OMP ≅ ∆OMQ [RHS congurence]
∴ PM = MQ [CPCT]
Or PQ = 2PM = 2MQ
Now, in right angled ∆OMQ.
Using Pythagoras Theorem,
OQ² = OM² + MQ²
(5)² = (3)² + (MQ)²
Or MQ = 25 – 9
Or MQ² = 16 = (4)²
Or MQ = 4 cm
Length of chord PQ = 2 MQ = 2 (4) cm = 8 cm
Hence, length of required chord is 8 cm.

Question 8.
A quadrilateral ABCD is drawn to the circumscribe a circle (see figure). Prove that AB + CD = AD + BC.

Solution:
Given: A Quadrilateral ABCD is drawn to circumscribe a circle.
To Prove: AB + CD = AD + BC
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside the circle and BP : BQ are tangents to the circle.
∴ BP = BQ ……………(1)
Similarly, AP = AS ………….(2)
and CR = CQ …………..(3)
Also, DR = DS ………….(4)
Adding (1), (2), (3) and (4), we get
(BP + AP) + (CR + DR) = (BQ + CQ) + (AS + DS)
AB + CD = BC + AD
is the required result.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given: A circle with centre O having two parallel tangents XY and X’Y’ and= another tangent AB with point of contact C
intersecting XY at A and X’Y’ at B.
To Prove: ∠AOB = 90°
Contruction: Join OC, OA and OB.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal.
Now, A is any point outside the circle from two tangents PA and AC are drawn to the circle.
∴ PA = AC
Also, in ∆ POA and ∆ AOC,
PA = AC (Proved)
OA = OA (common side)
OP = OC (radii of same circle)
∴ ∆POA ≅ ∆AOC [SSS congruence]
and ∠PAO = ∠CAO [CPCT]
Or ∠PAC = 2 ∠PAO = 2 ∠CAO ……………(1)
Similarly, ∠QBC = 2∠OBC = 2 ∠OBQ ………………(2)
Now, ∠PAC + ∠QBC = 180°
[Sum of the interior angles on the same side of transversal is 180°]
Or 2∠CAO + 2∠OBC = 180° [Using (1) & (2)]
Or ∠CAO + ∠OBC = 180 = 90° …(3)
Now, in ∆OAB,
∠CAO + ∠OBC + ∠AOB = 180°
90°+ ∠AOB = 180° [Using (3)]
Or ∠AOB = 180° – 90° = 90°
Hence, ∠AOB = 90°.

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre. [Pb. 20191
Solution:
Given. A circle with centre O. P is any point outside the circle PQ and PR are the tangents to the given circle from point P.

To Prove. ∠ROQ + ∠QPR = 180°
Proof. OQ is the radius and PQ is tangent from point P to the given circle.
∴ ∠OQP = 90° ………..(1)
[∵ The tangent at any point of a circle is perpendicular to the radius through the point of contact]
Similarly, ∠ORP = 90°
Now, in quadrilateral ROQP,
∠ROQ + ∠PRO + ∠OQP + ∠QPR = 360°
Or ∠ROQ + 90° + 90° + ∠QPR = 360° [Using (1) & (2)]
Or ∠ROQ + ∠QPR + 180 = 360°
Or ∠ROQ + ∠QPR = 360° – 180°
Or ∠ROQ + ∠QPR = 180°
Hence, the angle between the two tangents drawn from and external point to a circle is supplementary to angle subtended by the line segment joining the pnts of contact at the centre.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Given: A parallelogram ABCD circumscribed a circle with centre O.
To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thé circle and BE; BF are tangents to the circle.

To Prove: ABCD is a rhombus.
Proof: Since the lengths of tangents drawn from an external point to a circle are equal. Now, B is any point outside thè circle and BE; BF are tangents to the circle.
∴ BE = BF
Similarly AE = AH ……………(2)
and CG = CF
Also DG = DH
Adding (1), (2), (3) and (4), we get
(BE + AE) + (CG + DG) = (BF + CF) – (AH + DH)
Or AB + CD = BC +AD ……….(5)
Now, ABCD is a parallelogram, (Given)
∴ AB = CD and BC = AD …………(6)
From (5) and (6), we get
AB + AB = BC + BC
Or 2AB = 2BC or AB = BC
Or AB = BC = CD = AD
∴ ABCD is a rhombus.
Hence, parallelogram circumscribing a circle is a rhombus.

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig). Find the sides AB and AC.

Solution:
A triangle ABC is drawn to circumscribe a circle of radius 4 cm and the sides BC, CA, AB of ∆ABC touch the circle at
D, E, F respectively. Since the lengths of tangents drawn from an external point to a circle are equal.
∴ AE = AF = x cm(say)
CE = CD = 6 cm (Given)
and BF = BD = 5 cm
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OD ⊥ BC; OE ⊥ AC and OF ⊥ AB.
Also, OE = OD = OF = 4 cm.
Consider, ∆ABC
a = AC = (x + 6) cm ;
b = CB = (6 + 8) cm = 14 cm
c = BA = (8 + x) cm
S = \(\frac{a+b+c}{2}\)
∴ S = \(\frac{x+6+14+8+x}{2}\) = \(\frac{2 x+28}{2}\) = (x + 14)

area (∆ABC)= \(\sqrt{\mathrm{S}(\mathrm{S}-a)(\mathrm{S}-b)(\mathrm{S}-c)}\)

= \(\sqrt{\begin{array}{r}
(x+14)(x+14-\overline{x+6}) \\
(x+14-14)(x+14-\overline{8+x})
\end{array}}\)

= \(\sqrt{(x+14)(8)(x)(6)}\)

= \(\sqrt{48 x^{2}+672 x}\) cm² ………………(1)

area (∆OBC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × 14 × 4 = 28cm² …………….(2)

area(∆BOA)= \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (8 + x) × 4 = 28cm²…………….(3)

area (∆AOC) = \(\frac{1}{2}\) × base × altitude
= \(\frac{1}{2}\) × (6 + x) × 4 = 28cm² …………….(4)

From the figure, by addition of areas, we have
Or (∆ABC) = ar (∆OBC) + ar (∆BOA) + ar (∆AOC)
\(\sqrt{48 x^{2}+672 x}\) = 28 + 16 + 2x + 12 + 2x
Or \(\sqrt{48 x^{2}+672 x}\) = 4x + 56
Or 48x² + 672x =4[x+ 14]
Squaring both sides, we get
Or 48x² + 672x = 16 (x + 14)²
Or 48x (x + 14) = 16(x + 14)²
Or 3x = x + 14
Or 2x = 14
Or x = \(\frac{14}{2}\) = 7
∴ AC = (x + 6) cm = (7 + 6) cm= 13 cm
and AB = (x + 8) cm = (7 + 8) cm = 15 cm
Hence, AB = 15 cm and AC = 13 cm.

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
Given:
A quadrilateral PQRS circumscribing a circle having centre O. Sides PQ, QR, RS and SP touches the circles at L, M, N, T
respectively.

To Prove:
∠POQ + ∠SOR = 180°
and ∠SOP + ∠ROQ = 180°
Construction:
Join OP, OL, OQ, 0M. OR, ON, OS, OT
Proof: Since the two tangents drawn from an external point subtend equal angles at the centre.
∴ ∠2 = ∠3; ∠4 = ∠5 ; ∠6 = ∠7; ∠8 = ∠1
But, sum of all angles around a point is 360°
∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360°
Or ∠1 + ∠2 + ∠2 + ∠5 +∠5 + ∠6 + ∠6 + ∠1 = 360°
Or 2(∠1 + ∠2 + ∠5 + ∠6) = 360°
Or (∠1 + ∠2) + (∠5 + ∠6) = \(\frac{360^{\circ}}{2}\) = 180°
Or ∠POQ + ∠SOR = 180°
Similarly, ∠SOP + ∠ROQ = 180°
Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 3 Understanding Quadrilaterals Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 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 = ……………
Solution:
(i) AD = BC
(∵ Opposite sides are equal)

(ii) ∠DCB = ∠DAB
(∵ Opposite angles are equal)

(iii) OC = 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)

Solution:
□ ABCD is a parallelogram.
∴ ZB = ZD
∴ y = 100°
(∵ Opposite angles)
Now, y + z = 180° (∵ Adjacent singles are supplementary)
∴ 100° + z = 180°
∴ z = 180° – 100°
∴ z = 80°
Now, x = z (∵ Opposite angles)
∴ x = 80°
Thus, x = 80°, y = 100° and z = 80°.

(ii)

Solution:
It is a parallelogram.

∴ m∠P + m∠S = 180°
(∵ Adjacent angles are supplementary)
∴ x + 50° = 180°
∴ x = 180° – 50°
∴ x = 130°
x = y (∵ Opposite angles)
∴ y = 130°
Now, m∠Q = 50° (∵ ∠S and ∠Q are opposite angles)
m∠Q + z = 180° (∵ Linear pair of angles)
∴ 50° + z = 180°
∴ z = 180° – 50°
∴ z = 130°
Thus, x = 130°, y = 130° and z = 130°.

(iii)

Solution:
Vertically opposite angles are equal.

∴ m∠AMD = m∠BMC = 90°
∴ x = 90°
In ΔBMC
y + 90° + 30° = 180°
(Angle sum property of a triangle)
∴ y + 120° = 180°
∴ y = 180° – 120° = 60°
Here, ABCD is a parallelogram
∴ \(\overline{\mathrm{AD}} \| \overline{\mathrm{BC}}, \overline{\mathrm{BD}}\) is a transversal.
∴ y = z (∵ Alternate angles)
∴ z = 60° (∵ y = 60°)
Thus, x = 90°, y = 60° and z = 60°

(iv)

Solution:
□ ABCD is a Dparallelogram.

∠D = ∠B (∵ Opposite angles)
∴ y = 80°
m∠A + m∠D = 180° (∵ Adjacent angles are supplementary.)
∴ x + y = 180°
∴ x + 80° = 180°
∴ x = 180° – 80°
∴ x = 100°
m∠A = m∠BCD (∵ Opposite angles are equal)
∴ 100° = m∠BCD
Now, z + m∠BCD = 180° (∵ Linear pair of angles)
∴ z + 100° = 180°
∴ z = 180° – 100°
∴ z = 80°
Thus, x = 100°, y = 80° and z = 80°

(v)

Solution:
□ ABCD is a parallelogram.

∴ m∠B = m∠D (∵ Opposite angles)
∴ y = 112°
m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ (40° + z) + 112° = 180°
∴ 40° + z + 112° = 180°
∴ z + 152° = 180°
∴ z = 180° – 152°
∴ z = 28°
Now, \(\overline{\mathrm{DC}} \| \overline{\mathrm{AB}}\), \(\overleftrightarrow{\mathrm{AC}}\) is their transversal.
∴ z = x
∴ x = 28° (∵ z = 28°)
Thus, x = 28°, y = 112° and z = 28°

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
Solution:
In a quadrilateral ABCD,
∠D + ∠B = 180° (Given)
In parallelogram, the sum of measures of adjacent angles is 180°.
But here, the sum of measures of opposite angles is 180°.
∴ The quadrilateral may be a parallelogram.

(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
Solution:
In a quadrilateral ABDC,
AB = DC = 8 cm
AD = 4 cm
BC = 4.4 cm
∵ Opposite sides AD and BC are not equal.
∴ It cannot be a parallelogram.

(iii) ∠A = 70° and ∠C = 65° ?
Solution:
In a quadrilateral ABCD, m∠A = 70° and m∠C = 65°
Opposite angles ∠A ≠ ∠C
∴ It cannot be a parallelogram.

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

Here, ∠B = ∠D (see figure)
Yet, it is not a parallelogram.

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.
Solution:
Let ABCD be a parallelogram in which adjacent angles ∠A and ∠B are 3x and 2x respectively.

Adjacent angles are supplementary.
∴ m∠A + m∠B = 180°
∴ 3x + 2x – 180°
∴ 5x = 180°
∴ \(\frac{180^{\circ}}{5}\)
∴ m∠A = 3x = 3 × 36° = 108°
m∠B = 2x = 2 × 36° = 72°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 108° and m∠D = 72°
Thus, ∠A = 108°, ∠B = 72°, ∠C = 108° and ∠D = 72°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram such that adjacent angles ∠A = ∠B.

∴ m∠A + m∠B = 180°
(∵ Adjacent angles are supplementary)
∴ m∠A + m∠A = 180° (m∠B = m∠A)
∴ 2m∠A = 180°
∴ m∠A = \(\frac{180^{\circ}}{2}\) = 90°
∴ m∠B = 90°
Now, m∠A = m∠C and m∠B = m∠D (∵ Opposite angles)
∴ m∠C = 90° and m∠D = 90°
Thus, ∠A = 90°, ∠B = 90°, ∠C = 90° and ∠D = 90°.

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

Solution:
∠POA is the exterior angle of ΔHOE
∴ y + z = 70° (Exterior)
∴ m∠HOP = 180° – 70° = 110°
∠HEP = ∠HOP = 110° (Opposite angles of a parallelogram)
∴ x = 110°
∵ \(\overline{\mathrm{EH}} \| \overline{\mathrm{PO}}\), \(\overleftrightarrow{\mathrm{PH}}\) is a transversal.
∴ y = 40° (∵ Alternate angles are equal)
Now, y + z = 70°
∴ 40° + z = 70°
∴ z = 70° – 40° = 30°
Thus, x = 110°, y = 40° and z = 30°

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

Solution :
□ GUNS is a parallelogram.
∴ GS = NU and SN = GU (Y Opposite sides)
∴ 3x = 18 and 26 = 3y – 1
∴ x = \(\frac {18}{3}\)
∴ x = 6

∴ 3y – 1 = 26
∴ 3y = 26 + 1
∴ 3y = 27
∴ y = \(\frac {27}{3}\) = 9
Thus, x = 6 cm and y = 9 cm

(ii)

Solution:
□ RUNS is a parallelogram.
∴ Its diagonals bisect each other.
x + y = 16 ……(1)
and y + 7 = 20
y = 20 – 7 = 13 …..(2)
Substituting value of y in (1)
x + y = 16
x + 13 = 16
∴ x = 16 – 13
= 3
Thus, x = 3 cm and y = 13 cm

Question 9.

In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
□ RISK is a parallelogram.
∴ m∠R + m∠K = 180°
(∵ Adjacent angles are supplementary)
∴ m∠R + 120° = 180°
∴ m∠R = 180° – 120°
∴ m∠R = 60°
∠R and ∠S are opposite angles of parallelogram.
∴ m∠S = 60°
□ CLUE is a parallelogram.
∴ m∠E = m∠L = 70° (∵ Opposite angles)
In ΔESQ
∴ m∠E + m∠S + x = 180° (∵ Angle sum property of triangle)
∴ 70° + 60° + x = 180°
∴ 130° + x = 180°
∴ x = 180° – 130°
∴ x = 50°
Thus, x = 50°

Question 10.
Explain how this figure is a trapezium. Which of its two sides are parallel ? (Fig3.32)

Solution:
Since, 100° + 80° = 180°
i. e., ∠M and ∠L are supplementary.
∴ \(\overline{\mathrm{NM}} \| \overline{\mathrm{KL}}\)
(∵ Interior angles along on the same side of the transversal are supplementary)
One pair of opposite side of □ LMNK is parallel.
∴ □ LMNK is a trapezium.

Question 11.
Find m∠C in figure 3.33 if \(\overline{\mathbf{A B}} \| \overline{\mathbf{D C}}\).

Solution:
In □ ABCD, \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\) and BC is their transversal,
∴ m∠B + m∠C = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ 120° + m∠C = 180°
∴ m∠C = 180° – 120° = 60°
Thus, m∠C = 60°

Question 12.
Find the measure of ∠P and ∠S if \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) in figure 3.34. (If you find m∠R, is there more than one method to find m∠P ?)

Solution:
In □ PQRS \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
∴ □ PQRS is a trapezium.
\(\overleftrightarrow{\mathrm{PQ}}\) is a transversal of \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\)
m∠P + m∠Q = 180°
(∵ Interior angles on the same side of transversal are supplementary)
∴ mZP + 130° = 180°
∴ mZP = 180° – 130°
∴ mZP = 50°
In □ PQRS, ∠R = 90°
\(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\), \(\overleftrightarrow{\mathrm{RS}}\) is their transversal.
∴ m∠S + m∠R = 180°
∴ m∠S + 90° = 180°
∴ m∠S= 180°-90°
∴ m∠S = 90°
Yes, the sum of the measures of all angles of quadrilateral is 360°. ∠P and ∠R can be found.
m∠P + m∠Q + m∠R + m∠S = 360°
∴ m∠P + 130° + 90° + 90° = 360°
∴ m∠P + 310° = 360°
∴ m∠P = 360° – 310°
∴ m∠P = 50°