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Grade VIMathematics
Exam Preparation Booklet
Chapter WiseImportant Questions-Solutions
Basic Geometrical Ideas Important Questions
Topic: Lines and Curves
1. Name the point(s), line(s), line segment(s), and ray(s) that are there in the following figure:
(2 marks)
2. How many pairs of parallel lines and intersecting lines are there in the following figure? State them.
(2 marks)
3. Classify the following curves as open or closed curves. (a)
(2 marks)
Basic Geometrical Ideas (b)
(c)
(d)
4.
What are the vertex and the arms of the angle shown in the given figure?
(2 marks)
5. Two lines segments AB and CD intersect at a point O as shown in the given figure. AD and BC are the other line segments in the figure.
(2 marks)
Basic Geometrical Ideas
How many different angles are there inside the given figure? What are they?
6. Name the angles in the given triangle PQR. Name the arms and vertices related to these angles. Also, name the point(s) that lie in the interior, exterior, and on the boundary of ΔPQR.
(3 marks)
7.
(3 marks)
Basic Geometrical Ideas How can the region obtained by excluding the common region of the interior of the angles AOD and COF from the common region of the interior of the angles AOE and BOF be shaded?
Topic: Polygon and Triangles
8. How many curves are polygons in the given figure?
(i) (ii)
(iii) (iv)
(2 marks)
Basic Geometrical Ideas 9. How many diagonals are there in the following polygon? Name them.
(2 marks)
10.
(a) How many triangles are there with one of their vertices as A? What are they? (b) How many triangles are there with one of their sides as CD? What are they?
(2 marks)
11. Name the triangle in each of the following cases. (a) AB = 4 cm, BC = 3 cm, CA = 4 cm (b) PG = 2 cm, GR = 3 cm RP = 4 cm (c) ∠A = 90°, AB = 3 cm, AC = 4 cm
(3 marks)
Basic Geometrical Ideas Topic: Quadrilaterals and Circle
12.
(a) What is the name of the quadrilateral whose vertices are the given points? (b) What are the pairs of adjacent sides of the quadrilateral? (c) What are the pairs of opposite angles of the quadrilateral?
(3 marks)
13. Are the following statements correct? Justify your answer. (a) Every rectangle is a square. (b) Every square is a rhombus.
(3 marks)
14. (a) Find the length of radius of a circle whose diameter is 24 cm in length. (b) What is the length of the longest chord of this circle?
( 3 marks)
Basic Geometrical Ideas 15. The given figure shows a circle with centre O. Line segments AB, CD,
and EF are diameters of the circle. G is a point on the boundary of the circle.
(a) How many different semicircles are shown in the given figure? Name them (b) How can the segment of circle corresponding to the arc AFD be shaded? (c) How can the sector of circle corresponding to the arc CGEB be shaded?
(3 marks)
Basic Geometrical Ideas Solutions
Topic: Lines and Curves
1. Points: P, Q, R, and S
Line:
Line segments:
Rays:
2. There are 3 pairs of parallel lines. They are: p and q , q and r , r and p There are 7 pairs of intersecting lines in the given figure. They are: l and m , l and p , l and q , l and r , m and p , m and q , m and r
3. If the end points of a curve are visible in a curve, then it is called on open curve. Hence, curves (a) and (d) are open curves. If the end points of a curve are not visible in a curve, then it is called a closed curve. Hence, curves (b) and (c) are closed curves.
4. Two rays forming an angle are called the sides (or arms) of the angle. The common end point of the two rays that forms the angle is called the vertex. The angle shown in the given figures is ∠MLN. This angle has vertex L and its arms are
and .
5. There are six different angles inside the given figure. They are ∠AOD, ∠OAD, ∠ODA, ∠BOC, ∠OBC and ∠OCB.
6. The angles of the triangle PQR with the related arms and vertices are given in the table as shown below:
Angle Arms Vertex
∠PQR and
Q
Basic Geometrical Ideas
∠QRP and
R
∠RPQ and
P
∠PRS and
R
∠QRS and
R
In ΔPQR, (a) B and C are interior points (b) D is an exterior point (c) P, Q, R, and A are the points on the boundary of the triangle
7. Common region of the interior of the angles AOE and BOF is the region inside ∠BOE. Common region of the interior of the angles AOD and COF is the region inside ∠COD. Therefore, region obtained by excluding the common region of the interior of the angles AOD and COF from the common region of the interior of the angles AOE and BOF is same as the region inside ∠BOE excluding the region inside ∠COD. This region can be shaded as:
Topic: Polygon and Triangles
8. A closed curve that is entirely made up of line segments is called a polygon. In the figure, it is observed that: Curve (i) is not entirely made up of line segments. Curve (ii) is not closed. Curve (iv) is not closed. Curve (iii) is closed and is entirely made up of line segments. ∴ Curve (iii) is a polygon. Thus, there is only 1 polygon in the given figure.
Basic Geometrical Ideas
9. The diagonals of the given polygon are: PR, PS, PT, PU, QS, QT, QU, QV, RT, RU, RV, SU, SV, and TV There are 14 such diagonals.
10. (a) A triangle is a closed plane figure formed by three line segments. It is seen that there are only 6 triangles with one of their vertices as A. They are ΔABC, ΔACD, ΔACE, ΔADE, ΔAEP, and ΔAPC (b) There are only 3 triangles with one of their sides as CD. They are ΔACD, ΔECD, and ΔPCD.
11. (a) In ΔABC, two sides are of equal measure. Thus, it is an isosceles triangle. (b) All the sides of the given triangle are unequal. Thus, it is a scalene triangle. (c) In the given triangle, ∠A is 90°. Thus, it is a rightangled triangle.
Basic Geometrical Ideas Topic: Quadrilaterals and Circle
12. (a) The vertices of the quadrilateral are A, P, L, and X. It is known that a quadrilateral is named by naming the vertices in a cyclic manner. One of the names is APLX.
(b) In a quadrilateral, sides that share a common point (vertex) are called adjacent sides whereas sides that lie opposite to each other are called opposite sides. The pairs of adjacent sides of the quadrilateral are AP and AX, AP and PL, PL and LX, LX and AX. (c) Angles that share a common arm (side) are called adjacent angles whereas angles that lie opposite to each other are called opposite angles. The pairs of opposite angles of the quadrilateral are ∠A and ∠L, ∠P and ∠X.
13. (a) The statement, “every rectangle is a square”, is false. A rectangle is a type of quadrilateral in which opposites sides are equal and all the interior angles are of measure 90°. Therefore, in rectangle, it is not necessary that all sides are equal. Thus, every rectangle is not a square. (b) A rhombus is a type of quadrilateral in which all the sides are equal. Thus, the statement, “every square is a rhombus”, is true because in square, all the sides are of equal length. A square is a type of quadrilateral in which all sides are equal and all the interior angles are of measure 90°.
14. (a) We know that the diameter of a circle is double the size of the radius.
Basic Geometrical Ideas Thus, the length of radius of a circle is half the length of the diameter. ∴ Radius of the circle = 24 cm ÷ 2 = 12 cm (b) We know that the longest chord of a circle is its diameter. Thus, length of the longest chord of the circle = Length of diameter = 24 cm
15. (a) The diameter of a circle divides the circle into two equal parts and each part is called a semicircle i.e., a semicircle is the half of a given circle. Thus, the semicircles formed by the diameter AB are ACGEBOA and AFDBOA. The semicircles formed by the diameter CD are CGEBDOC and CAFDOC. The semicircles formed by the diameter EF are EBDFOE and EGCAFOE. Therefore, 6 semicircles are shown in the given figure. They are ACGEBOA, AFDBOA, CGEBDOC, CAFDOC, EBDFOE, and EGCAFOE. (b) The segment of a circle is the region in the interior of the circle enclosed by a chord and its corresponding arc. The segment of circle corresponding to arc AFD can be shaded as:
(c) The sector of a circle is the region in the interior of the circle enclosed by an arc on one side and a pair of radii on the other two sides. The sector of the circle corresponding to arc CGEB can be shaded as:
Understanding Elementary Shapes
Important Questions
Topic: Comparing Lengths of Line Segments
1. In the following figure, AB = 2 cm, AD = 9 cm, BC = 3 cm, and AE = 10 cm.
Find CD and DE. Also find the midpoint of .
(2 marks)
2. P, Q, R, and S are four points on a line such that ,
, and . Which two points are lying between the remaining two?
(2 marks)
Topic: Angles
3. Classify each of the following marked angles as right, straight, acute, obtuse, reflex, or complete. (a)
(b)
(c)
(3 marks)
Understanding Elementary Shapes
(d)
(e)
(f)
4. Where will the hour hand of a clock stop if (3 marks)
Understanding Elementary Shapes
(a) it starts at 1 and makes a revolution of angle 90° clockwise? (b) it starts at 4 and makes a revolution of angle 180° clockwise?
(c) starts at 8 and makes of a revolution clockwise?43
5. Using the protractor, find the difference between the measures of the angles ABC and PQR in the given figure?
(2 marks)
6. In the following figure, state with reason that is a perpendicular bisector
of whereas is not a perpendicular bisector of .
(2 marks)
7. The given figure shows a ΔABC where D is a point on BC such that AD is the perpendicular bisector of the side BC.
(2 marks)
Understanding Elementary Shapes
If BC = 10 cm, then what is the length of BD and the measure of ∠ADB?
Topic: Two Dimensional and Three Dimensional Shapes I
8. Classify the following triangles on the basis of angles or sides as given in the bracket. (a) In ΔPQR, PQ = 8 cm, ∠Q = 90°, and QR = 8 cm (both angle and side) (b) In ΔXYZ, XY = 13 cm, YZ = 13 cm, and ZX = 10 cm (side) (c) In ΔABC, AB = 5 cm, BC = 9 cm, and ∠ABC = 135° (angle) (d) In ΔLMN, ∠L = 49°, ∠M = 90°, and ∠N = 41° (angle) (e) In ΔABC, AB = 11 cm, BC = 9 cm, CD = 15 cm (side) (f) In ΔPQR, ∠P = 40°, ∠Q = 60°, and ∠R = 80° (angle)
(3 marks)
9. Name the following quadrilaterals, in which (a) all sides are equal and the diagonals are equal and perpendicular to each other (b) opposite sides are parallel (c) only one pair of opposite sides are parallel (d) opposite sides are equal and the diagonals are equal
(2 marks)
10. Name two quadrilaterals in which (a) all sides are equal (b) diagonals are of same length (c) all the interior angles are of equal measure
(3 marks)
11. What is/are the differences in the property/properties of a rectangle and a parallelogram?
(2 marks)
Topic: Two Dimensional and Three Dimensional Shapes II
12. Name the following polygons, whose (a) number of sides is equal to the sum of numbers of sides of a pentagon and
(3 marks)
Understanding Elementary Shapes
a triangle (b) number of sides is equal to the difference between seven times the numbers of sides of a triangle and twice the numbers of sides of an octagon Justify your answer with reasons.
13. What is the difference between a regular pentagon and a hexagon? (2 marks)
14. Write the number of faces, edges, and corners for the following shapes. (a) A hexagonal pyramid
(b) A pentagonal pyramid
( 3 marks)
15. What is the difference between the number of rectangular faces and triangular faces of a triangular prism?
(2 marks)
Understanding Elementary Shapes
Solutions
Topic: Comparing Lengths of Line Segments
1. We have AB = 2 cm, BC = 3 cm Since A, B, C, D, and E are on a line, AC = AB + BC = 2 cm + 3 cm = 5 cm CD = AD − AC = 9 cm − 5 cm = 4 cm DE = AE − AD = 10 cm − 9 cm = 1 cm It can be observed that AC = 5 cm and CE = CD + DE = 4 cm + 1 cm = 5 cm
This means point C divides the line segment into two equal parts. Therefore, C is the
midpoint of .
2. It is given that: and However, 12 cm = 3 cm + 7 cm + 2 cm
i.e., This shows that the points Q and S lie between the points P and R. This can be represented using figure as:
Topic: Angles
3. (a) The marked angle is an acute angle. (b) The marked angle is an obtuse angle. (c) The marked angle is a right angle. (d) The marked angle is a reflex angle. (e) The marked angle is a straight angle. (f) The marked angle is a complete angle.
4. (a) If the hour hand of a clock starts at 1 and makes a revolution of angle 90° clockwise, then it will stop at 4. This can be shown as:
Understanding Elementary Shapes
(b) If the hour hand of a clock starts at 4 and makes a revolution of angle 180° clockwise, then it will stop at 10. This can be shown as:
(c) It is known that of a revolution or a complete angle is 1 right angle.41
Thus, of a revolution or a complete angle is 3 right angles i.e., 3 × 90° = 270°43
Therefore, if the hour hand of a clock starts at 8 and makes of a revolution clockwise, then43
it will stop at 5. This can be shown as:
5. In order to measure the angles ABC and PQR, CB is produced to D and RQ is produced to S.
Understanding Elementary Shapes
∠DBC and ∠RQS are straight angles. ∴ ∠DBC = ∠RQS = 180° By using the protractor, the measures of the angles ABD and PQS are found as: ∠ABD = 60° and ∠PQS = 135° ∴Measure of the angle shown in the first figure = 180° + 60° = 240° Measure of the angle shown in the second figure = 180° + 135° = 315° Thus difference between the measure of the given angles = 315° − 240° = 75°
Error6. From the figure, we observed that and are perpendicular to each other at P and
whereas
Clearly, is a perpendicular bisector of whereas is not a perpendicular bisector
of .
7. In ΔABC, it is given that AD is the perpendicular bisector of the side BC. Thus, AD bisects BC. Also, AD is perpendicular to BC.
and ∠ADB = 90° Thus, the length of BD is 5 cm and the measure of ∠ADB is 90°.
Topic: Two Dimensional and Three Dimensional Shapes I
Understanding Elementary Shapes
8. (a) The given triangle has two equal sides (PQ = QR = 8 cm) with a right angle (∠Q = 90°). Therefore, ΔPQR is a rightangled isosceles triangle. (b) In the given ΔXYZ, two sides are of equal length (XY = YZ = 13 cm). Hence, it is an isosceles triangle. (c) In the given ΔABC, an angle is obtuse (∠ABC = 135°). Therefore, it is an obtuse triangle. (d) In the given ΔLMN, an angle is a right angle (∠M = 90°). Hence, it is a rightangled triangle. (e) In the given ΔABC, all sides are of different lengths. Therefore, it is a scalene triangle. (f) In the given ΔPQR, all angles are less than 90°. This means all angles are acute. Hence, it is an acute triangle.
9. (a) The quadrilateral in which all sides are equal and the diagonals are equal and perpendicular to each other is a square. (b) A parallelogram is a quadrilateral whose opposite sides are parallel. (c) A trapezium is a quadrilateral whose only one pair of opposite sides is parallel. (d) A rectangle is a quadrilateral whose opposite sides are equal and the diagonals are equal.
10. (a) In square and rhombus, all sides are equal.
Here, ABCD is a rhombus and AB = BC = CD = DA
Here, PQRS is a square and PQ = QR = RS = SP (b) In rectangle and square, the diagonals are of same length.
Understanding Elementary Shapes
Here, ABCD is a rectangle and AC = BD. PQRS is a square and PR = QS (c) In rectangle and square, all the interior angles are of equal measure, each measuring 90°.
ABCD and PQRS are square and rectangle respectively. Here, ∠A = ∠B = ∠C = ∠D = 90°, ∠P = ∠Q = ∠R = ∠S = 90°
11. (a) The common properties of a square and a rhombus can be listed as: (i) Opposite angles are equal. (ii) Diagonals are perpendicular to each other. (iii) All the sides are equal. (iv) Opposite sides are parallel. (b) There are two differences between a rectangle and a parallelogram, which are as follows. (i) The diagonals of a rectangle are equal whereas the diagonals of a parallelogram may or may not be equal. (ii) The measure of all angles of a rectangle is 90°, whereas the measure of all angles of a parallelogram may or may not be of 90°.
Topic: Two Dimensional and Three Dimensional Shapes II
12. (a) Pentagon has 5 sides and a triangle has 3 sides. Now, 5 + 3 = 8 We know that an octagon is a polygon of 8 sides. Clearly, an octagon is a polygon whose numbers of sides is equal to the sum of the numbers of sides of a pentagon and a triangle. (b) A triangle has 3 sides and 7 × 3 = 21
Understanding Elementary Shapes
An octagon is a polygon of 8 sides and 2 × 8 = 16 Now, 21 − 16 = 5 We know that a pentagon is a polygon of 5 sides. Clearly, a pentagon is a polygon whose numbers of sides is equal to the difference between seven times the numbers of sides of a triangle and twice the numbers of sides of an octagon.
13. A regular pentagon is a pentagon having 5 sides where all the sides are of equal length and all the angles are of equal measure. A hexagon is a polygon having 6 sides. In hexagon, all the sides may or may not be equal and all the angles may or may not be equal.
14. (a) Faces: 7 Edges: 12 Corners 7 (b) Faces: 7 Edges: 15 Corners: 10
15.
A triangular prism has 2 triangular faces and 3 rectangular faces as shown in the figure. Thus, the difference between the number of rectangular faces and triangular faces of a triangular prism is 3 − 2 = 1.
Data Handling Important Questions
Topic: Tally Chart
1. In a group of 40 students, 13 students like to play football and other students like to play cricket. How can the number of students that like to play cricket be represented using tally marks?
(2 marks)
2. The weights of 30 students in a class to the nearest kilogram are given below:
27 26 27 30 32 33 30 26
30 33 28 29 30 32 33 31
30 26 28 27 32 30 27 30
31 32 30 27 28 33
Prepare a table using tally marks for the given data.
(2 marks)
3. The following data shows the hobbies of 30 employees of a company: Dance, Music, Art, Art, Music, Sports, Reading, Dance, Dance, Music, Art, Reading, Reading, Sports, Music, Music, Sports, Dance, Sports, Reading, Reading, Music, Art, Dance, Sports, Sports, Reading, Dance, Art, Reading Organise the data in the form of a frequency distribution table.
(2 marks)
4. The following table of tally marks represents the marks awarded for an assignment set in a class of 20 students in a particular year.
Marks Tally Marks
4
5
6
(2 marks)
Data Handling
7
8
9
10
Answer the following questions regarding the given table. (a) How many students obtained marks above 6? (b) What is the difference between the number of students who obtained the highest and the lowest marks?
5. The following tally chart provides the information regarding the heights of all students of grade 6 to the nearest centimetre.
Height (in cm) Tally marks
135
136
137
138
139
(5 marks)
Data Handling
140
141
Read the tally charts and answer the following questions: (a) Write the number of students with heights 135 cm and 140 cm. (b) What is the total number of students in grade 6? (c) How many students are the tallest in grade 6? (d) How many students’ height is less than 138 cm? (e) How many students’ height is more than or equal to 139 cm?
Topic: Pictographs
6. Use the following information to answer the next question.
The given table shows the number of players selected for trial from a certain school for participating in interschool sports meet in different types of games − cricket, football, tennis, and badminton.
Game Number of players selected for trial
Cricket 50
Football 35
Lawn tennis 15
Badminton 10
(a) Represent the given data in the form of a pictograph by using one symbol to represent 10 players. (b) How many symbols are used for representing the number of football players selected for trial?
(3 marks)
Data Handling 7. How can the given information be represented using a pictograph assuming one
represents 4 students?
(3 marks)
8. How can the given information be represented in the form of a pictograph? Butterscotch Strawberry Butterscotch Chocolate Vanilla Strawberry
Butterscotch Vanilla Butterscotch Vanilla Butterscotch Butterscotch
Butterscotch Strawberry Chocolate Chocolate Butterscotch Butterscotch
Butterscotch Chocolate Vanilla Butterscotch Butterscotch Butterscotch
Vanilla Strawberry Chocolate Vanilla Vanilla Strawberry
Strawberry Vanilla Chocolate Chocolate Butterscotch Butterscotch
Vanilla Chocolate Butterscotch Vanilla
(2 marks)
9. The following pictograph represents the number of pizzas manufactured by a pizza company in different years.
Year Number of pizzas = 10000 pizzas
2004
(3 marks)
Data Handling
2005
2006
2007
2008
Observe the pictograph and answer the following questions. (a) In which year(s) was the highest number of pizzas manufactured? (b) In which year(s) was the least number of pizzas manufactured? (c) What is the total number of pizzas manufactured in all the five years?
10. The given pictograph shows the monthly earnings of a shopkeeper in the first four months of a particular year through selling eggs.
Month Earning (Rs)
January
February
March
April
Key : = Rs. 5,000
How can the months of the year be arranged according to the increasing or decreasing order of the monthly earnings of the shopkeeper?
(2 marks)
Data Handling 11. The given pictograph shows the number of fishes caught by four persons Joseph,
Bunty, Lalit, and Naveen from a lake on a particular evening.
Name Number of fishes caught
Joseph
Bunty
Lalit
Naveen
Key : 1 = 2 fishes
If the key of the pictograph is taken as 1 = 3 fishes, then how can the given pictograph be redrawn?
(2 marks)
Topic: Bar Graphs
12. Use the following information to answer the next question. Prepare a bar graph for the given data. Manish threw a dice 50 times and noted the number appearing each time. The given table shows the numbers appearing each time.
Prepare a bar graph for the given data.
(3 marks)
13. The given tally chart shows the number of car accidents in a city during a week of a particular month.
Day Tally marks
Sunday
(3 marks)
Data Handling
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
How can this information be represented using a bar graph?
14. The given bar graph shows the number of bikes sold by an authorised dealer during the years 2000 to 2003, where the bar corresponding to the year 2001 is missing.
If the dealer sold a total of 6,000 bikes during the years 2000 to 2003, then how many bikes were sold in the year 2001?
(2 marks)
Data Handling 15. A survey was conducted among sixty students of a school to know their favourite
flavours of potato chips with respect to onion and cheese, light salted, salt and vinegar, and chilly lemon. The result of the survey is displayed in the given bar graph.
How can the given bar graph be redrawn by taking the scale as 1 unit = 4 students?
(3 marks)
Data Handling Solutions
Topic: Tally Chart
1. Total number of students is 40 Number of students who like to play football is 13 ∴ Number of students who like to play cricket = 40 − 13 = 27 It is known that tally mark is used to represent 1 student each and is used to represent 5 students, where the fifth tally mark is marked across the remaining four to bunch them as five. Thus, the number of students that like to play cricket can be represented using tally marks as .
Data Handling 2. The given data can be represented in a table using tally marks as follows:
Weight Tally marks Frequency
26
3
27
5
28
3
29
1
30
8
31
2
32
4
33
4
30
3. The tally marks table is drawn by marking for each employee for his respective hobby. Here,
5 tally marks are represented by , where the fifth tally mark is marked across the remaining four tally marks to bunch them into five. It can be observed from the given data that: 6 employees prefer dance. 6 employees prefer art. 5 employees prefer music.
Data Handling 6 employees prefer sports. 7 employees prefer reading. Thus, the given data can be written in the form of a tally marks table as:
Hobby Tally marks Frequency
Dance
6
Art
6
Music
5
Sports
6
Reading
7
Total 30
4. The frequency table for the given data is:
Marks Tally Marks Frequency
4
2
5
2
6
4
7
5
Data Handling
8
4
9
2
10
1
(a) Number of students who obtained marks above 6 = 5 + 4 + 2 + 1 = 12 (b) Number of students who obtained the highest marks i.e., 10 = 1 Number of students who obtained the lowest marks i.e., 4 = 2 Difference between the number of students who obtained the highest and the lowest marks = 2 − 1 = 1
5.
Height (in cm) Tally marks Frequency
135
4
136
10
137 16
138
3
139
5
140
8
141 17
Data Handling (a) Number of students with height 135 cm = 4 Number of students with height 140 cm = 8 (b) Total number of students in grade 6 = 4 + 10 + 16 + 3 + 5 + 8 + 17 = 63 (c) Maximum height in class = 141 cm Number of tallest students in grade 6 = Number of students with height 141 cm = 17 (d) Number of students with height less than 138 cm = 4 + 10 + 16 = 30 (e) Number of students with height more than or equal to 139 cm = 5 + 8 + 17 = 30
6. (a) It is given that 10 players are to be represented by the symbol . This implies that 5
players can be represented by the symbol .
Therefore, 50 players can be represented by , 35 players by
,and so on. Thus, the given data can be represented in a pictograph as
Game Number of players selected for trial
Cricket
Football
Lawn tennis
Badminton
(b) Four symbols (3 complete and 1 half) are used for representing the number of football players selected for trial.
7. From the given bar graph, it is observed that the numbers of students consuming apple juice, pineapple juice, orange juice and mixed juice are 6, 16, 18 and 10 respectively.
The tally marks table is drawn by marking for each student. Here 5 students are represented
Data Handling
by tally marks , where the fifth tally mark is marked across the remaining four to bunch them as five. Thus, the given information can be represented in the form of a tally chart as:
Types of juices Tally marks Number of students
Apple 6
Pineapple 16
Orange 18
Mixed 10
(b) For preparing pictograph, it is given that one represents 4 students.
Therefore, two students are represented by .
So, 6 students can be represented as ; 16 students can be represented as
. Thus, the given information can be represented in the form of pictographs as:
Types of juices Number of students
Apple
Pineapple
Orange
Mixed
Key : = 4 students
Data Handling 8. From the given table, it is clear that the number of students preferring strawberry, vanilla,
chocolate and butterscotch flavour icecreams are 6, 10, 8 and 16 respectively.
Let 4 students be represented by . Then, 2 students are represented by .
∴ 6 students are represented by .
10 students are represented by .
8 students are represented by .
16 students are represented by . Thus, the given information can be represented in the form of a pictograph as:
Icecream flavour Number of students
Strawberry
Vanilla
Chocolate
Butterscotch
Data Handling
Key : = 4 students
9. From the pictograph, it is clear that the number of pizzas manufactured during the given years is as follows:
2004 − 40000
2005 − 25000
2006 − 35000
2007 − 45000
2008 − 30000
(a) The highest number of pizzas (i.e., 45000) was manufactured in 2007. (b) The least number of pizzas (i.e., 25000) was manufactured in 2005. (c) Total number of pizzas manufactured in all the five years = 40000 + 25000 + 35000 + 45000 + 30000 = 1,75,000
10. In the given pictograph, represents Rs. 5,000. Therefore, the given pictograph can be read as: Monthly earnings in the month of January = 5 × Rs. 5,000 = Rs. 25,000 Monthly earnings in the month of February = 3 × Rs. 5,000 =Rs. 15, 000 Monthly earnings in the month of March = 7 × Rs. 5,000 = Rs. 35,000 Monthly earnings in the month of April = 4 × Rs. 5,000 = Rs. 20,000 The numbers 25,000, 15,000, 35,000 and 20,000 are arranged in the increasing order as: 15,000 < 20,000 < 25,000 < 35,000 Thus, according to the increasing order of the monthly earnings of the shopkeeper, the four months can be arranged as February, April, January, March.
11. In the given pictograph, 1 represents 2 fishes. ∴ represents 1 fish.
Data Handling ∴Number of fishes caught by Joseph = 2 × 4 + 1 = 9 Numbers of fishes caught by Bunty = 6 × 2 = 12 Numbers of fishes caught by Lalit = 10 × 2 + 1 = 21 Numbers of fishes caught by Naveen = 9 × 2 = 18
If the key of the pictograph is taken as 1 = 3 fishes, then
9 fishes are represented by
12 fishes are represented by
21 fishes are represented by
18 fishes are represented by Thus, the required pictograph can be redrawn as:
Name Number of fishes caught
Joseph
Bunty
Lalit
Naveen
Key : 1 = 3 fishes
Topic: Bar Graphs
12. The number appearing each time and the corresponding number of times it appeared can be represented in a table as
Number appeared on the face
Number of times appeared
1 7
2 8
Data Handling
3 6
4 12
5 9
6 8
In order to draw the bar graph, an appropriate scale is chosen for the given data. As the greatest value of the data is 12 (number of times 4 appeared), the scale will start at 0 and end at 12. Equal divisions will be made on the vertical axis by making increments of 2. Therefore, the scale to be used is 1 unit = 2 times. Taking the number appeared on the face along the horizontal axis and the number of times it appeared along the vertical axis, the bar graph can be drawn as
13. The given table can be rewritten as:
Day Tally marks Number of car accidents
Sunday 9
Monday 12
Tuesday 6
Data Handling
Wednesday 15
Thursday 18
Friday 9
Saturday 21
In order to draw the bar graph for the given information, an approximate scale is chosen from the given data. As the greatest value of the given data is 21 (number of car accidents on Saturday), the scale will start at 0 and end at 21. Equal divisions can be made on the horizontal axis by making increments of 3. Therefore, the scale that we use is 1 unit = 3 car accidents. Now, taking the days along the vertical axis and the number of car accidents along the horizontal axis, the bar graph is obtained as:
14. Number of bikes sold in the year 2000 is 1,500 Number of bikes sold in the year 2002 is 1,200 Number of bikes sold in the year 2003 is 1,400 ∴The total number of bikes sold in the years 2000, 2002, and 2003 = 1,500 + 1,200 + 1,400 = 4,100 But, it is given that, a total of 6,000 bikes were sold by the dealer during the years 2000 to
Data Handling 2003. Hence, the number of bikes sold in the year 2001 is 6,000 − 4,100 = 1,900
15. From the bar graph, it is observed that: Number of students preferring onion and cheese flavoured potato chips = 12 Number of students preferring light salted flavoured potato chips = 24 Number of students preferring salt and vinegar flavoured potato chips = 16 Number of students preferring chilly lemon flavoured potato chips = 8 If the scale of the bar graph is taken as 1 unit = 4 students instead of taking 1 unit = 2 students, then equal divisions will be made on the horizontal line by making an increment of 4. Thus, the bar graph is redrawn as:
Mensuration Important Questions
Topic: Perimeter and Area of Closed Figures I
1.
What is the perimeter of the given figure?
(2 marks)
2. The length of a rectangle is 5 m and breadth is 4 m. What is the perimeter of the rectangle in centimetres?
(2 marks)
3. The perimeter of a triangle is 28 cm. If the two sides of this triangle are 11 cm and 10 cm, then what is the measure of the third side?
(2 marks)
4. The given figure is formed by joining 10 squares of side 3 cm. (2 marks)
Mensuration
What is the perimeter of the figure so formed?
5. The area and the breadth of a rectangular garden are 8925 m 2 and 85 m respectively. Find the cost of fencing the garden with two rounds of wire at the rate of Rs 5.50 per metre.
(3 marks)
6. What is the area and perimeter of the reflection of the given quadrilateral ABCD along the line l ?
(3 marks)
Mensuration Topic: Perimeter and Area of Closed Figures II
7. A wire which was bent in the shape of a regular octagon of side 14 cm is stretched into a straight line and then a piece of length 2 cm is cut out. The remaining wire is then bent into the shape of a regular decagon. What is the length of each side of the decagon so formed?
(3 marks)
8. Rahul runs one round around a rectangular park of length 240 m and breadth 160 m. Sahil runs one round around a squareshaped park of side 205 m. Who covers the longer distance and by how much?
(3 marks)
9. If the cost of fencing a squareshaped field at the rate of Rs 7 per m is Rs 840, then find the side of the field.
(2 marks)
Topic: Perimeter and Area of Closed Figures III
10. Ranbir bought a rectangular field of area 24000 square metres and width 120 m. He wants to fence it with two rounds of wire. Find the length of wire that will be required to fence the field.
(3 marks)
11. Find the area of the following figure.
(3 marks)
Mensuration 12. Use the following information to answer the next question .
Find the area of the given figure.
(3 marks)
13.
Find the area of the given figure.
(3 marks)
14. A carpet is to be laid in the living hall of Sneha’s house leaving a margin of 1.5 m all around. The living hall is 14 m long and 9 m wide. Find the area of the carpet.
(3 marks)
Mensuration 15. What is the area of hexagon ABCDEF?
(5 marks)
Mensuration Solutions
Topic: Perimeter and Area of Closed Figures I
1. Perimeter of a closed figure is the distance covered along the border forming the figure. ∴ Perimeter of the given figure = 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm + 3.5 cm = 35 cm
2. Length of rectangle = 5 m = 5 × 100 cm = 500 cm Breadth of rectangle = 4 m = 4 × 100 cm = 400 cm Perimeter = 2 (500 + 400) = 2 × 900 = 1800 cm
3. Perimeter of a triangle is the sum of the lengths of its three sides. Given,the perimeter of the triangle is 28 cm. Sum of two sides of the triangle = 11 cm + 10 cm = 21 cm ∴ Measure of the third side = Perimeter of the triangle − Sum of two sides of the triangle = 28 cm − 21 cm = 7 cm Thus, the measure of the third side of the triangle is 7 cm.
4. Side of each square is 3 cm. The dimensions of the given figure can be marked as:
Mensuration It is known that perimeter of a closed figure is the distance along the boundary of the closed figure. Thus, perimeter of the given figure = (3 + 6 + 3 + 3 + 6 + 3 + 3 + 3 + 6 + 3 + 3 + 3 + 6 + 3 + 6 + 3 + 6 + 9) cm = 78 cm
5. Area of rectangle = Length × Breadth ∴ 8925 cm 2 = Length × 85 m
Perimeter of rectangle = 2 × (Length × Breadth) = 2 × (105 m + 85 m) = 2 × 190 m = 380 m Length of wire required to fence the garden with two rounds = 2 × 380 m = 760 m ∴ Cost of fencing = Rs 5.50 × 760 = Rs 4180 Thus, the cost of fencing the garden with two rounds of wire is Rs 4180.
6. The reflection of rectangle ABCD about the line l looks exactly the same as ABCD. In that case, the breadth and the length of the reflection of ABCD will be 4 cm and 9 cm respectively. Hence, perimeter of reflection of ABCD about l = 2 (Length + Breadth) = 2 × (9 cm + 4 cm) = 2 × 13 cm = 26 cm Area of reflection of ABCD about l = Length × Breadth = 9 cm × 4 cm = 36 cm 2
Topic: Perimeter and Area of Closed Figures II
7. Since the wire was bent in the shape of the regular octagon of side 14 cm, the length of the wire is the perimeter of the octagon. Perimeter of the octagon = 8 × Side = 8 × 14 cm = 112 cm
Mensuration Therefore, original length of the wire is 112 cm. Since 2 cm long piece of wire is cut from the wire, length of remaining wire = 112 cm − 2 cm = 110 cm It is given that the remaining wire is rebent into the shape of a regular decagon. ∴Perimeter of regular decagon = 110 cm ⇒ 10 × Side = 110 cm
Thus, length of the side of the regular decagon is 11 cm.
8. Length and breadth of the rectangular park are respectively given as 240 m and 160 m. Side of the squareshaped park = 205 m Perimeter of the rectangular park = 2 (Length + Breadth) = 2 × (240 m + 160 m) = 2 × 400 m = 800 m Therefore, the distance covered by Rahul in one round around the rectangular park is 800 m. Perimeter of the squareshaped park = 4 × Side = 4 × 205 m = 820 m Therefore, the distance covered by Sahil in one round around the squareshaped park is 820 m. 820 m − 800 m = 20 m Thus, Sahil covers the longer distance by 20 m.
9. Cost of fencing the square field = Rs 840 Rate of fencing = Rs 7 per m It is known that, Cost of fencing the square field = Perimeter of the field × Rate of fencing Thus, perimeter of the field = m = 120 m 7
840 It is also known that, Perimeter of square = 4 × Side
Thus, the side of the field is 30 m.
Topic: Perimeter and Area of Closed Figures III
10 . Area of rectangle = Length ×Width Area = 24000 m 2 Width = 120 m
Mensuration
= 200 m Perimeter = 2 ×(Length + Width) = 2 × (200 + 120) m = 640 m Length of one round of wire, which is required to fence the field, is its perimeter. Thus, length of wire required to fence the field with two rounds = 2 × 640 m = 1280 m = 1000 m + 280 m = 1 km 280 m
11. The given figure can be disintegrated into 5 parts as
It is known that, Area of rectangle = Length × Breadth Therefore, we obtain Area of rectangle I = 6 × 1 cm 2 = 6 cm 2 Area of rectangle II = 8 × 1 cm 2 = 8 cm 2 Area of rectangle III = 10 × 2 cm 2 = 20 cm 2 Area of rectangle IV = 8 × 1 cm 2 = 8 cm 2 Area of rectangle V = 6 × 1 cm 2 = 6 cm 2 Thus, area of the given figure = Area of rectangle I + Area of rectangle II + Area of rectangle III + Area of rectangle IV + Area of rectangle V = 6 cm 2 + 8 cm 2 + 20 cm 2 + 8 cm 2 + 6 cm 2 = 48 cm 2
Mensuration 12.
It can be seen that,
Area of square AJHD = (AJ) 2 = (8 m) 2 = 64 m 2 Area of rectangle FEDC = FC × FE = 10 m × 6 m = 60 m 2
∴ Area of total figure = 6 m 2 + 64 m 2 + 60 m 2 + 70 m 2 + 20 m 2 = 220 m 2
13. From the given figure, it is observed that:
Covered area Number Area estimate (square units)
(i) Fully filled squares 17 17
(ii) Half filled squares 4 2
(iii) More than halffilled squares 3 3
(iv) Less than half filled squares 3 0
Thus, area of the given figure = (17 + 2 + 3) square units = 22 square units
Mensuration 14.
Length of the carpet = Length of the room − Area which was not carpeted = 14 m − 1.5 m − 1.5 m = 11 m Width of the carpet = Width of the room − 1.5 m − 1.5 m = 9 m − 1.5 m − 1.5 m = 6 m ∴Area of the carpet = Length × Width = 11 m × 6 m = 66 m 2
15. Area of the hexagon = Area of ΔABF + Area of rectangle BCDF + Area of ΔDEF On applying Pythagoras theorem to ΔABX, we obtain AB 2 = AX 2 + BX 2 ⇒ BX 2 = (5 cm) 2 − (4 cm) 2 ⇒ BX 2 = 25 cm 2 − 16 cm 2 ⇒ BX 2 = 9 cm 2 ⇒ BX 2 = (3 cm) 2 ⇒ BX = 3 cm Since AB = AF = 5 cm, ΔABF is isosceles. ∴ BX = XF = 3 cm BF = BX + XF = 3 cm + 3 cm = 6 cm
Area of ΔABF = × BF × AX = × 6 cm × 4 cm = 12 cm 221
21
Area of rectangle BCDF = BF × BC = 6 cm × 10 cm = 60 cm 2 Since ED = EF, ΔDEF is isosceles. Hence, perpendicular EY bisects side DF. ∴ DY = YF = 5 cm On applying Pythagoras theorem to ΔDEY, we obtain EY 2 = DE 2 − DY 2 ⇒ EY 2 = (13 cm) 2 − (5 cm) 2 ⇒ EY 2 = 169 cm 2 − 25 cm 2 ⇒ EY 2 = 144 cm 2
Mensuration ⇒ EY 2 = (12 cm) 2 ⇒ EY = 12 cm ∴ Area of ΔDEF = × DF × EY = × 10 cm × 12 cm = 60 cm 22
121
∴ Area of hexagon ABCDEF = 12 cm 2 + 60 cm 2 + 60 cm 2 = 132 cm 2
Thus, the area of hexagon ABCDEF is 132 cm 2 .
Symmetry
Important Questions
Topic: Symmetrical Figures and Lines of Symmetry
1. Identify the symmetrical figures from the following figures. Also, draw their lines of symmetry. (a)
(b)
(2 marks)
2. In the following figures, identify the figures that have horizontal line of symmetry or vertical line of symmetry or both or no line of symmetry. (a)
(b)
(3 marks)
Symmetry
(c)
3. Among the collection of letters E, U, V, C and L, which letters have horizontal lines of symmetry?
(2 marks)
4. On a squared paper, sketch the following. (a) Quadrilateral that shows 4 lines of symmetry (b) Quadrilateral that shows 2 lines of symmetry (c) Quadrilateral that shows 1 line of symmetry (d) Quadrilateral that shows no line of symmetry
(4 marks)
5. How many lines of symmetry does the triangle with given measures have? (a) ΔABC, with AB = 3 cm, BC = 4 cm, and CA = 5 cm (b) ΔPQR with ∠PQR = 90°, PQ = 4 cm, and QR = 4 cm (c) ΔXYZ with XY = 8 cm, YZ = 8 cm, and ZX = 8 cm
(3 marks)
Topic: Reflection of Figures
6. Complete each of the following figures such that the resulting figure has the given dotted lines as the lines of symmetry. (a)
(2 marks)
Symmetry
(b)
7. Use the following information to answer the next question.
In the given figure, a letter of the English alphabet is shown along with a vertical line.
(2 marks)
Symmetry
Find which letter of the English alphabet looks the same when it is reflected with respect to the vertical line. Give reason for your answer.
8. Complete each of the following figures such that the resulting figure has two dotted lines as two lines of symmetry. (a)
(b)
(3 marks)
Symmetry
(c)
9. How can the given figures drawn on a grid paper be completed by assuming the dotted lines as the lines of symmetry? (a)
(4 marks)
Symmetry
(b)
10. The given figure shows an equilateral triangle inscribed in a circle and a square.
What is the difference between the numbers of mirror lines of the two given figures?
(2 marks)
Symmetry
Solutions
Topic: Symmetrical Figures and Lines of Symmetry
1. (a) The given figure does not have any line of symmetry. Thus, this figure is not a symmetrical figure. (b) The given figure shows 4 lines of symmetry as shown below.
Thus, this figure is a symmetrical figure.
2. (a)
The above figure has only vertical line of symmetry. (b)
Symmetry
The above figure has vertical as well as horizontal line of symmetry. (c)
The above figure has no line of symmetry.
3. The line of symmetry of a figure is the line that divides the figure into two equal halves such that both halves are mirror images of each other. The lines of symmetry for the letters E, U, V, and C can be drawn as:
The letter L does not have a line of symmetry. Thus, the letters E and C have a horizontal line of symmetry.
Symmetry
4. (a)
(b)
(c)
Symmetry
(d)
5. (a) ΔABC with AB = 3 cm, BC = 4 cm and CA = 5 cm is a scalene triangle. It is clear that a scalene triangle has no lines of symmetry.
So, ΔABC has no line of symmetry. (b)ΔPQR with ∠PQR = 90°, PQ = 4 cm, and QR = 4 cm is an isosceles right angled triangle. It is clear that an isosceles triangle has one line of symmetry.
Symmetry
Thus, ΔPQR has only one line of symmetry. This can be shown as:
(c) ΔXYZ with XY = 8 cm, YZ = 8 cm and ZX = 8 cm is an equilateral triangle. It is clear that an equilateral triangle has three lines of symmetry. Thus, ΔXYZ has three lines of symmetry. These can be shown as:
6. The given figure can be completed such that the resulting figure has the given dotted lines as the lines of symmetry as (a)
Symmetry
Here, the dotted lines are the lines of symmetry of the obtained figure. (b)
Here, the dotted lines are the lines of symmetry of the obtained figure.
7. It can be seen that the letters, A, H, I, M, O, T, U, V, W, X, and Y, look the same when reflected with respect to the vertical line. These can be represented as
Symmetry
It is because these figures have vertical line of symmetry as
8. (a) By drawing the reflection of the given figure along the vertical line and then drawing the reflection of resulting figure along the horizontal line, we obtain the following figure.
(b) By drawing the reflection of the given figure along the vertical line and then drawing the reflection of resulting figure along the horizontal line, we obtain the following figure.
Symmetry
(c) By drawing the reflection of the given figure along the vertical line and then drawing the reflection of resulting figure along the horizontal line, we obtain the following figure.
9. The line that divides a figure into two equal halves such that the two halves are mirror images of each other, is called the line of symmetry of the figure. (a) Assuming line p as the line of symmetry, the given figure can be completed as:
Symmetry
Now, by assuming line q as the line symmetry, the obtained figure can be completed as:
This is the required completed figure. (b) Assuming line l as the line of symmetry, the given figure can be completed as:
Now, by assuming line m as the line of symmetry; the obtained figure can be completed as:
Symmetry
This is the required completed figure.
10. A line of symmetry of a figure divides the figure into two equal halves such that each half is the mirror image of the other. It is known that when a figure is folded along its mirror line, the mirror line becomes the line of symmetry of the figure. The mirror lines of the given figures can be drawn as:
It is observed that the first and second figures have 3 and 4 mirror lines respectively. Thus, difference between the numbers of mirror lines of given figures = 4 − 3 = 1
Practical Geometry
Important Questions
Topic: Construction of Circles of Given Radii
1. Draw two circles with centre P and Q of radii 5 cm each, where PQ = 8 cm.
Let them intersect at A and B. Measure the length of and examine
whether and are at right angles.
(2 marks)
2. Draw a circle of radius 4.2 cm whose centre is at O. Take a point A in the interior of the circle and a point B in the exterior of the circle. Compare the lengths of OA and OB with the radius of the circle.
(2 marks)
Topic: Construction of Line Segments Using Ruler and Compass
3. Draw a line segment of any length. Without measuring , draw
such that the length of is thrice that of .
(2 marks)
4. Given a line segment with unknown length, construct such that
the length of is four times that of .
(3 marks)
Topic: Construction of Perpendiculars to Lines
5. Draw a line segment of length 10.5 cm. Take a point X on it such that
BX = 2.5 cm. Through X, draw a line segment perpendicular to such that XY = XZ = 6 cm (use ruler and compasses). What is the perimeter of the quadrilateral AYBZ?
(3 marks)
Topic: Construction of Perpendicular Bisectors of Line Segments Using Ruler and Compass
6. Draw a circle with centre C and radius 5 cm. Draw a chord of length 8
cm. Construct the perpendicular bisector of that intersects it at point
M. Examine if this perpendicular bisector of passes through C. Find
(4 marks)
Practical Geometry
the length of .
Topic: Construction of Copy of Angles Using Ruler and Compass
7. Draw an angle of measure 144°. Copy this angle using ruler and compass only.
(3 marks)
Topic: Construction of Bisector of Angles Using Ruler and Compass
8. Construct an angle of measure 15° with ruler and compasses. Construct its bisector.
(4 marks)
9. Construct an angle of measure 75° with ruler and compasses and then construct its bisector.
(4 marks)
Topic: Construction of Angles of Special Measures
10. Construct an angle of 112.5° using a compass. (4 marks)
Practical Geometry
Solutions
Topic: Construction of Circles of Given Radii
1. The steps of construction are as follows: (1) Draw a line segment PQ of length 8 cm. (2) Taking P as centre, draw a circle of radius 5 cm. (3) Taking Q as centre, draw a circle of radius 5 cm. Let these two circles intersect at the points A and B.
Using the scale, we can measure the length of and it is found to be 6 cm. In quadrilateral APBQ, PA = PB = 5 cm (Radius of circle centred at P) QA= QB = 5 cm (Radius of circle centred at Q) ∴ PA = PB and QB = QA Hence, quadrilateral APBQ is a kite and in a kite, the diagonals intersect each other at 90°.
Hence, and are at right angles.
2. The required circle can be drawn as: (1) Firstly, open the compasses for the required radius 4.2 cm. (2) Mark a point ‘O’ at the centre of the circle. (3) Place the pointer of compasses on O. (4) Turn the compasses slowly to draw the circle. (5) Take a point A in the circle and B outside the circle.
Practical Geometry
By measuring the lengths of OA and OB, it can be observed that OA < 4.2 cm and OB > 4.2 cm. ∴ OA < Radius of the circle < OB
Topic: Construction of Line Segments Using Ruler and Compass
3. The following steps will be followed to draw the given line segment and then to
construct a copy of .
Step 1: Let be a given line segment.
Step 2: Adjust the compass up to the length of .
Step 3: Draw a line l and mark a point X on it. Step 4: Put the pointer on point X, and without changing the setting of the compass, draw on arc to cut the line l at P. Step 5: Put the pointer on point P and again draw an arc with the same radius as before, to cut
Practical Geometry
the line l at point Q. Step 6: Put the pointer on point Q and again draw an arc with the same radius as before, to cut the line l at point Y.
It can be observed that
Thus, is the required line segment that is thrice the length of .
4. The below given steps are followed to construct a line segment such that the length of
is four times that of the given line segment .
(1) Let be the given line segment.
(2) Adjust the compasses up to the length of .
(3) Draw any line l and mark a point X on it.
(4) Put the pointer on X and without changing the setting of compasses, draw an arc to cut the line l at point P.
(5) Put the pointer on point P and again draw an arc with the same radius as before to cut the line l at point Q.
(6) Put the pointer on point Q and again draw an arc with the same radius as before to cut the line l at point R.
Practical Geometry
(7) Put the pointer on point R and again draw an arc with the same radius as before to cut the line l at point Y.
Thus, is the required line segment that is four times the length of the line segment .
Topic: Construction of Perpendiculars to Lines
5. The steps of construction are as follows: (1) Draw AB = 10.5 cm
(2) Taking B as centre, draw an arc of radius 2.5 cm that cuts at point X.
(3) Taking X as centre and with a convenient radius, draw an arc intersecting line at two points P and Q.
.
(4) With P and Q as centre and a radius more than PX, construct two arcs intersecting each other at L and draw a line l through the points L and X.
Practical Geometry
(5) Taking X as centre, draw arcs of radii 6 cm that cut l to both sides of X at the points Y and Z. Join AY, AZ, BY, and BZ.
In the above figure, is the required perpendicular line segment to . By using a scale, we find that AY = AZ = 10 cm and BY = BZ = 6.5 cm Now, perimeter of quadrilateral AYBZ = AY + AZ + BZ + BY = 10 cm + 10 cm + 6.5 cm + 6.5 cm = 33 cm
Practical Geometry
Topic: Construction of Perpendicular Bisectors of Line Segments Using Ruler and Compass
6. The steps of construction are as follows: (1) Mark any point C on the sheet. (2) By adjusting the compasses up to 5 cm and by putting the pointer of the compasses at point C, turn the compasses slowly to draw the circle. It is the required circle of 5 cm radius.
(3) Now, mark any chord of length 8 cm in the circle.
(4) Taking A and B as centres, draw arcs on both sides of . Let these intersect each other at D and E.
(5) Join DE that intersects at point M.
Practical Geometry
Here, is the perpendicular bisector of .
When is extended, it will pass through point C.
Hence, the perpendicular bisector of chord passes through point C. In the above figure, we find that CM = 3 cm
Topic: Construction of Copy of Angles Using Ruler and Compass
7. The steps of drawing an angle of measure 144° and constructing the copy of this angle are:
Step 1: Draw a ray . Place the centre of the protractor at point O and the zero edge along
. Step 2: Mark a point P at 144°. Join OP. The measure of ∠POQ is 144°.
Step 3: Draw an arc of convenient radius in the interior of ∠POQ. Let it intersect the rays
and at points Y and X respectively. Step 4: Draw a line l and mark a point B on it. With the same radius used before, draw another arc while taking point B as centre. Let it cut the line l at point C. Step 5: Now adjust the compass up to the length of XY. With the length of XY as radius and taking C as centre, draw on arc which will intersect previously drawn arc at A.
Practical Geometry
Step 6: Join BA. is the required ray which makes the same angle with line l as the angle 144°.
Topic: Construction of Bisector of Angles Using Ruler and Compass
8. The steps of construction are as follows: We know that,
15° = 30°21 ×
Hence, we will construct the angle of measure 15°by bisecting the angle 30°. The below given steps are followed to construct an angle of 15°. (1) Draw a line l and mark a point P on it. Now taking P as centre and with convenient radius, draw an arc of a circle, which intersects line l at Q. (2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(3) Now, taking Q and R as centre and with radius more than RQ, draw arcs to intersect21
each other at S. Join PS that intersects the first drawn arc at T.
(4) Now, taking T and Q as centre and with radius more than TQ, draw arcs to intersect21
each other at U. Join PU that intersects the first drawn arc at V.
In the above figure, ∠UPQ = 15° To draw the bisector of this angle, this step should be followed.
Taking V and Q as centre and with radius more than VQ, draw arcs to intersect each other21
Practical Geometry
at W. Join PW. In the above figure,
∠WPQ = ∠UPQ21
Practical Geometry
9. It can be observed that: 75° = 60° + 15°
Hence, it is required to construct the angle of measure 75° by bisecting the angle between 90° and 60°. Then, the bisector of this angle can be drawn. The steps of construction are as follows: (1) Draw a line l and mark a point P on it. Taking P as centre and with a convenient radius, draw an arc of a circle, which intersects line l at Q. (2) Taking Q as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R. (3) Taking R as centre and with the same radius as before, draw an arc intersecting the arc at S (see figure). (4) Taking R and S as centre, draw arcs of same radius to intersect each other at T. (5) Join PT. Then, ray PT makes angle of 90° with line l . Let PT intersects the first drawn arc at U. (6) Taking U and R as centres, draw arc of radii more than half of UR to intersect each other at V. Let VP intersect the first drawn arc at W. The measure of ∠VPQ is 75°.
[∠VPQ = ∠RPQ + ∠VPR, ∠RPQ = 60°, ∠VPR = ∠TPR, ∠TPR = 30°]21
(7) Taking W and Q as centres, draw arcs of radius more than half of WQ in the interior of angle of 75º. Let these intersect each other at X. Join PX.
Thus, PX is the required bisector of the angle of 75°.
Topic: Construction of Angles of Special Measures
10. The angle 112.5° can be written as
Practical Geometry
Therefore we first draw angles of 90° and 135°, and then bisect the angle between them. The steps of the construction are as follows: Step 1: Draw a line AB and mark a point C on it. Step 2: Draw an angle of 90° at C using a compass. The angle obtained is ∠QCB = 90º and ∠ACQ = 180° − 90º = 90º. Step 3: With M and Q as centres, and radius more than half of MQ, draw arcs intersecting each other at R. The angle obtained is ∠RCB = 135º. Step 4: With K and Q as centres and radius more than half of KQ, draw arcs intersecting each other at P. Join PC.
Thus, ∠PCB = 112.5° is the required angle.
