Vedanta Excel in Mathematics TG Book - 6 Final

9. List out the following points.
(i) Pair of angles having the common vertex and a common arm are called adjacent angels.
(ii) Adjacent angles on the straight line are supplementary.
(iii) The vertically opposite angles are always equal.

Solution of selected questions from EXERCISE 13.2

1. State with reason whether ∠a and ∠b are adjacent angles or not in the following
figures.

a) Solution:
Here, a and ∠b have a common vertex O and common
arm OB. So, ∠a and ∠b are adjacent angles.

b) Solution:
Here, a and ∠b have a common vertex O and
but they don’t have common arm. So, ∠a and
∠b are not adjacent angles.

c) Solution:
Here, a and ∠b have a common arm BC and but they don’t have
common vertex. So, ∠a and ∠b are not adjacent angles.

d) Solution:
Here, a and ∠b have a common vertex I but they don’t
have common arm between them. So, ∠a and ∠b are not
adjacent angles.

2. Make equations and solve them to find unknown angles.
a) If 2x° and 3x° are adjacent angles in linear pair, find them.
Solution:
Here, 2x°+3x° = 180° [The sum of adjacent angles in linear pair is 180°]
or, 5x° = 180°
or, x° = 36°
Hence, 2x° = 2×36° = 72° and 3x° = 3×36° = 108°

b) If y° and (y + 10)° are complementary angles, find them.
Solution:
Here, y°+(y + 10)° = 90° [Being complementary angles]
or, 2y° = 90° - 10°

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or, 2y° = 80°
or, y° = 40°
Hence, y° = 40° and (y + 10°) = 40° + 10° = 50°

c) If (a + 10)° and (a + 20°) are supplementary angles, find them.
Solution:
Here, (a + 10)°+(a + 20)° = 180° [Being supplementary angles]
or, 2a° +30° = 180°
or, 2a° = 180° - 30°
or, 2a° = 150°
or, a° =75°
Hence, (a + 10)°= 75°+10° = 85° and (a + 20°) = 75° + 20° = 95°

d) If two complementary angles are equal, find them.
Solution:
Suppose that each of the complementary angles be x
Then, x°+x° = 90° [Being complementary angles]
or, 2x° = 90°
or, x° = 45°
Hence, the requires complementary angles are 45° and 45°.

3. a) Find the supplement of complement of 50°.
Solution:
Here, the complement of 50° = 90° - 50° = 40°
Also, the supplement of 40° = 180° - 40° = 140°
Hence, the supplement of complement of 50° is 140°.

b) Find the complement of supplement of 100°.

Solution:

Here, the supplement of 100° = 180° - 100° = 80°

Also, the complement of 80° = 90° - 80° = 10°

Hence, the complement of supplement of 100° is 10°.

4. Find the sizes of unknown angles.

a) Solution:

Here,

(i) y° = 85° [Being vertically opposite angles]

(ii) x° + 85° = 180° [Being linear pair]

or, x° = 180° - 85° = 95°

(iii) z° = x° = 95° [Being vertically opposite angles]

b) Solution: [Being linear pair]
Here,

(i) x° +4x° = 180°
or, 5x° = 180°

or, x° = 36°

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So, 4x° = 4×36° = 144°

(ii) y° = x° =36° [Being vertically opposite angles]

(iii) z° = 4x° =144° [Being vertically opposite angles]

c) Solution:

Here,

(i) x° +2x° + 60°= 180° [Being the parts of a straight line]

or, 3x° = 180°- 60°

or, 3x° = 120°

or, x° = 40°

So, 2x° = 2×40° = 80°

(ii) y° = 60° [Being vertically opposite angles]

(iii) z° +60° =180° [Being linear pair]

or, z° = 180° - 60° = 120°

Activity 3 Pairs of angles made by a transversal with parallel line segments

1. For warm up, ask about the parallel lines.
2. Draw a transversal to a pair of parallel lines and explain about the interior and exterior

angles.

3. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out
the angles in the Z-shape and rotate it to fit the pair of angles. Then call it as the alternate

angles.

4. Define alternate angles as the pair of non-adjacent interior angles lying in the opposite
sides of the transversal.

5. Similarly, define alternate exterior angles as the pair of non-adjacent exterior angles lying
in the opposite sides of the transversal.

6. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out
the angles in the F-shape and translate it to fit the pair of angles. Then call it as the

corresponding angles.

7. Under discussion, conclude that the corresponding angles as the pair of non-adjacent
angles, one interior and another exterior, lying in the same side of the transversal.

8. Construct the parallel lines and a transversal line upon them on a chart paper. Cut out the
angles in the C-shape and then call it as the co-interior angles.

9. Under discussion, define the co-interior angles as the pair of interior angles lying in the
same side of the transversal.

10. Also, define the co-interior exterior angles as the pair of exterior angles lying in the same
side of the transversal.

11. Note down the following points.
(i) The alternate angles are always equal.
(ii) The alternate exterior angles are always equal.
(iii) The corresponding angles are always equal.

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Solution of selected questions from EXERCISE 13.3

1. Find the sizes of unknown angles.
Solution:
Here,
(i) 2x0 +x° = 180° [Being the sum of co-interior angles]
or, 3x° = 180°
or, x° = 60° and 2x°=2×60° = 120°
(ii) y° = x° = 60° [Being alternate angles between parallel lines]
(iii) z°=2x° =120° [Being corresponding angles]

2. Find the sizes of unknown angles.

Solution:

Here,

(i) 4x0 +x° = 180° [Being linear pair]

or, 5x° = 180°

or, x° = 36° and 4x°=2×36° = 144°

(ii) p° = x° = 36° [Being alternate angles between parallel lines]

(iii) q°=4x° =144° [Being corresponding angles]

(iv) r°=x° =36° [Being corresponding angles]

3. Find the sizes of unknown angles.

Solution:

Here,

(i) w° = 82° [Being vertically opposite

angles]

(ii) x° = 82° [Being alternate angles

within parallel lines]

(iii) x°+y° =180° [Being linear pair]

or, 82° + y° = 180°

or, y° = 98°

(iv) z°=y° =98° [Being corresponding angles]

4. Find the sizes of unknown angles.
Solution:
Here,
(i) b°+150° =180° [Being co-interior angles]
or, b°=180°- 150°
or, b° = 30°
(ii) (a + b) ° =45° [Being alternate angles between parallel lines]
or, a° + 30° = 45°
or, a° = 15°

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5. From the figure given alongside, show that

a) ∠w = ∠c b) ∠x = ∠s c) ∠y = ∠g

d) ∠a = ∠r e) ∠d = ∠s f) ∠p = ∠c

Solution:

a) Here, need to show: ∠w = ∠c

(i) ∠w = ∠a [Being corresponding angles]

(ii) ∠a = ∠c [Being vertically opposite

angles]

From (i) and (ii), we get

∠w = ∠c Proved

b) Here, need to show: ∠x = ∠s

(i) ∠x = ∠q [Being corresponding angles]

(ii) ∠q = ∠s [Being vertically opposite angles]

From (i) and (ii), we get

∠x = ∠s Proved

c) Here, need to show: ∠y = ∠g

(i) ∠y = ∠c [Being corresponding angles]

(ii) ∠c = ∠g [Being corresponding angles]

From (i) and (ii), we get

y = ∠g Proved

d) Here, need to show: ∠a = ∠r

(i) ∠a = ∠e [Being corresponding angles]

(ii) ∠e = ∠r [Being alternate angles]

From (i) and (ii), we get

a = ∠r Proved

e) Here, need to show: ∠d = ∠s

(i) ∠d = ∠h [Being corresponding angles]

(ii) ∠h = ∠s [Being corresponding angles]

From (i) and (ii), we get

d = ∠s Proved

f) Here, need to show: ∠p = ∠c

(i) ∠p = ∠e [Being corresponding angles]

(ii) ∠e = ∠c [Being alternate angles]

From (i) and (ii), we get

∠p = ∠c Proved

6. In the adjoining figure, if ∠x + ∠a + ∠y = 180°,

show that ∠a + ∠b + ∠c = 180°.

Solution:

Here,

(i) ∠x = ∠b [Being alternate angles]

(ii) ∠y = ∠c [Being alternate angles]

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Now, ∠x + ∠a + ∠y = 180°

or, ∠b + ∠a + ∠c = 180° [From (i) and (ii)]

Hence, ∠a + ∠b + ∠c = 180°. Proved

7. In the figure alongside, PQ//RS. Find the values
of x° and y°.
Solution:
Here,
(i) x° + 40° = 130° [Being corresponding
angles]
or, x° = 130° - 40°
or, x° = 90°
(ii) y° = x° = 90° [Being alternate angles]

Extra Questions

1. Use the protractor to measure the angle. Then, write down the name
and size of the angle.

2. List separately the following angles as acute, obtuse, right, straight or
reflex angles.
70°, 110°, 55°, 180°, 200°, 90°, 300°, 360°, 160°, 270°

3. If x° and 40° make a right angle, find the size of x°. [Ans: 50°]
4. From the given figure, find the value of x°. [Ans: 60°]
5. Find the complement of 50°. [Ans: 40°]
6. Find the supplement of 160°. [Ans: 20°]

7. If 4x° and 5x° are a pair of complementary angles, find them. [Ans: 40°, 50°]

8. Find the value of x from the given figure. [Ans: 125°]


9. Find the sizes of unknown angles.

[Ans: x = y = 30°, z =150°]

10. Find the sizes of unknown angles.
[Ans: x = y = 60°, 2x = z =120°]

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Unit Geometry: Triangles and Polygons

14

Allocated teaching periods: 9
Competencies
- To classify the triangles according to the length of sides and measure of angles.
- To solve the simple problems related to the sum of interior angles of triangle.
- To solve the simple problems related to the exterior angle of triangle is equal to sum of two

opposite interior angles.
- To identify the special types of quadrilateral and sum of their interior angles.
- To identify the parts of circle

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To tell the types of triangles on the basis of side lengths
- To identify the type of triangle on basis of angles.
- To tell the sum of interior angles of triangle
- To tell the sum of interior angles of quadrilateral.
- To identify and define the parts of circle.

- To solve the simple problems based on sum of angles of
triangle

2. Understanding (U) - To solve the simple problems related on sum of angles of
quadrilateral.

- To solve the problems based on sum of angles of triangle

3. Application (A) - To solve the simple problems related on sum of angles of

quadrilateral.

- To make project work in classification of triangle and
4. High Ability (HA) present in the class.

- To make the tangram and present different shapes in class.

Required Teaching Materials/ Resources
Ruler, protractor, strips, sticks, chart paper, scissors, threads, ICT tools like Geo-Gebra etc.

Pre-knowledge: triangle, measurement of segment and angle.

Teaching Activities

Activity 1 Types of triangles

1. Show some models of triangles made up of strips, sticks etc. and discuss on the types of

triangles on the basis of sides and angles.

2. Under discussion, classify the triangles based on side lengths as equilateral, isosceles and

scalene triangles.

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3. Recall the measurement of any angles by using protractor
4. Classify the triangles according to the measurements of angles as acute angled, right angled

and obtuse angled triangles.

5. List out the following points under discussion.
(i) In an equilateral triangle, the lengths of its all sides are equal.
(ii) The size of each angle of an equilateral triangle is 60°.
(iii) In an isosceles triangle, the lengths of its any two sides are equal.
(iv) The angles opposite to the two equal sides of an isosceles triangle are always equal.

These equal angles are also called the ‘base angles’ of the isosceles triangle.

6. Use triangular paper sheet or visualize in Geo-Gebra to show that the sum of angles of
triangle is 1800.

7. Use Geo-Gebra to show that the exterior angle of triangle is equal to sum of its two opposite
interior angles.

Solution of selected questions from EXERCISE 14.1 and 14.2

1. Let’s tell and write the answers as quickly as possible.

a) How many acute angles are there in an acute angled triangle? Three

b) How many obtuse angles are there in an obtuse angled triangle? Only one

c) Are two obtuse angles possible in a triangle? No

d) How many right angles and acute angles are there in a right angled triangle? Only

one right angle and two acute angles

e) Is an obtuse angle possible in a right angled triangle? No

f) If one of the acute angles of a right angled triangle is 50°, the size of remaining acute

angle is 40°

2. Find the sizes of unknown angles of these triangles.

a) Solution:

Here, x°+ 40°+ 90° = 180° [The sum of angles if triangle is 180°]

or, x° + 130° = 180°

or, x° = 180° - 130°

or, x° = 50°

b) Solution:

Here, (x + 20)°+ 30°+ 90° = 180° [The sum of angles if

triangle is 180°]

or, x° + 140° = 180°

or, x° = 180° - 140°

or, x° = 40°

Hence, (x + 20)° = (40° + 20)° = 60°

c) Solution:

Here,

(i) m°+ 108° = 180° [Being linear pair]

or, m°= 180° - 108°=72°

(ii) m° = n°=72° [The base angles of isosceles triangle equal]

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3. a) a°, 2a° and 60° are the angles of a triangle. Find the size of a° and 2a°.
Solution:

Here,
a° + 2a° + 60° = 180° [The sum of angles of triangle is 180°]
or, 3a° = 180° - 60°
or, 3a° = 120°
or, a° = 40°

Hence, a° = 40° and 2a° = 2×40° = 80°.

b) If p°, 2p° and 3p° are the angles of a triangle, find them.

Solution:

Here,

p° + 2p° + 3p° = 180° [The sum of angles of triangle is 180°]

or, 6p° = 180°

or, p° = 30°

Hence, p° = 30°, 2a° = 2×30° = 60°and 3p° = 3×30° = 90°.

c) If the angles of a triangle are in the ratio 2:3:4, find them.

Solution:

Let the required angles of the tringle be 2x°, 3x° and 4x°.

Now, 2x° + 3x° + 4x° = 180° [The sum of angles of triangle is 180°]

or, 9x° = 180°

or, x° = 20°

Hence, 2x° = 2×20° = 40°, 3x° = 3×20° = 60°and 4x° = 4×20° = 80°.

4. a) If x° is the exterior angle and 30° and 50° are the two opposite interior angles, find
the value of x°.
Solution:

x° =30° + 50° [The ext. angle of ∆ is equal to the sum of two opposite angles]
= 80°

b) If 110° is the exterior angle and 70° and y° are the opposite interior angles, find
the value of y°.
Solution:

110° =70° +y° [The ext. angle of ∆ is equal to the sum of two opposite angles]
or, 110° -70° =y°

Hence, y° =40°

c) If 2x° is the exterior angle and x° and 45° are the opposite interior angles, find the
value of x°.
Solution:

2x° =x° +45° [The ext. angle of ∆ is equal to the sum of two opposite angles]
or, 2x° -x° =45°

Hence, x° =45°

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5. Find the sizes of unknown angles of these triangles.
Solution:

Here,
(i) a°=b° [The base angles of isosceles triangle are equal]
(ii) a°+ b°+ 48° = 180° [The sum of angles if triangle is 180°]
or, a° + a° = 180°-48°
or, 2a° = 132°
or, a° = 66°
(iii) c°= a°+48° [The ext. angle of ∆ is equal to the sum of two opposite
angles]
= 66°+48° = 114°

Hence, a°=b°=66° and c° = 114°

Activity 2 Quadrilaterals

1. Make the groups of students and give 4/4 pencils or sticks or strews. Then, ask the figure
so formed by joining all the four pencils or sticks or strews.

2. Give some real life examples of objects in quadrilateral shape such as mat, a piece of
carpet, face of books and so on.

3. Visualize the quadrilaterals by using Geo-Gebra or use rubber bands to make quadrilaterals
in geoboard.

4. Draw the quadrilaterals in the board and discuss about the vertices, sides and angles of
the quadrilaterals.

5. Divide the students into groups and give the group activities.
Group A: Draw the rectangles of different sizes. Define rectangle.
Group B: Draw the square of different sizes. Define square.
Group C: Draw the parallelogram of different shape and sizes. Define parallelogram.
Group D: Draw the rhombus of different shape and sizes. Define rhombus.
Group E: Draw the trapezium of different shape and sizes. Define trapezium.
Then, give opportunity to each group to present one after other groups.
6. Discuss about some properties of rectangle, square, parallelogram, rhombus and

trapezium.
7. Give the 7 pieces of tangram to the students and ask to make a rectangle, square,

parallelogram, trapezium and rhombus by using the 7 polygonal pieces.
8. Show that the sum of angles of quadrilateral is equal to 360°.
9. Discuss about the polygons having 3, 4, 5, 6, 7 or 8 sides. i.e., triangle, quadrilateral,

pentagon, hexagon, heptagon and octagon.


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10. Also, divide the polygons into the triangles by joining the diagonals and verify that the
sum of interior angles is (n – 2) ×360°.

Solution of selected questions from EXERCISE 14.3

1. From the figure given alongside, write down

(i) name of polygon
(ii) name of 4 triangles
(iii) name of 3 quadrilaterals
(iv) name of 2 pentagons.
Solution:

Here,

(i) The name of polygon is hexagon RAMESH.
(ii) The name of 4 triangles are ∆RAM, ∆RME, ∆RES and ∆RHS
(iii) The name of 3 quadrilaterals are RAME, RMES and RESH
(iv) The name of 2 pentagons are RAMES and RMESH

2. If x°, 2x°, 3x° and 150° are the angles of a quadrilateral, find the size of x°, 2x° and 3x°.
Solution:
Here, x° + 2x° + 3x° + 150°=360° [The sum of angles of quadrilateral is always
360°]
or, 6x° = 360° - 150°
or, 6x° = 210°
or, x° = 35°
Hence, x° = 35°, 2x° = 2×35° = 70° and 3x° = 3×35° = 105°

3. Find the unknown sizes of angles of these quadrilaterals.
Solution:
a) (i) x°=66° [Being vertically opposite angles]
(ii) x° + y° + 2y° + 3y°=360° [The sum of angles of quadrilateral]
or, 66° +6y°=360°
or, 6y°=360°- 66°
or, 6y°=294°
or, y°=49°
Hence, y° = 49°, 2y° = 2×49° = 98° and 3y° = 3×49° = 147°

b) (i) p°+90°=180° [Being linear pair]
or, p°=180° - 90° = 90°

(ii) q°+40°=180° [Being linear pair]
or, q°=180° - 40° = 140°

(iii) p° + q° + r° +(r°+10°) =360° [The sum of angles of quadrilateral]
or, 90° +140°+2r°+10°=360°

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or, 240°+2r°=360°
or, 2r°=360° - 240°
or, 2r° = 120°
or, r° = 60°
Hence, p° = 90°, q° = 140°, r°=60° and (r° + 10°) = (60° + 10°) = 70°

4. Find the sum of the interior angles of the following polygons.
a) pentagon b) hexagon c) heptagon d) octagon
Solution:
a) In pentagon, number of sides (n) = 5
So, the sum of interior angles of pentagon = (n – 2) ×180°
= (5 – 2) ×180°
= 3×180°
=540°
b) In hexagon, number of sides (n) = 6
So, the sum of interior angles of hexagon = (n – 2) ×180°
= (6 – 2) ×180°
= 4×180°
=720°
c) In heptagon, number of sides (n) = 7
So, the sum of interior angles of heptagon = (n – 2) ×180°
= (7– 2) ×180°
= 5×180°
=900°
d) In octagon, number of sides (n) = 8
So, the sum of interior angles of octagon= (n – 2) ×180°
= (8 – 2) ×180°
= 6 ×180°
=1080°

5. If p°, 2p°, 3p°, 4p° and 5p° are the angles of a pentagon, find them.
Solution:
In pentagon, number of sides (n) = 5
The sum of interior angles of pentagon = (n – 2) ×180°= (5 – 2) ×180°= 3×180°=540°
Now, p° + 2p° + 3p° + 4p° + 5p° = 540°
or, 15p° = 540°
or, p° = 36°
Hence, p° = 36°, 2p° = 2×36°=72°, 3p° = 3×36°=108°, 4p° = 4×36°=144° and 5p° =
5×36°=180°


6. If y°, 4y°, 100°, 150°, 120° and 160° are the angles of a hexagon, find the sizes of
unknown angles.

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Solution:
In hexagon, number of sides (n) = 6
The sum of interior angles of hexagon = (n – 2) ×180°= (6 – 2) ×180°= 4×180°=720°
Now, y° + 4y° + 100° + 150° + 120° +160°= 720°
or, 5y° +530°= 720°
or, 5y° = 720° - 530°
or, 5y° = 190°
or, y° = 38°
Hence, y° = 38° and 4y° = 4×38°=152°

Activity 3 Circle

1. Take a square sheet of paper. Fold it to get the following shapes. Cut out the folded paper
of the fifth step along the dots shown in (v) and unfold it. Ask the name of shape so
formed to the students.


2. Discuss with a few real life examples having circular faces.
3. Under discussion, list out the circular shapes which are available in the surrounding.
4. Draw the outline figure of coin and ask about the shape so observed.
5. Take a compass with a pencil and fix the compass needle at a point

in the board. Then, take an arc of suitable radius in the compass and
rotate it about the fixed point through a complete turn. Then ask the
name of the shape to the students.
6. Give a classwork to draw circle on the paper by using compass.
7. Define circle under discussion.
8. Draw a circle and show its various parts such as centre,
circumference, radius, chord, diameter, arc, segment, sector etc. in chart paper using
compass or using Geo-Gebra or other construction tools and present in the class.
9. With measurements, discuss about
(i) the relation of the radii
(ii) the relation between the radius and diameter
(iii) the relation of diameters of the circle.

Solution of selected questions from EXERCISE 14.4

1. Let’s write ‘True’ or ‘False’ for the following statements.
a) Circle is a plane figure. True
b) A circle has exactly one centre. True
c) Radius is the line segment that joins any two points of the circumference. False

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d) All the radii of a circle are equal. True
e) Every chord is a diameter. False
f) Diameter divides the circle into two halves. True
g) The diameter of the circle is double of its radius. True
h) A diameter is the longest chord in a circle. True
i) Sector is the part of circle enclosed within two radii and their corresponding arc.

True
j) The segment of a circle is the region between the chord and corresponding arc.

True

2. Differentiate between sector and segment of a circle with an
appropriate diagram.
Solution:
Here,
The sector is the part of circle enclosed within two radii and their corresponding arc.
In the given figure, the shaded region is the minor sector and the unshaded part is the
major sector.
On the other hand, the segment of a circle is the region between the chord and
corresponding arc. In the given figure, the shaded region ACB is the minor segment
and the unshaded part is the major segment.

3. a) How many radii can you draw in a circle?
Solution:
Infinitely many radii can be drawn in a circle.

b) How many chords can be drawn in a circle?
Solution:
Infinitely many chords can be drawn in a circle.

c) How many circles can be drawn through two given points?
Solution:
Infinitely many circles can be drawn through two given points.

Extra Questions

1. Find the value of x from the given figure. [Ans: 35°]

2. If the angles of a triangle are in the ratio 1:2:3, find them. [Ans: 30°, 60°, 90°]

3. If x°, 75°, 125° and 60° are the angles of a quadrilateral, find the size of x°. [Ans: 100°]

4. Find the sum of interior angles of hexagon. [Ans: 720°]

5. Define sector of circle and show it in a diagram.

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Unit Geometry: Construction

15

Allocated teaching periods: 10
Competencies
- To draw the angles by using protractor and compass
- To construct the perpendicular line by using set-square and compass.
- To construct the parallel line by using set-square and compass.
- To construct equilateral triangle, square and rectangle.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To recall the types of angles
- To define perpendicular line
- To recall parallel lines

- To construct the angle by using protractor
- To construct the angle by using compass
2. Understanding (U) - To construct the perpendicular/parallel line

- To construct an angle and bisect it by using compass
- To construct equilateral triangle
3. Application (A) - To construct square when the side length is given
- To construct rectangle when length and breadth are given

Required Teaching Materials/ Resources
Ruler, protractor, compass, ICT tools like GeoGebra etc.

Pre-knowledge: types of angles, sum of angles of triangle etc.

Teaching Activities

Activity 1 Construction of angle by using protractor

1. Ask to the students about the use of protractor.

2. Construct an angle by using protractor.

For example:

To construct ∠AOB = 70°. Follow these

steps

(i) Draw an arm OA, and place your

protractor as shown.

(ii) Count round the edge from 0° to 70°, and

mark B.

(iii) Remove the protractor, and join OB.

Now, you have constructed ∠AOB = 70°

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3. Similarly, construct an obtuse angle under discussion.
4. Give classwork to construct a few angles.

Activity 2 Construction of angles 30°, 45°, 60° and 90° by using set-squares
1. Take the set-squares and outline triangles by using them. Measure the angles of the triangles

and label the measurement of each of the corners of set squares.

2. Under discussion, ask the following questions to the students.
(i) Which angles can be drawn by using the longer set-square?
(ii) Which angles can be drawn by using the shorter set-square?
3. Follow these steps to construct the angles by using set-squares.
(i) Draw a line segment
(ii) Place the 30° corner of a (30°, 60°, 90°) set-square at the

point of the line segment OA.
(iii) Draw line segment OB. ∠AOB = 30°.
(iv) Similarly, draw the angles of 60°and 90°by using the same set-squares.
(v) Also, draw the angles of 45°and 90° by using another set-squares.

Activity 3 Construction of angles by using compass

1. The standard angles like 30°, 45°, 60°, 90°, 120°, 135°, 150° etc. and other angles like 75°,
105°, 165° can be drawn by using compass.

2. Under discussion, construct the angles by using compass.
(a) Construction of 60°
(i) Draw an arm OA and place the pointed metal end of the compasses at O.
(ii) Draw an arc taking suitable radius to cut OA at C.
(iii) Place the pointed metal end of the compasses at C and draw an arc of the same

radius to cut the first arc at D.
(iv) Join O, D and produce it to B. Now, ∠AOB = 60°

Vedanta Excel in Mathematics Teachers' Manual - 6 116

(b) Construction of 120°
(i) Repeat the process of construction of 60° till steps (iii)
(ii) Place the metal end of the compasses at D and draw an arc of the same radius to cut

the first arc at E.
(iii) Join O, E and produce it to F. Here, ∠AOF = 120°.



(c) Construction of 90°
(i) Repeat the process of construction of 120° till step (ii)
(ii) Place the metal end of the compasses at D and E respectively and draw two arcs of

the same radius cutting each other at G.
(iii) Join O, G. Here, ∠AOG = 90°


3. Similarly, describe the process of constructing other angles which are possible by using

compass, metal end of the compasses should place in suitable angles.

Angle required to construct Places on the first arc Checking

30° 0° and 60° (0°+60°)÷2 = 60°÷2 =30°

45° 0° and 90° or 30° and 60° (0°+90°)÷2 = 90°÷2 =45°

75° 60° and 90° (60°+90°)÷2 = 150°÷2 =75°

105° 90° and 120° (90°+120°)÷2 = 210°÷2 =105°

150° 120° and 180° (120°+180°)÷2 = 300°÷2 =150°

135° 120° and 150° (120°+150°)÷2 = 270°÷2 =135°

165° 150° and 180° (150°+180°)÷2 = 330°÷2 =165°

4. Similarly, discuss about the bisecting the angles.

Activity 4 Construction of perpendiculars

1. Recall perpendicular lines.
2. Make a discussion on the construction of perpendicular to a line.

117 Vedanta Excel in Mathematics Teachers' Manual - 6

a) Construction of perpendicular by using set–squares
Case I: Construction of a perpendicular to a line from a point outside the line.

Steps:
(i) Draw a line segment AB and mark a point P just above it.
(ii) Place the two set squares as shown in the figure and draw a line from the

point P on AB. Here, PQ is perpendicular to AB at Q.
Case II: Construction of a perpendicular to a line at a point on the line.

Steps:
(i) Draw a line segment AB and mark a point P on it.
(ii) Place the two set squares like before as shown in the figure and draw a

line on the point P. Here, QP is perpendicular to AB at P.
b) Construction of perpendicular by using compasses
Case I: Construction of a perpendicular to a line from a point outside the line.

Vedanta Excel in Mathematics Teachers' Manual - 6 118

Steps:

(i) Draw a line segment AB and mark a point P just above it.

(ii) Place the metal end of the compasses at P and draw an arc to cut AB at X and
Y.

(iii) Now, place the metal end of the compasses at X and Y respectively and draw
two intersecting arcs just below the line. Let the point of intersection of these
arcs be O.

(iv) Join P, O that meets AB at Q. Here, PQ is perpendicular to AB at Q.

Case II: Construction of a perpendicular to a line at a point on the line.

Steps: Draw a line segment AB and mark a point P on it.
(i)
(ii) Place the metal end of the compasses at the point P and draw an arc to
(iii) cut AB at X and Y.
(iv)
Place the metal end of the compasses at X and Y respectively and draw
(v) two arcs of the same radius to cut the first arc at M and N.

Now, place the metal end of the compasses at M and N respectively and
draw two intersecting arcs. Suppose, the point of intersection of these
two arcs be Q.

Join, P, Q. Here, PQ is perpendicular to AB at P.

c) Construction of perpendicular bisector of line segment
Steps:

(i) Draw a straight line segment AB of any length.
(ii) Place the metal end of the compasses at the point A and B respectively

and draw two intersecting arcs above and below the line. Suppose, P and
Q are the points of intersecting of these arcs.

(iii) Join P, Q. Here, PQ is the perpendicular bisector of the line AB.

119 Vedanta Excel in Mathematics Teachers' Manual - 6

Activity 5 Construction of parallel lines by using set-squares

1. Recall parallel lines

2. Draw the parallel lines using set square with discussion.

Steps: Draw a line segment AB and mark a point P just above it.
(i) Place an edge of a set square along the line.
(ii) Place an edge of another set square along the edge of the first set square
(iii) and the point as shown in the figure.
Move the first set square slowly up to the point P holding the second set
(iv) square firmly.
Now, draw PQ as shown in the figure. Here PQ is parallel to AB.
(v)

Activity 6 Construction of equilateral triangle

1. Ask the definition of equilateral triangle to the students.
2. Construct the equilateral triangle with an example.

Example:
Construct an equilateral triangle in which each side is 4.5 cm.
Method-I:
Steps
(i) Draw AB = 4.5 cm
(ii) Construct ∠A = 60o and

∠B = 60o at A and B respectively.
(iii) Name the point of intersection of the

arms of ∠A and ∠B as C.
Here, 'ABC is the required equilateral

triangle.
Method-II:
Steps

(i) Draw AB = 4.5 cm
(ii) From A and B, draw an arc with radius

AB = 4.5 cm.

(iii) Name the point of intersection of these
arcs as C. Join A, C and B, C. Then, ABC

is the required triangle.

Vedanta Excel in Mathematics Teachers' Manual - 6 120

3. Make the group of students and give the questions to construct the equilateral
triangle.

Activity 7 Construction of square

1. Recall square.
2. With discussion on certain properties of square, take an example and construct

the square. Example: Construct a square in which each side is 3.4 cm.
Steps
(i) Draw PQ = 3.4 cm
(ii) Construct ∠P = 90oand ∠Q = 90o at P

and Q respectively.
(iii) On PX and QY, mark PS = QR = 3.4 cm
(iv) Join S, R. Here, PQRS is the required

square

Activity 8 Construction of rectangle

Opposite sides of a rectangle are equal and its
each angle is 90°.

Construct a rectangle whose length is 5.5 cm
and breadth is 4 cm.
Steps
(i) Draw AB = 5.5 cm
(ii) Construct ÐA = 90° and ÐB = 90°

at A and B respectively.
(iii) On AX and BY, mark AD = BC = 4 cm.
(iv) Join C and D
Here, ABCD is the required rectangle.

121 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Perimeter, Area and Volume

16

Allocated teaching periods: 11
Competencies

- To find the relation between inch and centimetre, foot and centimetre, inch and
meter, foot and meter, inch and foot etc.

- To find the perimeter of plane figures
- To find the area of area of plane figures
- To draw the net of prisms and pyramids
- To find the area and volume of cube and cuboid
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To tell the relation between foot and inch, inch and

centimetre

1. Knowledge (K) - To define perimeter of plane figures.
- To define area of plane figures.

- To identify the solid figure when its net is given

- To tell formula of finding the volume of cube and cuboid

- To find the perimeter of plane figures (square, rectangle,

triangles, circle etc.)

2. - To calculate the area of plane figures (square, rectangle,

Understanding (U) triangles, circle etc.)

- To draw the net for the given prims and pyramids

- To calculate the area and volume of cube and cuboid.

- To solve the word problems based on conversion of units

of measurements

3. Application (A) - To solve the problem related to perimeter of plane figures
- To solve the problems related to area of plane figures

- To solve some word problems based on volume of cube

and cuboid.

4. Higher Ability (HA) - To compare the area covered by different shapes formed
from same wire and analyse it.

Required Teaching Materials/ Resources
Cardboard paper, scissors, scale, measuring tape, ICT tools like Geo-Gebra etc
Pre-knowledge: perimeter and area of rectangle and square

Teaching Activities

Activity 1 Length and Distance

1. Ask the following questions related to length and distance.

Vedanta Excel in Mathematics Teachers' Manual - 6 122

(i) Can you guess the length and breadth of your mathematics book, by using a ruler?
(ii) Can you guess the height of your best friend? Measure it by using a ruler or a

measuring tape.

(iii) Can you guess the length and breadth of your classroom? Measure them by using a
measuring tape.

(iv) Can you guess the distance between your house and school?
2. Under discussion, define the length as measurement of farness (or closeness) between

two points (or two ends).

3. Also, define the distance as the length of the space between two points (or places).
4. Ask the units of measuring length or distance. Then, say the units of measuring lengths

such as millimetre (mm), centimetre (cm), meter (m), inch (in), foot (ft.) etc. And the

units of measurement of distance in meter (m), kilometre (km) or in mile.

5. Recall that 1 cm = 10 mm 1 m = 100 cm 1 km = 1000 m
6. Make the group of students and give them the ruler or measuring tapes. Give the

following group-works

a) Group A: Measure the length of bench in (i) metre and cm (ii) feet and inches
b) Group B: Measure the height of one the group member in (i) metre and cm (ii) feet and

inches

c) Group C: Measure the length of whiteboard in (i) metre and cm (ii) feet and inches
d) Group D: Measure the length of class in (i) metre and cm (ii) feet and inches

7. Take a ruler or a measuring tape to observe the scales
in cm, inch and foot. Then establish the following
relationship among cm, inch, foot and meter under
discussion.

(i) 1 inch (in) = 2.54 cm
(ii) 1 foot (ft) = 12 inch (in)
(iii) 1 foot (ft) = 12 ×2.54 cm = 30.48 cm
(iv) 1 meter (m) = 100 cm = 100÷2.54 inch = 39.37 inch (in)
(v) 1 meter (m) = 39.37 inch = 39.37÷12 feet = 3. 28 feet (ft)

Solution of selected questions from EXERCISE 16.1

1. Convert each of the following measurements in to cm.

a) 5 m b) 18 m c) 20 m 25 cm d) 37 m 40 cm e) 6 ft f)

14 ft g) 6 in h) 10 in i) 5 ft 3 in j) 8 ft

9 in

Solution:

a) 1 m = 100 cm ∴5 m = 5×100 cm = 500 cm

b) 1 m = 100 cm ∴18 m = 18×100 cm = 1800 cm

c) 1 m = 100 cm ∴20 m 25 cm = 20×100 cm + 25 cm = 2000 cm + 25 cm = 2025

cm

d) 1 m = 100 cm ∴37 m 40 cm = 37×100 cm+40 cm = 3700 cm+40 cm= 3740 cm

123 Vedanta Excel in Mathematics Teachers' Manual - 6

e) 1 ft = 30.48 cm ∴6 ft = 6×30.48 cm = 182.88 cm
f) 1 ft = 30.48 cm ∴14 ft = 14×30.48 cm = 426.72cm
g) 1 inch = 2.54 cm ∴6 in = 6×2.54 cm = 21.24cm
h) 10 inch = 2.54 cm ∴10 in = 10×2.54 cm = 25.4 cm
i) 1 ft = 30.48 cm and 1 in = 2.54 cm

Now, 5 ft = 5×30.48 cm = 152.4 cm and 3 in = 3 ×2.54 cm = 7.62 cm
So, 5ft 3 in = 152.4 cm + 7.62 cm = 160.02 cm
j) 1 ft = 30.48 cm and 1 in = 2.54 cm
Now, 8 ft = 8×30.48 cm = 243.84 cm and 9 in = 9 ×2.54 cm = 22.86 cm
So, 8ft 9 in = 243.84 cm + 22.86 cm = 266.7cm

2. Change each of the following measurements in to inch.

a) 4 m b) 20 m c) 16 m 50 cm d) 48 m 25 cm e) 7 ft

f) 24 ft g) 35.5 ft h) 48. 75 ft i) 25 ft 7 in j) 53

ft 11 in

Solution:

a) 1 m = 39.37 inch ∴4 m = 4×39.37 inch = 157.48 inch

b) 1 m = 39.37 inch ∴20 m = 20×39.37 inch = 787.4 inch

c) 16 m 50 cm =16m + 50 cm = 16 m + 50÷100 m = 16 m + 0.5 m = 16.5m

We have, 1 m = 39.37 inch ∴16.5 m = 16.5×39.37 inch = 649.61 inch

d) 48 m 25 cm = 48 m + 25 cm = 48 m + 25÷100 m = 48 m + 0.25 m = 48.25 m

We have, 1 m = 39.37 inch ∴48.25 m = 48.25×39.37 inch = 1,899.6 inch

e) 1ft = 12 in ∴7 ft = 7×12 in = 84 in

f) 1ft = 12 in ∴24 ft = 24×12 in = 288 in

g) 1ft = 12 in ∴35.5 ft = 35.5×12 in = 426 in

h) 1ft = 12 in∴48.75 ft = 48.75×12 in = 585 in

i) 1ft = 12 in ∴25 ft 7 in= 25 ft + 7 in = 25 ×12 in + 7 in = 300 in + 7 in = 307

in

j) 1ft = 12 in ∴53 ft 11 in= 53 ft + 11in=53×12 in + 11 in = 636in + 11in =

647 in

3. Change the following measurements in to ft.

a) 15 m b) 28 m c) 40 m 20 cm d) 55 m 55 cm e) 381 cm

f) 76.2 cm g) 24 in h) 66 in i) 7 ft 6 in

j) 28 ft 3 in

Solution:

a) 1 m = 3.28 ft ∴15 m = 15×3.28 ft = 49.2 ft

b) 1 m = 3.28 ft ∴28 m = 28×3.28 ft = 91.84 ft

c) 40 m 20 cm =20 m + 20 cm = 40 m + 20÷100 m = 40 m + 0.2 m = 40.2m

We have, 1 m = 3.28 ft ∴40.2 m = 40.2×3.28 ft = 131.86 ft

d) 55 m 55 cm =55 m + 55 cm = 55 m + 55÷100 m = 55 m + 0.55 m = 55.55 m

We have, 1 m = 3.28 ft ∴55.55 m = 55.55×3.28 ft = 182.2 ft

Vedanta Excel in Mathematics Teachers' Manual - 6 124

e) 381 cm = 381÷100 m = 3.81 m
We have, 1 m = 3.28 ft ∴3.81 m = 3.81×3.28 ft = 12.5 ft

f) 76.2 cm = 76.2÷100 m = 0.762 m
We have, 1 m = 3.28 ft ∴0.762m = 0.762×3.28 ft = 2.5 ft

g) 12 in = 1 ft ∴24 in= 24÷12ft = 2 ft
h) 12 in = 1 ft ∴66 in= 66÷12ft =5.5 ft
i) 12 in = 1 ft ∴6 in= 6÷12ft =0.5 ft. So, 7 ft 6 in = 7 ft + 0.5 ft = 7.5 ft
j) 12 in = 1 ft ∴3 in= 3÷12ft =0.25 ft. So, 28 ft 3 in = 28 ft + 0.25 ft = 28.25 ft

4. Express the following measurements in to m.

a) 12 m 60 cm b) 35 m 25 cm c) 32.8 ft d) 53.3 ft

e) 3937 in f) 727.4 in g) 25 ft 6 in h)

36 ft 9 in

Solution:

a) 100 cm = 1 m ∴60 cm = 60÷100 m = 0.6 m.

So, 12 m 60 cm = 12m + 60 cm = 12 m + 0.6 m = 12.6 m

b) 100 cm = 1 m ∴25 cm = 25÷100 m = 0.25 m.

So, 35 m 25 cm = 35 m + 25 cm = 35 m + 0.25 m = 35.25 m

c) 3.28 ft = 1 m ∴32.8 ft = 32.8÷3.28 m = 10 m

d) 3.28 ft = 1 m ∴53.3 ft = 53.3÷3.28 m = 16.25 m

e) 39.37 in = 1 m ∴3937 in = 3937÷39.37 m = 100 m

Alternatively, 3937 in=3937÷12 ft=328.08 ft and 328.08 ft

=328.08÷3.28m=100m

f) 39.37 in = 1 m ∴727.4 in = 727.4÷39.37 m = 18.48 m

g) 25 ft 6 in = 25 ft + 6÷12 ft = 25 ft + 0.5 ft = 25.5 ft

So, 25.5 ft = 25.5÷3.28m = 7.77 m

h) 36 ft 9 in = 36 ft + 9÷12 ft = 36 ft + 0.75 ft = 36.75 ft

So, 36.75 ft = 36.75÷3.28m = 11.2m

5. a) Bishwant is 5 feet 6 inch tall. Find his height in inch, centimetre and metre.
Solution:
Here, height of Bishwant = 5 ft 6 in
We have; 1 ft = 12 in ∴5 ft = 5×12 in = 60 in.
So, his height in inch = 5ft 6 in = 60 in + 6 in = 66 in
Also, 1 inch=2.54 cm. Thus, his height in cm = 66×2.54 cm = 167.64 cm
Again, 100 cm = 1 m. So, his height in metre = 167.64÷100 m = 1.68 m

b) The length of a kitchen garden is 10 m 50 cm and breadth is 6 m 20 cm. Find
the length and breadth of the kitchen garden in ft.
Solution:
Here, the length of garden = 10 m 50 cm = 10m + 50÷100 m = 10.5 m
The breadth of the garden =6 m 20 cm = 6m + 20÷100 m = 6.2 m

125 Vedanta Excel in Mathematics Teachers' Manual - 6

Converting the length and breadth into ft.
We have, 1 m = 3.28 ft.

So, the length of the garden = 10.5×3.28 ft = 34.44 ft
The breadth of the garden = 6.2×3.28 ft = 20.34 ft

c) The height of Mt. Everest is 8,848 m 86 cm and the height of Kanchenjunga is

8,586 m, find their heights in feet.
Solution:
Here, the height of Mt. Everest=8,848 m 86 cm=8848m+86÷100 m=
8848.86m
The height of Kanchenjunga=8,586 m
We have, 1 m = 3.28 ft.

So, the height of Mt. Everest=8848.86×3.28 ft =29,024.26 ft
The eight of Kanchenjunga =8,586 m= 8,586×3.28 ft = 28,162.08 ft

d) The length and breadth of a school’s play-ground are 220 feet 9 inch and 150

feet 3 inch respectively. Find the length and breadth of the ground in m.
Solution:
Here, the length of play-ground= 220 ft. 9 in= 220 ft+9÷12 ft =220.75 ft
The breadth of play-ground=150 ft. 3 in=150 ft.+3÷12 ft=150.25 ft
We have, 1 m = 3.28 ft.

So, the length of the play-ground=220.75÷3.28m=67.2m
The breadth of the play-ground=150.25÷3.28m=45.81m

6. a) Ram is 137.16 cm tall and Sita is 4 ft 3 in tall. Who is taller and by how many
feet? Find.

Solution:
Here, Ram’s height= 137.16 cm = 137.16÷30.48 ft=4.5 ft
Sita’s height = 4 ft 3 in = 4 ft + 3÷12 ft=4 ft+0.25 ft = 4.25 ft
Now, 4.5 ft – 4.25 ft = 0.25 ft.
So, Ram is 0.25 ft. taller than Sita.

b) If the length of whiteboard of class-VI is 2m 80 cm and breadth is 4 ft 9 in,
by how many meters is the length of the whiteboard more than the breadth?
Find.

Solution:
Here, length of whiteboard = 2 m 80 cm = 2 m + 80÷100 m = 2.8 m
The breadth of whiteboard = 4 ft 9 in=4 ft + 9÷12 ft = 4.75 ft
We have, 3.28 ft =1 m ∴the breadth = 4.75÷3.28 m=1.45m
Now, 2.8m – 1.45 m = 1.35 m
Hence, the length of the whiteboard is 1.35 m more than the breadth.

Vedanta Excel in Mathematics Teachers' Manual - 6 126

c) For school’s uniform, Shashwat needs 1 m 50 cm cloth for a shirt and 5 ft 8
in cloth for a pant. How long cloth does he require for the uniform? Find in
meter.

Solution:
Here, the cloth required for a shirt=1 m 50 cm = 1 m + 50÷100 m = 1.5 m
The cloth required for a pant =5 ft 8 in=5 ft + 8÷12 ft = 5.67 ft
We have, 3.28 ft =1 m ∴the cloth required for a pant = 5.67÷3.28 m=1.73 m
So, the cloth required for the uniform =1.73m + 1.5 m = 3.23 m

d) In a village, there are three vertical electric poles. If the distance between the

first and the second poles is 98 ft 3 in and the distance between the second

and the third poles is 48 m 60 cm. Estimate the shortest length of electrical

cable wire in ft. that joins these three poles.

Solution:

Here, the distance between the first and second poles =98 ft 3 in

=98 ft+3÷12 ft

=98.25 ft

The distance between the second and third poles =48 m 60 cm

=48 m+60÷100 m

=48.6 m

=48.6×3.28 ft=159.41 ft

The shortest length of electrical cable wire that joins these three poles= 98.25

ft.+ 159.41 ft = 257.66 ft.

Note: Please! Focus on Project Work

Activity 2 Perimeter of plane figures

1. Ask the following questions related to perimeter.

(i) What is the length of wire required to fence a garden for one round?

(ii) What distance do you cover by walking around a park?

2. Discuss about the plane figures and their perimeter with proper examples

3. Ask the definition of perimeter.

4. Measure the total length of boundaries of the garden/ ground. Discuss about the

perimeter and cost estimation of fencing at the certain rates.

5. Discuss the formulae to find the perimeter of triangle, rectangle and square and

regular polygons by drawing the figures.

(i) Perimeter of triangle = sum of lengths of all three sides = a + b + c

(ii) Perimeter of an equilateral triangle = a + a + a = 3a where ‘a’ is the length of

its each side.

(iii) Perimeter of an isosceles triangle = a + b + a = 2a + b where ‘a’ is the length

of its each of equal sides and ‘b’ is the length of base.

(iv) Perimeter of a rectangle = l + b + l + b = 2l + 2b = 2 (l + b) where ‘l’ is the

length and ‘b’ is the breadth of rectangle.

127 Vedanta Excel in Mathematics Teachers' Manual - 6

(v) Perimeter of a square = l + l + l + l = 4l where ‘l’ is the length of its each side.
(vi) Perimeter of a regular pentagon = 5l where ‘l’ is the length of its each side.
(vii) Perimeter of a regular hexagon = 6l where ‘l’ is the length of its each side.

Solution of selected questions from EXERCISE 16.2

1. The perimeter of a rectangular field is 180m. Find the length of a wire required to
fence it with 4 rounds.

Solution:

Here, the perimeter of a rectangular field (P) = 180 m

Now,

The length of a wire required to fence the field with 4 rounds = 4P =4×180 m=720m

2. You are running around a rectangular ground of length 42 m and breadth 24 m.
How many metres do you cover when you complete 6 rounds around it?

Solution:

Here, the length of a rectangular ground (l) = 42 m

The breadth of the ground (b) = 24 m

Now, the perimeter of a rectangular ground (P) = 2 (l + b) = 2 (42 m + 24 m) =132 m

Now,

The distance covered in 6 complete rounds around the ground =6P=6×132 m=792 m

3. Your friend Anuradha has drawn a triangle whose length of sides are in the ratio of
2:3:4 and perimeter is 18 cm, find: (i) the length of each side. (ii) difference between
the longest and shortest sides.

Solution:

Let the length of three sides of the triangle be 2x cm, 3x cm and 4x cm.

Now, perimeter (P) = 18 cm

or, 2x + 3x + 4x = 18

or, 9x = 18

or, x =2

∴2x = 2×2 cm = 4 cm, 3x = 3×2 cm = 6 cm and 4x = 4×2 cm = 8 cm

(i) The length of sides of the triangle are 4 cm, 6 cm and 8 cm.

(ii) difference between the longest and the shortest sides = 8 cm – 4 cm = 4 cm.

4. Once in a classroom, teacher gave 4 square tiles
to Samriddhi and Surav each. Then ask them
to arrange them in rectangular or square form.
Samriddhi arranged the tiles in square form and
Saurav arranged in the rectangular form, whose
perimeter is more and by how much?

Vedanta Excel in Mathematics Teachers' Manual - 6 128

Solution:
For Samriddhi; side length of square (l) = 2 cm
∴Perimeter (P) = 4l = 4×2 cm = 8 cm
For Saurav, length of rectangle (l) = 4 cm and breadth (b) = 1 cm
∴Perimeter (P) = 2 (l + b) = 2 (4 cm + 1 cm) = 2×5 cm = 10 cm
Thus, the perimeter of Saurav’s rectangle is 10 cm – 8 cm = 2 cm more than the
perimeter of Samriddhi’s square.

5. Find the perimeter of each figure formed by using

6 square tiles. By what percentage is the maximum

perimeter more than the minimum one?

Solution:

Perimeter of the first rectangle =2 (l + b) = 2 (3 cm + 2 cm) = 2×5 cm = 10 cm

Perimeter of the second rectangle =2 (l + b) = 2 (6 cm + 1 cm) = 2×7 cm = 14 cm

Difference = 14 cm – 10 cm = 4 cm

Percentage in difference = 4 × 100% = 40%
10
Hence, the maximum perimeter more than the minimum perimeter by 40%.

6. A wire is in the shape of rectangle. Its length is 6 cm and breadth is 4 cm. If the
same wire is rebent in the shape of square, what will be the length of each side?
Solution:
Perimeter of the first rectangle =2 (l + b) = 2 (6 cm + 4 cm) = 2×10 cm = 20 cm
Also, perimeter of square = perimeter of rectangle
or, 4l = 20 cm
or, l = 5 cm

Hence, the length of each side of square is 5 cm.

7. A rope is in the square shape. It’s each side is 6 cm. If the same wire is rebent in the
rectangular shape with length 8 cm, what would be its breadth?
Solution:
Here, the perimeter of square (P) = 4l = 4×6 cm = 24 cm
Length of rectangle (l) = 8 cm and breadth (b) =?
Now, the perimeter of the rectangle = perimeter of the square
or, 2 (l + b) = 24 cm
or, 2 (8 cm + b) = 24 cm
or, 16 + 2b = 24
or, 2b = 8
or, b = 4
Hence, the breadth of the rectangle is 4 cm.
Note: Please! Focus on Project Work

129 Vedanta Excel in Mathematics Teachers' Manual - 6

Activity 3 Area of plane figures

1. Ask the following questions related to area.

(i) How much paper do you need to cover a rectangular wall of your classroom?

(ii) How much carpet do you require to cover the floor of your room?

2. Discuss about the plane figures and their area with proper examples

3. Ask the definition of area.

4. Draw the regular figures on a graph board and ask the area of the figures by counting

the number of square boxes covered by the figures.

5. Draw the irregular figures on a graph board and ask the area of the figures by counting

the number of square boxes covered by the figures.

6. Discuss the formulae to find the area of rectangle and square by drawing the figures or

using GeoGebra tool.

(i) Area of a rectangle = l × b where ‘l’ is the length and ‘b’ is the breadth.

(ii) Area of a square = l 2 where ‘l’ is the length of its each side.

Solution of selected questions from EXERCISE 16.3

1. A rectangular hall is 10 m long and 8.5 m broad. (i) Find the area of carpet required to
cover its floor. (ii) If the cost of 1 m2 of carpet is Rs 75, find the cost of carpeting the floor.
Solution:
Here, the length of rectangular hall (l) = 10 m and the breadth of the hall (b) = 8.5 m
Now,
(i) The area of carpet required to cover the floor = the area of the hall
=l×b = 10m ×8.5 m = 85
m2
(ii) The cost of carpeting the floor = Area ×Rate = 85×Rs 75 = Rs 6,375

2. The area of a square is 49 cm2. (i) Find its length (ii) Find its perimeter.
Solution:
Here,
(i) The area of square = 49 cm2 or, l2 = 49 cm2 or, l = 7 cm
Thus, the length of the square is 7 cm.
(ii) The perimeter of the square (P) = 4l = 4×7 cm = 28 cm

3. The area of a rectangular floor is 32 m2. If its breadth is half of its length, find its
perimeter.
Solution:
Let, the length of the rectangular floor (l) = x m then breadth of the floor (b) = x/2 m
Now, the area of the floor (A) = 32 cm2
or, l × b = 32
or, x × = 32
or, x2 = 64
or, x = 8

Vedanta Excel in Mathematics Teachers' Manual - 6 130

So, the length (l) = x m = 8 m and breadth (b) = 8/2 m =4 m
Again, the perimeter of the floor = 2 (l + b) = 2 (8 m + 4 m) = 24 m

4. Find the area of the following shaded regions.
Solution:
a) Area of the bigger rectangle = 12 cm × 8 cm = 96 cm2
Area of the smaller rectangle = 10 cm × 6 cm = 60 cm2
Now,
Area of the shaded region = Area of the bigger rectangle – Area of the smaller
rectangle = 96 cm2 – 60 cm2 = 36 cm2

b) Area of the bigger rectangle = 10 cm × 6 cm = 60 cm2
Area of the smaller rectangle =5 cm × 1 cm = 5 cm2
Now,
Area of the shaded region = Area of the bigger rectangle – Area of the smaller
rectangle = 60 cm2 – 5 cm2 = 55 cm2

c) Area of the bigger rectangle = 9 cm × 7 cm = 63 cm2
Area of the smaller rectangle = 2 cm × 1 cm = 2 cm2
Now,
Area of the shaded region
= Area of the bigger rectangle – Area of the smaller rectangle
= 63 cm2 – 2 cm2 =61 cm2

d) Area of the bigger rectangle = 14 cm × 10 cm = 140 cm2
Area of the smaller rectangle =6 cm × 2 cm = 12 cm2
Now,

Area of the shaded region = Area of the bigger rectangle – Area of the smaller
rectangle = 140 cm2 – 12 cm2 =128 cm2

e) Area of the bigger rectangle = 15 cm × 12 cm = 180 cm2
Area of the smaller rectangle = 15 cm × 2 cm = 30 cm2
Now,

Area of the shaded region = Area of the bigger rectangle – Area of the smaller
rectangle = 180 cm2 – 30 cm2 =150 cm2

f) Area of the bigger rectangle (A1) = 6 cm × 7 cm = 42 cm2
Area of the 4 smaller rectangles (A2) = 4 (2 cm × 3 cm) = 24 cm2
Now,
Area of the shaded region = A1– A2= 42 cm2 – 24 cm2 =18 cm2

5. The students of a school with two banners of equal area were participating in a
rally on ‘Children’s Day’. The first banner was 8 ft. long and 3 ft. wide. If the second
banner was 6 ft. long, what was its width?
Solution:
Here, the area of the first banner = l × b =8 ft. × 3 ft. = 24 sq. ft.
The area of the second banner = l × b =6 ft. × b = 6b sq. ft.

131 Vedanta Excel in Mathematics Teachers' Manual - 6

Now, the area of the first banner = the area of the second banner
or, 24 = 6b

or, b = 4
Hence, the breadth of the banner is 4 ft.

6. Mr. Khadka has bought two pieces of carpet for his room. Carpet-X has area 48 sq. ft

and width 6 ft. The area of carpet - Y is one-half area of carpet-X. If both the carpets

have same length, what is the width of carpet-Y?

Solution:

Here, the area of the carpdet-X = 48 sq. ft

or, l × b =48

or, l × 6=48

or, l =8 ft. 1
2
Also, the area of the carpet-Y = ×48 sq. ft =24 sq. ft. and length (l) = 8 ft.

Again, the area of the carpet-Y =24 sq. ft

or, l × b = 24

or, 8 × b=24

or, b =3 ft.

Hence, the width of the carpet –Y is 3 ft.

7. The area of a squared lawn and rectangular vegetable garden are equal. The
perimeter of the lawn is 24 m and its side is twice the breadth of the garden. Find the
length of the garden.
Solution:
Here, the perimeter of a squared lawn = 24 m
or, 4l = 24 m
or, l = 6 m
Also, the area of the lawn = l 2 = (6 m)2 = 36 m2
And, the side of lawn = twice the breadth of the rectangular garden
or, 6 m = 2b
or, b = 3 m
Again, the area of the garden = the area of the lawn
or, l × b = 36
or, l × 3=36
or, l= 12 m.
Hence, the length of the garden is 12 m.

8. The area of a rectangular playground and a squared park are equal. The perimeter
of the park is 80 m and its side is half of the length of the ground. Find the width of
the ground.
Solution:
Here, the perimeter of a squared park = 80 m

Vedanta Excel in Mathematics Teachers' Manual - 6 132

or, 4l = 80 m
or, l = 20 m
Also, the area of the lawn = l 2 = (20 m)2 = 400 m2
And, the side of park= half of the length of the rectangular playground
or, 20 m = l/2
or, l= 40 m
Again, the area of the ground = the area of the park
or, l × b = 400

or,40 × b =400
or, b =10 m.
Hence, the width of the ground is 10 m.

9. The following figures have equal
perimeters.
(i) Do they have equal areas?
(ii) Which one has the more area
and by how much?
Solution:
(i) The area of the rectangle = l × b = 12 cm × 6 cm =72 cm2
The area of the square = l 2 = (9 cm)2 = 81 cm2
So, they have not equal areas.
(ii) Difference in area = 81 cm2 – 72 cm2 = 9 cm2
Hence, the square has the 9 cm2 more area than the rectangle.

Note: Please! Focus on Project Work

Activity 4 Faces, edges and vertices of solid figures

1. Give the examples of some of the solid objects such as books, marker, pencil, jar,
bucket, ball, ice-cream, dice, tent etc.

2. Discuss about cuboid, sphere, cylinder, cone and pyramid with their proper examples.
3. Discuss about the vertex, faces and edges of cuboid.
4. Similarly, make a discussion on the numbers of

vertices, faces and edges of triangular prism and

pyramid. Then, verify the relation for polyhedron,

F+V–E=2

133 Vedanta Excel in Mathematics Teachers' Manual - 6

Solid figure Number of Number of Number of F+V–E
faces (F) vertices (V) edges (E)
Cuboid …………… ……………
Triangular prism …………… …………… ……………
Pyramid …………… ……………
…………… ……………
……………
…………… ……………

5. Divide the students into a few number of groups and give the group work to make
skeleton models by of cube, tetrahedron, pyramid by using match-sticks, pieces of
straws etc.

6. Also, draw the nets of the solid figures (cube, cuboid, tetrahedron, pyramid) on the
separate hard-paper. Cut the nets andfold along the dotted lines and paste the edges
of the folded faces with glue. Then form the solids as wished.

Activity 5 Area of solid

1. With real life examples, discuss about the solid objects and their dimensions.

2. With a cuboidal object or rectangular classroom, ask the following questions.

(i) How many faces does it have?

(ii) Are all of its faces rectangular?

(iii) What is the area of its base?

3. Discuss about the total surface area (T.S.A) of cube and cuboid and list the following

formulae.

(a) In any Cuboid,

(i) Area of its base = l ×b

(ii) Total surface area = 2(lb + bh + lh)

(b) In any Cube,

(i) Area of its each face = l 2

(ii) Total surface area = 6l 2

Activity 6 Volume of Solids

1. To warm up, ask the following questions.

(i) How much water can a jug hold?

(ii) How much air can the classroom contain?

2. Discuss about the meaning of volume of solid objects.

3. Ask the units of measuring volumes.

4. Make a discussion on the prism with proper examples and make clear that the volume

of prism is obtained by multiplying the base area and the height.

5. Under discussion, list out the following formulae

(i) Volume of cuboid = Area of base × height = l×b×h

(ii) Volume of cube = Area of base × height = l2 × l = l3

6. Give proper guidance to solve the problems as given in the exercise.

Vedanta Excel in Mathematics Teachers' Manual - 6 134

Solution of selected questions from EXERCISE 16.5

1. A rectangular tank is 2 m long, 1.5 m broad and 1 m high.
(i) What is the volume of the tank?
(ii) If it is completely filled with water, what is the volume of water?
(iii) If 1 m3 = 1000 litres, how many litres of water does it hold?
Solution:
Here, length of the tank (l) = 2 m,
breadth of the tank (b) = 1.5 m
height of the tank (h) = 1 cm
(i) Volume of the tank (V) = l×b×h = 2 m×1.5 m×1 m=3 m3
(ii) When the tank is completely filled with water,
the volume of water = volume of the tank =3 m3
(iii) Since, 1 m3 = 1000 litres. The capacity of the tank = 3×1000 litre = 3,000 litres
Hence, the tank can hold 3,000 litres of water.

2. A rectangular tank is 80 cm long, 60 cm broad and 50 cm high.
(i) What is the volume of the tank?
(ii) What is the volume of water when it is completely filled?
(iii) If 1000 cm3 = 1 litre, how many litres of water does it hold?

Solution:
Here, length of the tank (l) = 80 cm,
breadth of the tank (b) = 60 cm
height of the tank (h) = 50 cm
(i) Volume of the tank (V) = l×b×h = 80 cm×60 cm×50 cm=240000 cm3
(ii) When the tank is completely filled with water,
the volume of water = volume of the tank =2,40,000 m3
(iii) Since, 1000 cm3 = 1 litres. The capacity of the tank = 2,40,000÷1000 litre = 240
litres
Hence, the tank can hold 240 litres of water.

3. If the total surface area of a cube is 24 cm2,
(i) find its length
(ii) find its volume.
Solution:
Here,
(i) Total surface area of a cube = 24 cm2
or, 6l 2 = 24 cm2
or, l 2 = 4 cm2
∴l = 2 cm
(ii) The volume of the cube (V) = l 3 = (2 cm)3 = 8 cm3


4. The length of a cuboid is two times its breadth and its height is 2 cm. If the volume of

135 Vedanta Excel in Mathematics Teachers' Manual - 6

the cuboid is 36 cm3, find its length and breadth.
Solution:
Let the breadth (b) of the cuboid be x cm. So, its length (l) will be 2x cm.
The height of the cuboid (h) = 2 cm
Now,
Volume of the cuboid = 36 cm3

or, l × b × h = 36 cm3
or, 2x × x × 2 cm = 36 cm3
or, 4x2 cm = 36 cm3
or, x2 = 9 cm2
or, x = 3 cm
Hence, the length= 2x = 2×3 cm = 6 cm and the breadth of the cuboid = x cm =3 cm.

5. The breadth of a rectangular block is half of its length and its height is 4 cm. If the

volume of the block is 200 cm3. Find its (i) length and breadth (ii) total surface area

Solution:

Let the length (l) of the rectangular block be x cm. So, its breadth (b) will be x/2 cm.

The height of the block (h) = 4 cm

Now,

Volume of the cuboid = 200 cm3

or, l × b × h = 200 cm3

or, x × x/2 × 4 cm = 200 cm3

or, 2x2 cm = 200 cm3

or, x2 = 100 cm2

or, x = 10 cm

(i) The length of the block (l)=x cm= 10 cm and the breadth (b)= x cm = 10 cm
2 2
= 5 cm

(ii) Total surface area = 2 (l×b + b×h + l×h) = 2 (10×5 + 5×4 + 10×4) =220 cm2

6. The ratio of length, breadth and height of a cuboid is 5:3:2. If its volume is 240 cm3,
calculate its total surface area.
Solution:
Let the length (l) of the cuboid= 5x cm, the breadth (b) = 3x and height (h) = 2x
Now,
Volume of the cuboid = 240 cm3
or, l × b × h = 240 cm3
or, 5x × 3x × 2x cm = 240 cm3
or, 30 x3 = 240 cm3
or, x3 = 8cm3 or, x = 2 cm
So, the length (l)=5x cm= 5×2 cm = 10 cm and the breadth (b)= 3x cm= 3×2 cm = 6
cm and height (h) =2x cm= 2×2 cm = 4 cm
Again, Total surface area = 2 (l×b + b×h + l×h) = 2 (10×6 + 6×4 + 10×4) =248 cm2

Vedanta Excel in Mathematics Teachers' Manual - 6 136

Extra Questions

1. Mr. Ghising is 5 ft 6 in. convert his height in terms of metre and centimetre. [Ans1 m 68
cm]

2. The height of Mt. Everest is 8,848 m 86 cm, find its heights in feet. [Ans: 29,024.26 ft]

3. The perimeter of a rectangle is 56 cm and its length is 18 cm. Find its area. [Ans: 180 cm2]

4. A wire is in the shape of rectangle. Its length is 24 cm and breadth is 16 cm. If the same
wire is rebent in the shape of square, what will be the length of each side. [Ans: 20 cm]

5. The area of a rectangular floor is 50 m2. If its length is two times of its breadth,

a) find its length and breadth

b) find its perimeter. [Ans: (i) 10 cm, 5 cm (ii) 30 cm]

6. Draw a net of a cube.

7. The length of a cuboid is two times of its breadth and its height is 2.5 cm. If the volume

of the cuboid is 45 cm3, find its length and breadth. [Ans: 6 cm, 3 cm]

137 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Symmetrical Figures, Design of Polygons and

17 Tessellations

Allocated teaching periods: 4
Competencies
- To draw the possible axis/ axes of symmetry of the figures
- To design the tessellations from triangle, square and rectangle
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To define axis of symmetry
- To define tessellations

- To draw the possible axis/ axes of symmetry of the
2. Understanding (U) figures

- To design the tessellations from triangle, square and
rectangle

Required Teaching Materials/ Resources

Ruler, pencil, colourful paper, double alphabets, models of figures etc.
Pre-knowledge: design
Teaching Activities
Activity 1 Axis of symmetry

1. Take some figures and ask how can we divide the figure into two congruent
shapes.


2. Draw double (English or Nepali) letters and discuss about the dotted line which

divide the letters in to two halves.
3. Make a discussion on line or axis of symmetry.
4. Discuss about the number of axes of symmetry of the figures.
5. Tell the students to draw the axes of symmetry of the figures which are given in

the exercise.

Vedanta Excel in Mathematics Teachers' Manual - 6 138

Activity 2 Tessellations

1. Give some real life examples of simple designs/ patterns such as patterns in tiles
in house/school’s ground/ walls, carpet and cloths etc.

2. Describe about tessellations with examples.
3. Tell the students to draw and colour the tessellations from rectangle or square.

139 Vedanta Excel in Mathematics Teachers' Manual - 6

Unit Statistics

18

Allocated teaching periods: 10

Competencies
- To construct frequency table by using tally bars
- To represent the given data in bar graph
- To solve simple problems based on mean

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
- To define statistics
2. Understanding (U) - To tell the meaning of frequency
3. Application (A) - To define bar graph
4. High Ability (HA) - To tell the formula of finding the mean

- To draw the frequency distribution table
- To find the mean of individual series

- To construct the bar graph
- To calculate the mean of discrete series

- To collect the obtained marks in mathematics in an exam
and present in a bar graph.

Required Teaching Materials/ Resources
Colourful chart-paper, colourful markers, chart paper with required formulae, graph-board,
graph paper, highlighter etc.

Pre-knowledge: Bar graph average etc.

Teaching Activities

Activity 1 Frequency table

1. Explain about statistics

2. Collect the heights (in cm) of students and arrange the data in ascending/descending order.

3. Discuss about the equal heights (repetition of heights) and frequency of heights.

4. Under discussion, draw the frequency table using tally bars.

5. Similarly, divide the students into groups and provide the data such as ages, marks, daily

pocket money etc. and say to draw the frequency table using tally bars for the corresponding

data.

6. Give a project work as “Collect the marks obtained by your friends in the recently conducted

Vedanta Excel in Mathematics Teachers' Manual - 6 140

mathematics examination. Group the marks into the class interval of 10 and show them in
a cumulative frequency distribution table.”
7. Solve the problems which are given in the exercise.

Activity 2 Bar graph­

1. Take a bar graph representing the number of students of various classes of school or bar

graph representing the number of SEE graduated students since last 5 years and ask the

questions from the bar graph.

2. Discuss about the bar graph.

3. With a proper data, draw the bar graph along with the students.

4. Solve the problems which are given in the exercise.

5. Give some project works based on bar graph.

Activity 3 Average

1. Collect the weights of 10 students and find their average weight.

2. Make the groups of students and tell to measure the heights of the students of each group.

Then, tell them to calculate the average height of students of own group.

3. With such more examples and activities, discuss about the average.

4. With discussion, define average as the value obtained by dividing the sum of all data by the

number of data.

Solution of selected questions from EXERCISE 18.3

1. The average marks of Ram, Hari, Shyam and Gopal in Mathematics is 65. If Ram, Hari,

and Shyam obtained 60, 70 and 80 respectively, how many marks did Gopal obtain?

Solution:

Let the marks obtained by Gopal be x.

Then, total sum of marks of Ram, Hari, Shyam and Gopal = 60 + 70 + 80 + x = 210 + x

Average marks = 65

Now,

Average marks = Total sum of marks
or, 65 = 21N0o+. oxf students

4
or, 210 + x = 260

or, x = 260 – 210

or, x = 50

So, the marks obtained by Gopal is 50.

2. The average expenditure of a family in the last week was Rs. 111. If the expenditures

on Sunday, Monday, Tuesday, Wednesday, Thursday and Friday were Rs 140, Rs 126, Rs

105, Rs 70, Rs 84 and Rs 98 respectively, find the expenditure on Saturday.

Solution:

Let the expenditure on Saturday be Rs x.

Total expenditure of the week =Rs 140+Rs 126+Rs 105+Rs 70+Rs 84+Rs 98+Rs x

=Rs (623+x)

141 Vedanta Excel in Mathematics Teachers' Manual - 6

Average marks =Rs 111

Now, Total expenditure of the week
623 + x No. of days
Average expenditure =
or, 111 = 7

or, 623 + x = 777

or, x = 777 – 623

or, x = 154

So, the expenditure on Saturday is Rs 154.

3. The average height of 3 children in a class is 80 cm and the average height of 5 children
is 84 cm. Find the average height of all of these children.

Solution:

Here, average height of 3 children = 80 cm

Now,

Average height = Total sum of height of 3 children
3
Total sum of height of 3 children
or, 80 = 3

or, Total sum of heights of 3 children = 240 cm

Also, average height of 5 children = 84 cm

or, 84 = Total sum of height of 5 children
5

or, Total sum of heights of 5 children = 420 cm

Again, total sum of heights of 8 children = 240 cm + 420 cm = 660 cm

Average height of 8 children = Total sum of height of 8 children = 82.5 cm
8

So, the average height of 8 children is 82.5 cm

4. Shashwat obtained an average marks of 15 in 3 subjects. When the marks obtained in
another subject is also included, the average marks is increased by 1. Find the marks

obtained in the fourth subject.

Solution:

Here, average marks in 3 subjects = 15

Now, Total obtained marks in 3 subjects
3
Average marks =

or, 15 = Total obtained marks in 3 subjects
3

or, Total obtained marks in 3 subjects = 45

Also, average marks of 4 subjects = 15 + 1 = 16

The obtained marks in fourth subject = x (say), total marks in 4 subjects = 45 + x
Total obtained marks in 4 subjects
We have, average marks = 4

or, 16 = 45 + x
4

Vedanta Excel in Mathematics Teachers' Manual - 6 142

or, 45 + x = 64
or, x = 19
So, the marks obtained by Shashwat in the fourth subject is 19.

5. The average marks of English, Nepali, Science and Social Studies obtained by Sunayana
is 70. If the marks obtained in Mathematics is also included, the average marks is

increased by 5. Find her marks in Mathematics.

Solution:

Here, average marks in English, Nepali, Science and Social Studies = 70

Now, Total obtained marks in 4 subjects
4
Average marks =

or, 70 = Total obtained marks in 4 subjects
4

or, Total obtained marks in English, Nepali, Science and Social Studies = 280

Also, average marks of 5 subjects including Mathematics = 70 + 5 = 75

The obtained marks in Mathematics = x (say), total marks in 5 subjects = 280 + x

We have, average marks = Total obtained marks in 5 subjects
5
280 + x
or, 75 = 5

or, 280 + x = 375

or, x = 95

So, Sunayana got 95 marks in Mathematics.

Extra Questions

1. The data given below represents the weights in kilograms of 30 students of a class.
Present the data in a frequency table by using tally bars.

25, 30, 35, 25, 30, 40, 35, 35, 45, 40, 30, 25, 30, 35, 40,
30, 35, 25, 30, 45, 30, 35, 40, 45, 30, 25, 35, 45, 30, 25
2. The number of students who passed the S.E.E. examination from a school in different

years are given in the table. Draw a bar graph to represent the data.

Year (in B.S.) 2072 2073 2074 2075 2076 2077

No. of students 25 40 36 48 50 55

3. Calculate the average of the following marks obtained by 7 students of a class in

mathematics in an examination.

87, 63, 45, 72, 35, 95, 79 [Ans: 68]

4. The marks obtained by 6 students of class six are 14, 15, 13, 17, 19 and x. If their average

marks is 16, find the value of x. [Ans: 18]

5. In a class, the average height of 4 children is 90 cm and the average height of 6 children

is 95 cm. Find the average height of all of these children. [Ans: 93 cm]

143 Vedanta Excel in Mathematics Teachers' Manual - 6