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AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.2

Question 1.
Look at the given triangle and answer the following questions.
i) Which points are marked in the exterior of the triangle?
ii) Which points are marked on the triangle?
iii) Which points are marked in the interior of the triangle?

Solution:
i) The points marked in the exterior of the triangle are X, Y and Z.
ii) The points marked on the triangle are A, I, B, J and C.
iii) The points marked in the interior of the triangle are K, L and O.

Question 2.
Look at the adjacent figure. Answer the following questions.
i) How many sides are there in the triangle? What are they?
ii) How many vertices lie there on the triangle? What are they?
iii) What is the side opposite to the vertex P?
iv) What is the vertex opposite to \(\overline{\mathrm{PR}}\) ?

Solution:
In the given ∆PQR,
i) It has three sides. They are \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}\) and \(\overline{\mathrm{PR}}\)
ii) It has three vertices. They are P, Q and R.
iii) The side opposite to the vertex P is \(\overline{\mathrm{QR}}
iv) The vertex opposite to the side [latex]\overline{\mathrm{PR}}\) is Q.

Question 3.
Look at the given triangle and answer the following questions.
i) How many angles are there in the triangle? What are they?
ii) What is the angle opposite to \(\overline{\mathrm{MN}}\) ?
iii) Where is the right angle in the given triangle ?

Solution:
In the given ∆MON,
i) It has three angles, they are ∠MNO, ∠NOM and ∠OMN.
ii) Angle opposite to the side \(\overline{\mathrm{MN}}\) is ∠MON or ∠O.
iii) In the triangle ∠MON is the right angle.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.1

Question 1.
What is the name of four sided polygon? Draw the rough sketch.
Solution:
A four sided polygon is called a quadrilateral.

ABCD is a quadrilateral.

Question 2.
Draw a rough sketch of pentagon.
Solution:
A five sided polygon is called a pentagon.
So, pentagon has five sided figure.

ABCDE is a pentagon.

Question 3.
Write all the sides of the given polygon ABCDEF.
Solution:
Given polygon ABCDEF is a Hexagon.
Its six sides are \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}}, \overline{\mathrm{DE}}, \overline{\mathrm{EF}} \text { and } \overline{\mathrm{FA}}\)
So, it is called a Hexagon.

Question 4.
Write the interior angles of the polygon PQRST.
Solution:
Given the polygon PQRST has five sides.
So, it is called a Pentagon.
It has five interior angles.
They are ∠TPQ, ∠PQR, ∠QRS, ∠RST and ∠STP.
They can also be written as ∠P, ∠Q, ∠R, ∠S, ∠T.

Question 5.
Measure the length of the sides of the polygon PQRST.
Solution:
The given polygon has five sides.
They are \(\overline{\mathrm{PQ}}\) = 2cm; \(\overline{\mathrm{QR}}\) = 2.5 cm;
\(\overline{\mathrm{RS}}\) .= 2.4 cm; \(\overline{\mathrm{ST}}\) = 2.2 cm, \(\overline{\mathrm{PT}}\) = 2.5 cm
It’s a pentagon.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Unit Exercise Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Unit Exercise

Question 1.
In the given figure, measure the length of AC. Check whether
i) AB + AC > AC
ii) AC > AD – DC
Solution:
In the given figure, AB = 4.2 cm; BC = 5.5 cm
AC = 5.4 cm; CD = 3 cm; AD = 4 cm.
i) AB + AC = 4.2 + 5.4 = 9.6 cm > 5.4 cm
AB + AC > AC
ii) AD – DC = 4 – 3 = 1 cm < 5.4 cm
AD – DC < AC (or) AC > AD – DC

Question 2.
Draw a line segment \(\overline{\mathbf{A B}}\). Put a point C in between \(\overline{\mathbf{A B}}\). Extend \(\overline{\mathbf{C B}}\) upio D such that CD > AB. Now check whether AC and BD are equal length.
Solution:

Let draw a line \(\overline{\mathbf{A B}}\) = 5 cm and mark a point C on \(\overline{\mathbf{A B}}\) such that \(\overline{\mathbf{A C}}\) = 3 cm (or) \(\overline{\mathbf{BC}}\) = 2 cm Extend \(\overline{\mathbf{C B}}\) up to D such that \(\overline{\mathbf{C D}}\) = 5 cm (\(\overline{\mathbf{C B}}\) = 2 cm, \(\overline{\mathbf{B D}}\) = 3 cm)
∴ \(\overline{\mathbf{A C}}\) = 3cm, \(\overline{\mathbf{C B}}\) = 2cm, \(\overline{\mathbf{B D}}\) = 3cm
∴ \(\overline{\mathrm{AC}}=\overline{\mathrm{BD}}\) = 3cm.

Question 3.
Draw an angle ∠AOB as m∠AOB = 40°. Draw an angle ∠BOC such that ∠AOC = 90°
Check whether m∠AOB + m∠BOC = m∠AOC.
Sol. Draw an angle ∠AOB = 40° and ∠AOC = 90° on the same ray \(\overrightarrow{\mathrm{OA}}\)
Now, measure ∠AOB = 40° and ∠BOC = 50°
∴ m∠AOB + m∠BOC = 40° + 50 = 90°
m∠AOB + m∠BOC = 90° = m∠AOC
∴ m∠AOB + m∠BOC = m∠AOC.

Question 4.
Draw an angle ∠XYZ as m∠XYZ = 62°. Measure the exterior angle ∠XYZ.
Solution:

Draw angle ∠XYZ = 62° (Interior)
Now, measure the exterior angle ∠XYZ.
∴ Exterior angle of ∠XYZ = 298°

Question 5.
Match the following.
1. Set square — A) to measure angles
2. Protractor — B) to measure the lengths of line segments
3. Divider — C) to draw parallel lines
Sol. 1) Set square — C) to draw parallel lines
2) Protractor — A) to measure angles
3) Divider — B) to measure the lengths of line segments

Question 6.
List out the letters of English alphabet (capital letters) which consist of right angles.
Solution:
The letters which consists of right angles in English alphabet are

Question 7.
Measure the angles ∠AQP, ∠CPR, ∠BRQ.

Find m∠AQP, m∠CPR, m∠BRQ.
Solution:
In the given figure, measure the angles ∠AQR, ∠CPR, ∠BRQ
∠AQP = 120°
∠CPR = 120°
and ∠BRQ = 120°
∴ m∠AQP = 120°; m∠CPR = 120°; m∠BRQ = 120°

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.4

Question 1.

Measure all the angles in the above figures.

∠1 = 70°
∠3 = 70°
∠5 = 70°
∠7 = 70°

∠2 = 110°
∠4 = 110°
∠6 = 110°
∠8= 110°

∠a = 60°
∠c = 60°
∠e = 50°
∠g = 50°

∠b = 120°
∠d = 120°
∠f = 130°
∠h = 130°

Question 2.
Sum of which two angles is 180° in each figure ?
Solution:
From above question,
i) ∠1 + ∠2 = 180°

i) ∠1 + ∠2 = 180°
∠2 + ∠3 = 180°
∠1 + ∠4 = 180°
∠3 + ∠4 = 180°

∠5 + ∠6 = 180°
∠6 + ∠7 = 180°
∠7 + ∠8 = 180°
∠5 + ∠8 = 180°

∠1 + ∠8 = 180°
∠4 + ∠5 = 180°
∠2 + ∠7 = 180°
∠3 + ∠6 = 180°

ii) ∠a + ∠b = 180°
∠b + ∠c = 180°
∠c + ∠d = 180°
∠a + ∠d = 180°

∠e + ∠f = 180°
∠f + ∠g = 180°
∠g + ∠h = 180°
∠e + ∠h = 180°

Question 3.
In the given figure measure ∠FOG and draw the same in your note book.

Solution:

Question 4.
In the given figure measure the angles ∠AOB, ∠BOC.

Solution:

∠AOB = 110°
∠BOC = 60°
∠AOC = 50°

Question 5.
Write some acute, obtuse and reflexive angles atleast 2 for each.
Solution:
Acute angles : 10°, 30°, 45°, 60°, 89° (< 90°)
Obtuse angles : 110°, 150°, 160°, 172°, 178° (90° < obtuse < 180°) Reflex angles : 210°, 270°, 300°, 345°, 359° (reflex > 180°)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.3

Question 1.
Given ” \(\overline{\mathrm{AB}} / / \overline{\mathrm{CD}}\), l ⊥ m”. Which are perpendicular? Which are parallel?
Solution:
are parallel lines (// is the symbol for parallel),
l, m are perpendicular lines (⊥ is the symbol for perpendicular).

Question 2.
Write the set of parallels and perpendiculars in the given by using symbols.

Solution:
a) \(\overline{\mathrm{AB}}, \overline{\mathrm{DC}}\) are parallel lines, we denote this by writing \(\overline{\mathrm{AB}} / / \overline{\mathrm{DC}}\) and can be read as \(\overline{\mathrm{AB}}\) is parallel to \(\overline{\mathrm{DC}}\)
b) \(\overline{\mathrm{AD}}, \overline{\mathrm{BC}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{AD}} / / \overline{\mathrm{BC}}\) and can be read as \(\overline{\mathrm{AD}}\) is parallel to \(\overline{\mathrm{BC}}\)
c) \(\overline{\mathrm{AQ}}, \overline{\mathrm{PC}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{AQ}} / / \overline{\mathrm{PC}}\) and can be read as \(\overline{\mathrm{AQ}}\) is parallel to \(\overline{\mathrm{PC}}\).
d) \(\overline{\mathrm{AB}}, \overline{\mathrm{AD}}\) are perpendicular lines. We denote this by writing \(\overline{\mathrm{AB}} \perp \overline{\mathrm{AD}}\) and can be read as \(\overline{\mathrm{AB}}\) is perpendicular to \(\overline{\mathrm{AD}}\).
e) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}\) are perpendicular lines. We denote this by writing \(\overline{\mathrm{AB}} \perp \overline{\mathrm{BC}}\) and can be read as \(\overline{\mathrm{AB}}[latex] is perpendicular to [latex]\overline{\mathrm{BC}}\) .
f) \(\overline{\mathrm{BC}}, \overline{\mathrm{CD}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{BC}} \perp \overline{\mathrm{CD}}\) and can be read as \(\overline{\mathrm{BC}}\) is perpendicular to \(\overline{\mathrm{CD}}\).
g) \(\overline{\mathrm{CD}}, \overline{\mathrm{AD}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{CD}} \perp \overline{\mathrm{AD}}\) and can be read as \(\overline{\mathrm{CD}}\) is perpendicular to \(\overline{\mathrm{AD}}\).

ii) \(\overline{\mathrm{PX}}, \overline{\mathrm{QR}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{PX}} \perp \overline{\mathrm{QR}}\) and can be read as \(\overline{\mathrm{PX}}\) is perpendicular to \(\overline{\mathrm{QR}}\).

iii) a) \(\overline{\mathrm{LM}}, \overline{\mathrm{KN}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{LM}} / / \overline{\mathrm{KN}}\) and can be read as \(\overline{\mathrm{LM}}\) is parallel to \(\overline{\mathrm{KN}}\).
b) \(\overline{\mathrm{MN}}, \overline{\mathrm{LK}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{MN}} / / \overline{\mathrm{LK}}\) and can be read as \(\overline{\mathrm{MN}}\) is parallel to \(\overline{\mathrm{LK}}\).
c) \(\overline{\mathrm{ON}}, \overline{\mathrm{LP}}\) are parallel lines. We denote this by writing \(\overline{\mathrm{ON}} / / \overline{\mathrm{LP}}\) and can be read
as \(\overline{\mathrm{ON}}\) is parallel to \(\overline{\mathrm{LP}}\)
d) \(\overline{\mathrm{LM}}, \overline{\mathrm{ON}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{ON}} \perp \overline{\mathrm{LM}}\) and can be read as \(\overline{\mathrm{ON}}\) is perpendicular to \(\overline{\mathrm{LM}}\).
e) \(\overline{\mathrm{LP}}, \overline{\mathrm{KN}}\) are perpendicular lines, we denote this by writing \(\overline{\mathrm{LP}} \perp \overline{\mathrm{KN}}\) and can be read as \(\overline{\mathrm{LP}}\) is perpendicular to \(\overline{\mathrm{KN}}\).

Question 3.
From the given figure find out the Intersecting lines and concurrent lines.

Solution:
i) Intersecting lines : (l, m); (l, n); (n, o); (m, o); (l, o); (m, n)
ii) Intersecting lines: (p, q); (p, r); (p, s); (q, r); (q, s)
Concurrent lines : (p, q, s)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.2

Question 1.
Measure the lengths of the given line segments.

Solution:
i) \(\overline{\mathrm{AB}}\) = 2.4 cm
ii) \(\overline{\mathrm{PQ}}\) = 1.5 cm
iii) \(\overline{\mathrm{KL}}\) = 1cm, \(\overline{\mathrm{LM}}\) = 1 cm
\(\overline{\mathrm{KM}}=\overline{\mathrm{KL}}+\overline{\mathrm{LM}}\) =1 + 1 = 2 cm

Question 2.
Draw the following line segments.
i) AB = 6.3 centimeters ii) MN = 3.6 centimeters

Question 3.
Draw PQ = 4.6 cm and extend upto R such that PR = 6 cm.
Solution:

\(\overline{\mathrm{PR}}=\overline{\mathrm{PQ}}+\overline{\mathrm{QR}}\) = 4.6 + 1.4 = 6 cm

Question 4.
Draw a line segment \(\overline{\mathrm{OP}}\) with certain length and mark a point Q on it.
Check whether \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\)
Solution:

Given, \(\overline{\mathrm{OP}}\) = 8 cm; \(\overline{\mathrm{PQ}}\) = 3 cm
\(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}\) = 8cm – 3cm = 5 cm = \(\overline{\mathrm{OQ}}\)
∴ \(\overline{\mathrm{OP}}-\overline{\mathrm{PQ}}=\overline{\mathrm{OQ}}\) = 5cm

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts Ex 8.1

Question 1.
In the given figure there are some points marked. Name them.

Solution:

Question 2.
Join the points given below. Name the line segments so formed in the figure.

Solution:

\(\overline{\mathrm{AD}}\) read as segment AD
\(\overline{\mathrm{BC}}\) read as segment BC

\(\overline{\mathrm{AB}}\) read as segment AB

\(\overline{\mathrm{DC}}\) read as segment DC

\(\overline{\mathrm{AB}}\) read as segment AB
\(\overline{\mathrm{BC}}\) read as segment BC
\(\overline{\mathrm{CD}}\) read as segment CD
\(\overline{\mathrm{DE}}\) read as segment DE
\(\overline{\mathrm{EF}}\) read as segment EF
\(\overline{\mathrm{FA}}\) read as segment FA

Question 3.
Identify the following from the given figure.
(i) Any six points.
(ii) Any six line segments. (Starts with G)
(iii) Any six rays. (With initial point I)
(iv) Any three lines.

Solution:
Points are : A, B, C, D, E, F
G, H, I, and J
ii) Line segments are : \(\overline{\mathrm{GF}}, \overline{\mathrm{GE}}, \overline{\mathrm{GD}}, \overline{\mathrm{GH}}, \overline{\mathrm{GC}} \text { and } \overline{\mathrm{GA}}\)
iii) Rays are: \(\overrightarrow{\mathrm{IA}}, \overrightarrow{\mathrm{IC}}, \overrightarrow{\mathrm{IB}}, \overrightarrow{\mathrm{ID}}, \overrightarrow{\mathrm{JA}}, \overrightarrow{\mathrm{JC}}, \overrightarrow{\mathrm{JE}}\)
iv)Lines are :

Question 4.
Write ‘True’ or ‘False’.
(i) A line has two end points.
(ii) Ray is a part of line.
(iii) A line segment has two end points.
(iv) We can draw many lines through two points.
Solution:
i) A line has two end points. (False)
ii) Ray is a part of line.(True)
iii) A line segment has two end points.(True)
iv) We can draw many lines through two points.(False)

Question 5.
Draw and Name :
(T) Line containing point K.
QS) Draw a circle and a line such that the line intersects the circle.
a) at no points b) at one point c) at two points
(iii) Can we draw a line and a circle with 3 intersecting points ?
Solution:

iii) No, we can’t draw a line and a circle with 3 intersecting points. As a line can intersect the circle at two points only.

Question 6.
List out all capital letters in English alphabet, that you can write with 3 line segments by using all.
Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 12 Data Handling InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Data Handling InText Questions

Let’s Do (Page No. 159)

Question 1.
Take a die. Throw it and record the number. Repeat the activity 40 times and record the numbers. Represent the data in a frequency distribution table using tally marks.
Solution:

(Page No. 159)

Question 1.
In what way is the bar graph better than the pictograph ?
Solution:
Bar graphs are better indicative as they show exact numerical value. Also, to indicate negative values and positive values, it looks easier in a bar graph.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area InText Questions

Check Your Progress (Pg.No. 147)

Question 1.
Find the perimeters of the given figures.

In the above figures (I) and (ii), find the perimeter of ΔKLM, ΔKMN and □ KLMN.
(a) Compare the perimeters of ΔKLM and □ KLMN
Compare the perimeters of ΔKMN and □ KLMN.
What can you say? –
Solution:
i) In ΔKLM, given KL = 2cm; LM = 2.6 cm; MK = 3.8 cm
Perimeter of ΔKLM = KL + LM + MK
= 2 + 2.6 + 3.8
= 8.4 cm

In ΔKMN, given KM = 3.8 cm; MN = 2cm; KN = 2.6 cm
Perimeter of ΔKMN = KM + MN + KN
= 3.8 + 2 + 2.6 = 8.4 cm

ii) In □ KLMN, given KL = 2cm, LM = 2.6cm, MN = 2cm and KN = 2.6 cm
Perimeter of □ KLMN = KL + LM + MN + KN = 2 + 2.6 + 2 + 2.6 = 9.2 cm

a) By comparing the perimeter of ΔKLM and □ KLMN
8.4 cm < 9.2 cm
∴ Perimeter of ΔKLM < Perimeter of □ KLMN
By comparing the perimeter of ΔKMN and □ KLMN
8.4 cm < 9.2 cm
∴ Perimeter of ΔKMN < Perimeter of □ KLMN.
By comparing the perimeters of triangle with quadrilateral ‘
Perimeter of the triangle is always less than the perimeter of quadrilateral.

Let’s Explore (Pg.No. 150)

Question 1.
If the radius of a circle is doubled, then, what is the change in its circumference?
Solution:
If the radius is r, then its circumference of the circle is 2πr.
If radius is doubled means r = 2r (new)
Write r = 2r in 2πr.
Then, the circumference = 2π x 2r
= 4πr = 2.27πr
Circumference is doubled.
So, if the radius of a circle is doubled, then the circumference is also doubled.

Question 2.
If the radius is halved, then, what is the change in its circumference?
Solution:
If the radius is r, then its circumference of the circle is 2πr.
If radius is halved means r = \(\frac{\mathbf{r}}{2}\) (new)
Write r = \(\frac{\mathbf{r}}{2}\) in 2πr.
Then, the circumference = 2π x \(\frac{r}{2}=\frac{1}{2}\). 2πr
Circumference is halved.
So if the radius of a circle is halved, then the circumference is also halved.

Check Your Progress (Page No. 151)

Question 1.
Find the area of a square with side 16 cm.
Solution:
Given side of the square (s) = 16 cm
Area of the square A = s x s = 16 x 16 = 256 sq.cm

Question 2.
Length and breadth of a rectangle are 16 cm and 12 cm respectively. Find its area.
Solution:
Given length of the rectangle l = 16 cm
breadth of the rectangle b = 12 cm
Area of the rectangle A = l x b = 16 x 12 = 192 sq.cm

Let’s Think (Page No. 151)

Question 1.
Find the perimeter and area of a square of side 4 cm. Are these same? Give some examples to support your answer.
Solution:

From the above table,
If side is 4 cm, then only its perimeter and areas are equal.
In other cases perimeter and areas are not same.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry InText Questions

Check Your Progress (Page No. 139)

Question 1.
Measure the lengths of \(\overline{\mathrm{AP}}\) and \(\overline{\mathrm{PB}}\) in both the constructions. What do you observe?
Solution:
In the construction we observe that \(\overline{\mathrm{AP}}=\overline{\mathrm{PB}}\)
That is P can divide AB into two equal parts.

Let’s Think (Page No. 140)

Question 1.
In the construction of perpendicular bisector in step 2, what would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\) ?
Solution:
If we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\)
Arcs cannot intersect each other. So, we can’t construct the perpendicular bisector to the given line \(\overline{\mathrm{AB}}\)

(Page No. 144)

Question 1.
Construct angles of 180°, 240°, 300°.
Solution:

i) ∠AOB = 180°
Steps of construction:

• Draw any ray \(\overline{\mathrm{OA}}\) of any length.
• Place the pointer of the compasses at ‘O’ with ‘O’ as centre any convinient radius draw an arc cutting OA at M.
• With M as centre and without altering radius draw an arc which cuts the previous arc at P.
• Draw an arc with the same radius from P which cuts the previous arc at Q and from Q draw another arc which meets at R.
• Join OR. ( \(\overline{\mathrm{OB}}\) ). Then ∠AOB is the required angle.
  Hence the required angle ∠AOB = 180° is constructed.

ii) ∠PQR = 240°
Steps of construction:

• Draw any ray \(\overline{\mathrm{QP}}\) of any length.
• Place the pointer of the compasses at O with ‘O’ as centre any convinient radius draw an arc cutting QP at A.
• With A as centre and without altering radius draw an arc which cuts the previous arc at B.
• Draw an arc with the same radius from B which cuts the previous arc at C and from C draw another arc which meets at D.
• Draw an arc with the same radius from D draw an arc which cuts the previous arc at E.
• Join \(\overrightarrow{\mathrm{QE}}(\overrightarrow{\mathrm{QR}})\). Then ∠PQR is the required angle.
  Hence the required angle ∠PQR = 240° is constructed.

iii) ∠XYZ – 300°
Steps of construction :

• Draw any ray \(\overline{\mathrm{YZ}}\) of any length.
• Place the pointer of the compasses at Y with Y as centre any convinient radius draw an arc cutting YZ at P.
• With P as centre and without altering radius draw arc which quts the previous arc at Q and from Q draw another arc which cuts the previous arc at R.
• Draw an arc with the same radius from R which cuts the previous arc at S and from S draw another arc which meets at T.
• Draw an arc with the same radius from T which cuts the previous arc at U.
• Join \(\overrightarrow{\mathrm{YU}}(\overrightarrow{\mathrm{YX}})\). Then ∠XYZ is required angle.
  Hence the required angle ∠XYZ = 300° is constructed.
  

(Page No. 145)

Question 1.
Construct an angle of 45° by using compasses.
Solution:
45°
Steps of construction :
i) Draw any ray \(\overline{\mathrm{OP}}\) of any length.
ii) Draw arcs with the same radius from A to B and from B to C which cuts the previous arc at B and C respectively.
iii) Draw arcs from B and from C with same radius which can intersect at X.
iv) Join \(\overrightarrow{\mathrm{OX}}(\overrightarrow{\mathrm{OD}})\) i.e., ∠POD = 90°.
v) Draw the bisector to ZPOD which is \(\overrightarrow{\mathrm{OQ}}\)
vi) Now, ∠POQ = ∠QOD = \(\frac{\angle \mathrm{POD}}{2}=\frac{90}{2}\) = 45°
∴ ∠POQ = 45° (Q.E.D)

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes InText Questions

(Page No. 125)

Question 1.
Draw six different types of rough sketches of polygons in your notebook. In which case, it is not possible to form a polygon ?
Hence, what is the least number of sides needed to form a polygon ? Obviously three.
Solution:

From the above figure we conclude that with one side and two sides cannot form a polygon. So, atleast three sides needed to form a polygon.

Check Your Progress (Page No. 127)

Look at the adjacent figure.

Question 1.
Which points are marked in the interior of A GHI ?
Solution:
The points marked in the interior of AGHI are A, B and 0.

Question 2.
Which points sure marked on the triangle?
Solution:
The points marked on the triangle are G, P, H, I and Y.

Question 3.
Which points are’marked in the exterior of A GHI ?
Solution:
The points marked in the exterior of AGHI are M, R, S, X and Z.

(Page No. 130)

Question 1.
Take a rectangular sheet (like a post-card). Fold it along its length so that one half fits exactly over the other half. Is this fold a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the folded line is the line of symmetry.

Question 2.
Open it up now and again fold along its width in the same way. Is this second fold also a line of symmetry ? Why ?
Solution:
Yes. Because the two parts of the rectangular sheet (post-card) coincide exactly with each other.
So, the second folded line also a line of symmetry.

Question 3.
Do you find that these two lines are the lines of symmetry ?
Solution:
Yes. These two lines are the lines of symmetry. One is line of symmetry along length and the other is line of symmetry along width.

Project (Page No. 130)

Question 1.
Collect symmetrical figures from your surroundings and prepare a scrap book.
Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 8 Basic Geometric Concepts InText Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 8th Lesson Basic Geometric Concepts InText Questions

(Page No. 111)

i) How many rays are there ?
Solution:
Four

ii) How many coins are close to player Q ?
Solution:
B, C and D coins are close to player Q.

iii) While striking with striker there is a possibility of a coin touches with any other. Draw all such possibilities in the given picture by means of the line segments.
Solution:
\(\overline{\mathrm{CB}}\) and \(\overline{\mathrm{DE}}\) .

iv) How many such line segments can be drawn in the picture ?
Solution:
\(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{ED}}, \overline{\mathrm{AC}}, \overline{\mathrm{AE}}, \overline{\mathrm{CD}}, \overline{\mathrm{CE}}, \overline{\mathrm{AD}}, \overline{\mathrm{BD}} \text { and } \overline{\mathrm{BE}}\)

Let’s Do (Page No. 112)

Question 1.
Observe the table and their notations and fill the gaps.

Solution:

Lets Explore (Page No. 114)

Question 1.
Measure the lengths of all line segments in the given figures by using divider and scale. Then compare the sides of the given figures.

Solution:
i) In ΔABC; \(\overline{\mathrm{AB}}\) = 2.2 cm; \(\overline{\mathrm{BC}}\) = 2 cm and \(\overline{\mathrm{AC}}\) = 2.2 cm ‘
2.2 cm = 2.2 cm > 2 cm i.e., \(\overline{\mathrm{AB}}=\overline{\mathrm{AC}}>\overline{\mathrm{BC}}\)
Two sides are equal and one side is different in length.

ii) In PQRS rectangle \(\overline{\mathrm{PS}}=\overline{\mathrm{QR}}\) = 2.7 cm; \(\overline{\mathrm{P Q}}=\overline{\mathrm{RS}}\) = 1.8 cm and \(\overline{\mathrm{PR}}=\overline{\mathrm{QS}}\) = 3.2 cm.
Opposite sides are equal and diagonals are equal in length,

iii) In KLMN square \(\overline{\mathrm{KL}}=\overline{\mathrm{LM}}=\overline{\mathrm{MN}}=\overline{\mathrm{KN}}\) = 1.8 cm
All sides are equal in length.

Let’s Do (Page No. 115)

Question 1.
(i) Find the parallel lines in the below figure. Name, write and read them.

Solution:
a) l, m are parallel lines.
We denote this by writing l || m and can be read as l is parallel to m.
b) n, o are parallel lines.
We denote this by writing n||o and can be read as n is parallel to o.
c) p, q are parallel lines.
We denote this by writing p||q and can be read as p is parallel to q.

ii) Find the intersecting lines in the above figure. Name, write and read them.
Solution:
Intersecting lines are (l, q); (m,q); (m, r); (n,q); (p, r); (o, r); (o, q); (q, r);

iii) Find the concurrent lines in the above figure. Name, write and read them.
Solution:
Three or more lines passing through the same point are called concurrent lines. Concurrent lines are (l, o, p) & (m, n, p).

iv) Find the perpendicular lines in the above figure. Name, write and read them.
Solution:
a) p, o are perpendicular lines.
We denote this by writing p ⊥ o and can be read as p is perpendicular to o.
b) p, n are perpendicular lines.
We denote this by writing p ⊥ n and can be read as p is perpendicular to n.
c) n, q are perpendicular lines.
We denote this by writing n ⊥ q and can be read as n is perpendicular to q.
d) q, o are perpendicular lines.
We denote this by writing q⊥ o and can be read as q is perpendicular to o.

(Page No. 119)

Question 1.
Measure the angles at the vertices.

Solution:

In triangle ABC,
m∠BAC = 60°
m∠ABC = 60°
m∠ACB = 60°

In triangle XYZ,
m∠YXZ = 40°
m∠XYZ = 70°
m∠XZY = 70°

In triangle PQR,
m∠QPR = 35°
m∠PQR = 38°
m∠PRQ = 107°