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

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Unit Exercise

Question 1
Find the area of the square whose perimeter is 48 cm.
Solution:
Given perimeter of the square = 4 x s = 48 cm
Divide with 4 on both sides, \(\frac{4 \times \mathrm{s}}{4}=\frac{48}{4}\)
side (s) = 12 cm
Area of the square A = s x s = 12 x 12 = 144 sq. cm

Question 2.
If the length of a rectangle is 14cm and its perimeter is 3 times of its length. Find its area.
Solution:
Given the length of a rectangle l = 14 cm
Breadth of a rectangle b = ? cm
Perimeter of the rectangle = 2(l + b) = 3 times of length.
2(14 + b) = 3 x 14
Divide with 2 on both sides,
\(\frac{2(14+b)}{2}=\frac{3 \times 14}{2}\)
14 + b = 21
Subtract with 14 on both sides.
14 + b – 14 = 21 – 14
Breadth b = 7cm
Area of the rectangle A = l x b = 14 x 7 = 98 sq.cm

Question 3.
Find circumference of the circle whose diameter is 14 cm.
Solution:
Given the diameter of circle d = 14 cm
Circumference of the circle C = π.d
\(\frac{22}{7}\) x 14 =44 cm

Question 4.
14cm and 12cm are the length and breadth of a rectangle. If the breadth is increased by 6cm and length is decreased by 6cm, find the difference in areas.
Sol. Given length of the rectanglel = 14 cm
Breadth of the rectangle b = 12 cm
Area of the rectangle A₁ = l x b
= 14 x 12 = 168 sq. cm
If length decreased by 6 cm, then length l = 14-6 = 8 cm
If breadth increased by 6cm, then breadth b = 12 + 6 = 18 cm
Then, the area of the rectangle A2 = l x b = 8 x 18 = 144 sq. cm
Difference of the areas = A₁ – A₂ = 168 – 144 = 24 sq. cm

Question 5.
Find the perimeter of the following figures. What did you observe?

Solution:
i) Perimeter =12 cm + 8cm + 12cm + 8 cm = 40 cm
ii) Perimeter = 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 3cm + 2cm + 12cm + 8 cm = 40 cm
iii) Perimeter = 5cm + 3cm + 2cm + 3cm + 5cm + 2cm + 5cm + 3cm + 2cm + 3cm + 5cm + 2cm = 40 cm
By observing the perimeters of the above figures perimeters are same for the different shaped figures.

Question 6.
A square sheet of 8cm side was taken and made into 64 equal small squares. Find the perimeter of square sheet and also find the sum of the perimeters of all 64 small squares. What did you observe?
Solution:
Side of the square sheet s = 8cm
Perimeter of the square sheet = 4 x s = 4 x 8 = 32 cm
Side of the small square = 1cm
Perimeter of each small square = 4 x side = 4 x 1 = 4 cm
Perimeter of 64 small squares = 4 x 64 = 256 cm.
By observing sum of perimeters of all 64 small squares = 8 x perimeter of all squares.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.3

Question 1.
Find the area of the rectangle of measurements 15 cm and 8 cm as length and breadth respectively.
Solution:
Given length of the rectangle l = 15 cm
breadth of the rectangle b = 8 cm
Area of the rectangle A = l x b
= 15 x 8 = 120 sq.cm

Question 2.
Find the area of a square whose perimeter is 64 m.
Solution:
Given perimeter of a square = 4 x side = 64 m
Divide with 4 on both sides,
\(\frac{4 \times \operatorname{side}}{4}=\frac{64}{4}\)
side (s) = 16 m
Area of the square = s.s = 16xl6 = 256 sq.m

Question 3.
Perimeters of a rectangle and square are equal. If the length of the rectangle is 14 cm and the perimeter of the square is 44 cm, find the area of the rectangle.
Solution:
Given the length of the rectangle l = 14 cm
breadth of the rectangle b = ?
Perimeter of the square = 44 cm
Given perimeter of the rectangle = Perimeter of the square ‘
2(l + b) = 44
2(14 + b) = 44
Divide with 2 on both sides,
\(\frac{2(14+b)}{2}=\frac{44}{2}\)
14 + b = 22
Subtract 14 on both sides,
14 + b – 14 = 22 – 14
Breadth of the rectangle b = 8 cm
Area of the rectangle = l x b = 14 x 8 = 112 sq. cm

Question 4.
Find the perimeters and areas of the following and answer the questions.
A) A rectangle with length and breadth as 16 cm and 8 cm respectively.
B) A rectangle with length and breadth as 14 cm and 10 cm respectively.
C) A square with side 12 cm.
(i) Which of the above perimeters are equal ?
(ii) Are all these areas equal? If not, which one has the bigger area?
Solution:
A) Given length of the rectangle l =16 cm
– Breadth of the rectangle b = 8 cm
Perimeter of the rectangle P = 2(1 + b)
= 2(16 + 8) = 2×24 = 48 cm
Area of the rectangle A = l x b
= 16 x 8 = 128 sq. cm

B) Given length of the rectangle l = 14 cm
Breadth of the rectangle b = 10 cm
Perimeter of the rectangle P = 2(l + b)
= 2(14 + 10) = 2 x 24 = 48 cm
Area of the rectangle A = l x b
= 14 x 10 = 140 sq.cm

C) Given side of the square s = 12 cm
Perimeter of the square P = 4 x s = 4 x 12 = 48 cm
Area of the square A = s x s = 12 x 12 = 144 sq. cm
i) Perimeters of rectangle A and rectangle B and square are equal.
ii) No. Areas are not equal and area of the square is greater than the areas of rectangles.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.2

Question 1.
Find the circumferences of the circles with the radius given below.
A) 7 cm B) 3.5 cm C) 14 cm
Solution:
A) Given radius (r) = 7 cm
Circumference of a circle C = 2πr (∵ π = \(\frac{22}{7}\))
= 2 x \(\frac{22}{7}\) x 7 = 44 cm
∴ Circumference of a circle = 44 cm ‘

B) Given radius (r) = 3.5 cm
Circumference of a circle C = 2πr (∵ π = \(\frac{22}{7}\))
= 2 x \(\frac{22}{7}\) x 3.5
∴ Circumference of a circle = 22 cm

C) Given radius (r) = 14 cm
Circumference of a circle C = 2πr (∵ π = \(\frac{22}{7}\))
= 2 x \(\frac{22}{7}\) x 14
∴ Circumference of a circle = 88 cm

Question 2.
Given below are the circumferences of different circles. Find the radius of each circle.
A) 4.4 m B) 176 cm C) 1.54 cm
Solution:
A) Given circumference of a circle = 2πr = 4.4 m
2 x \(\frac{22}{7}\) x r = 4.4
Divide with 2 x \(\frac{22}{7}\) on both sides.

∴ Radius of the circle (r) = 0.7 m = 70 cm.

B) Given circumference of a circle = 2πr =176 cm
2 x \(\frac{22}{7}\) x r = 176
Divide with 2 x \(\frac{22}{7}\) on both sides.

∴ Radius of the circle (r) = 28 cm.

C) Given circumference of a circle is C = 2πr = 1.54 cm
2 x \(\frac{22}{7}\) x r = 1.54
Divide with 2 x \(\frac{22}{7}\) on both sides.

∴ Radius of the circle (r) = 0.245 cm

Question 3.
A gold smith has 8.8m of gold wire with him. He has to make gold rings of 2cm radius. How many such rings he can make with it?
Solution:
Given the radius of the gold ring r = 2 cm
Length of the gold wire = 8.8 m (or) = 880 cm
To find the? number of rings we have to divide the length of the gold wire by circumference of the gold ring.
Now, circumference of the gold ring C = 2πr

Number of rings made by gold smith = 70

Question 4.
A wire was bent in the shape of a circle with radius 7cm. If the same wire was again used to make a square, then find its side.
Solution:
Given radius of the circle r = 7 cm
Circumference of the circle C = 2πr
= 2 x \(\frac{22}{7}\) x 7 = 44 cm
Given length of the wire same.
So, perimeter of square = circumference of the circle
4 x side = 44 cm
Divide with 4 on both sides = \(\frac{4 \times \text { side }}{4}=\frac{44}{4}\)
Side = 11 cm
∴ Side of square = 11cm.

Question 5.
In a chemical factory two wheels of different radius were connected with a belt. Radius of the bigger wheel is 21cm and radius of the smaller wheel is 7cm. If the bigger wheel rotates completely 100 times, find out the number of times that the smaller wheel rotates. .
Solution:
Given the radius of the bigger wheel R = 21 cm
Circumference of the bigger wheel C = 2πR = 2 x \(\frac{22}{7}\) x 21 = 132 cm
If the bigger wheel completes 100 rotations,
Distance covered by the bigger wheel = Number of rotates x circumference
= 100 x 132 = 13,200 cm
Now, radius of smaller wheel r = 7 cm.
Circumference of the smaller wheel C = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44cm
If the smaller wheel completes ‘n’ rotations,
Distance covered by the smaller wheel = number of rotations x circumference
= n x 44 = 44n cm
Distance covered by the small wheel = Distance covered by the bigger wheel
44n = 13200 cm
Divide with 44 as both sides,
\(\frac{44 n}{44}=\frac{13200}{44}\) = 300
n = 300
∴ Number of rotations made by the smaller wheel = 300.

Question 6.
Mohan is playing with a ring of diameter 14 cm, which is made up of metallic wire. When his brother asked, Mohan stretched the wire and made it as two equal parts. With those parts, he made another two small rings. Find the radius of smaller ring. Sol. Given the diameter of the bigger ring
Solution:
Given the diameter of the bigger ring = 14 cm
Radius of the bigger ring R = \(\frac{\text { diameter }}{2}\)
R = \(\frac{14}{2}\) = 7 cm
Length of the wire = Circumference of the bigger ring
= 2πR = 2 x \(\frac{22}{7}\) x 7 = 44 cm
If Mohan stretched the wire into’ halves.
Then the length of half of the wire = circumference of the bigger ring ÷ 2
= \(\frac{44}{2}\) = 22 cm
Circumference of the smaller ring = 2πr = 22
2 x \(\frac{22}{7}\) x r = 22
Divide with 2 x \(\frac{22}{7}\) on both sides

∴ Radius of the smaller ring = 3.5 cm

Question 7.
In designing an iron gril a black smith needed 70 rings with radius of 7cm each. Find how much length of the rod he required, if the wastage is 20 cm.
Solution:
Given radius of the ring r = 7 cm
Circumference of the ring C = 2πr = 2 x \(\frac{22}{7}\) x 7 = 44 cm
Length of rod required to make one ring = 44 cm
Length of rod required to make 70 rings = 44 x 70 = 3080 cm
Wastage = 20 cm
Required length of rod to make 70 rings = 3080 + 20 = 3100 c

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 11th Lesson Perimeter and Area Ex 11.1

Question 1.
Find the perimeters of the following figures.

i) Check whether the perimeter of Δ XYZ = 3 x Length of the side?
ii) Check whether the perimeter of □ ABCD = 4 x Length of the side?
iii) Check whether the perimeter of □ PQRS = 4 x Length of the side?
i) In ΔXYZ, XY = 2cm, YZ = 2cm and XZ = 2cm
Perimeter of ΔXYZ = XY + YZ + XZ = 2 + 2 + 2 = 6cm
Perimeter of ΔXYZ = 6 cm
By observing that perimeter of ΔXYZ = 6cm = 3 x 2cm
= 3 x length of the side in ΔABC
So, perimeter of ΔXYZ = 3 x length of the side

ii) Given in quadrilateral ABCD, AB = 3cm, BC = 3cm, CD = 3cm and AD
Perimeter of □ ABCD = AB + BC + CD + AD = 3 + 3 + 3 + 3
Perimeter of □ ABCD = 12 cm
By observing that perimeter of □ ABCD = 12cm = 4 x 3cm
= 4 x length of the side in □ ABCD
So, perimeter of □ ABCD = 4 x length of the side

iii)In quadrilateral PQRS, .
Given PQ = 2cm, QR = 2cm, RS = 2cm and PS = 2cm
Perimeter of PQRS = PQ + QR + RS + PS = 2 + 2 + 2 + 2 = 8cm
Perimeter of PQRS = 8cm.
By observing that perimeter of PQRS = 8cm = 4×2 cm
= 4 x length of the side in PQRS
So, perimeter of PQRS = 4 x length of the side.

Question 2.
Measurements of two rectangular fields are 50m x 30m and 60m x 40m. Find their perimeters. Check whether the perimeters are 2 x length + 2 x breadth.
Solution:

Given, in rectangle ABCD, AB = DC = 50 m and BC = DA = 30 m
Perimeter of rectangle ABCD = AB + BC + CD + DA
= 50 + 30 + 50 + 30
= 2 x 50 + 2x 30 = 160 m
∴ Perimeter of rectangle ABCD = 2 x length + 2 x breadth
Given in rectangle PQRS, PQ = RS = 60 m and QR = SP = 40 m
Perimeter of rectangle PQRS = PQ + QR + RS + SP
= 60 + 40 + 60 + 40
= 2 x 60 + 2 x 40 = 200 m
∴ Perimeter of rectangle PQRS = 2 x length + 2 x breadth

Question 3.
Find the perimeter of
a) An equilateral triangle whose side is 3.5cm.
b) A square whose side is 4.8cm.
Solution:
a) Given side of an equilateral triangle is 3.5 cm.
We know that, perimeter of an equilateral triangle
= 3 x length of the side = 3 x 3.5 = 10.5 cm .
∴ Perimeter = 10.5 cm

b) Given side of a square = 4.8 cm
We know that, perimeter of a square
= 4 x length of the side 4 x 4.8
∴ Perimeter = 19.2 cm

Question 4.
Length and breadth of top of one table is 160cm and 90cm respectively. Find how . much length of beading is required for each table.
Solution:
Given the length of top of table =160 cm
breadth of top of table = 90 cm g
To find the length of beading we have to find the ®
perimeter of the table top. .
So, perimeter of table top = 160 + 90 + 160 + 90 = 500 cm 160 cm
Required length of beading of table top = 500 cm.

Question 5.
Manasa has 24cm of metallic wire with her. She wanted to make some polygons with equal sides whose sides are integral without milling into pieces values. Find how many such polygons she can make with the length of 24cm metallic wire?
Solution:
Given length of metallic wire = 24 cm
We know that 24 = 1 x 24
= 2 x 12
= 3 x 8
= 4 x 6
∴ 24 can be divided into 1,2,3,4,6,8,12 & 24
But we can’t form polygons with sides 1 & 2.
∴ The polygons with sides 3, 4, 6, 8, 12 and 24 can be formed.
The length of the sides are equal so \(\) and \(\) units.
They are 8 cm, 6cm, 4cm, 3cm, 2cm and 1cm respectively

Question 6.
Find the perimeter of the following figures. (i) and (ii)

i) Perimeter of the given polygon is sum of the lengths of its all sides.
Perimeter = 5cm + 3cm+ 1cm + 2cm + 1 cm + 1 cm + 1 cm + 1 cm + 1cm + 2cm + 1 cm + 3cm
Perimeter = 22 cm

ii) Perimeter of the given polygon is sum of the lengths of its all sides.
Perimeter = 1cm + 5cm + 1 cm + lcm+ 2cm + 1cm + 1cm + 1cm + 1cm + 1cm + 2cm + 1cm
Perimeter = 18cm

Question 7.
Statement P : So many rectangles exists with the same perimeter.
Statement Q : So many squares exists with the same perimeter.
Which option is correct?
A) P wrong Q correct
B) P correct Q wrong
C) P and Q are correct
D) P and Q are wrong
Solution:
B) P correct Q wrong.
Lengths and breadths of a rectangle can change for the same perimeter. But, the side of a square cannot change for the same perimeter.

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

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Unit Exercise

Question 1.
Construct a circle with centre X and diameter 10 cm. Sol. Given diameter of circle d = 10 cm
Solution:
We know that radius of circle r = \(\frac{d}{2}=\frac{10}{2}\) = 5
So, radius of the circle is 5 cm.
Now, draw a circle with radius 5 cm.

Question 2.
Draw four circles of radius 2 cm, 3 cm, 4 cm and 5 cm with the same centre P.

Solution:
Radii of four circles are PA = 2cm, PB = 3cm, PC = 4cm and PD = 5cm with the same centre P was constructed.

Question 3.
Draw the angles given below using a protractor.
(i) 75°
(ii) 15°
(iii) 105°
Solution:
(i) 75°

∠ADI = 75°
Steps of construction :

1. Draw a ray \(\overrightarrow{\mathrm{DI}}\) with vertex D.
2. Place the centre point of the protractor at D and the line be aligned with DI.
3. Mark a point A at 75°.
4. Join DA. ∠ADI = 75° is formed.
   Hence the required angle ∠ADI = 75° is constructed.

ii) 15°

∠KIS = 15°
Steps of construction:

1. Draw a ray \(\overrightarrow{\mathrm{IS}}\) with vertex I.
2. Place the centre point of the protractor at I and the line be aligned with \(\overrightarrow{\mathrm{IS}}\).
3. Mark a point K at 15°.
4. Join IK.∠KIS = 15° is formed.
   Hence the required angle ∠KIS = 15° is constructed.

(iii) 105°

∠MAD = 105°
Steps of construction:

1. Draw a ray \(\overrightarrow{\mathrm{AD}}\) with initial point A.
2. Place the centre point of the protractor
   at A and the line be aligned with \(\overrightarrow{\mathrm{AD}}\).
3. Mark a point M at 105°.
4. Join AM.∠MAD = 105° is formed.
   Hence the required angle∠MAD = 105° ¡s constructed.

Question 4.
Construct ∠ABC = 50° and then draw another angle ∠XYZ equal to ∠ABC without using a protractor.
Solution:

Given ∠ABC = 50° and ∠XYZ = 50°.
Steps of construction:

1. Construct ∠ABC = 50°by using protractor.
2. By taking any radius draw arcs from B on AB and BC at P and Q respectively.
3. Draw a ray \(\overrightarrow{\mathrm{YX}}\) with initial point Y.
4. By taking BP as radius draw an arc on \(\overrightarrow{\mathrm{YX}}\) from Y which meets at K.
5. Draw arc from Y by taking PQ as radius in the ∠BAC which cuts the previous arc and mark it as L. Now draw \(\overrightarrow{\mathrm{YZ}}\). So, ∠XYZ = 50° is formed.
   Hence ∠XYZ = 50° is constructed which is equal to ∠ABC = 50°.

Question 5.
Construct ∠DEF = 60°. Bisect it, measure each half by using a protractor.
Solution:

Given ∠DEF = 60°
Draw EX as the bisector of ∠DEF.
So, ∠DEX = ∠XEF = \(\frac{\angle \mathrm{DEF}}{2}=\frac{60^{\circ}}{2}\) = 30°
∴ ∠DEX = ∠XEF = 30°

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.4

Question 1.
Construct ∠ABC = 60° without using protractor.
Solution:
Steps of construction :

1. Draw a line l and mark a point B on it.
2. Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line l at a point C.
3. With the pointer at C as centre and with the same radius as above. Now draw an arc that passes through B.
4. Cut intersecting point of two arcs is A. Join BA. ∠ABC = 60° is formed.
   Hence the required angle ∠ABC = 60° is constructed.
   

Question 2.
Construct an angle of 120° with using protractor and compasses.
Solution:
Steps of construction :

1. Draw a ray \(\overline{\mathrm{OA}}\) of any length.
2. Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OA at M.
3. With M as centre and without altering radius draw an arc which cuts the previous arc at P.
4. With P as centre and without changing radius draw an arc which cuts the first arc at Q.
5. Join OQ. Then ∠AOQ is the required angle.
   Hence the required angle ∠AOQ = 120° is constructed.
   

Question 3.
Construct the following angles using ruler and compasses. Write the steps of construction in each case.
i) 75°
ii) 15°
iii) 105°
Solution:
i) 75°
Steps of construction:

1. Construct ∠RAM 900
2. Construct ∠PAM = 600 AM as common arm
3. Now, ∠PAR = ∠RAM – ∠PAM = 90° – 60° = 30°
4. Draw bisector to ∠PAR which is \(\overrightarrow{\mathrm{AQ}}\).
5. Now, ∠MAQ = ∠PAM + ∠PAQ = 60° + \(\frac{30^{\circ}}{2}\)
   = 60° + 15° = 75° (QED)
   

ii) 15°
Steps of construction:

1. Construct ∠BQR = 60°
2. Draw the bisector to ∠BQR which is \(\overrightarrow{\mathrm{QC}}\)
3. Now,∠CQR = \(\frac{60^{\circ}}{2}\) = 30°
4. Draw the bisector \(\overrightarrow{\mathrm{QP}}\) to∠CQR.
5. ∠PQR = \(\frac{1}{2}\) x 30° = 15° (Q.E.D)
   

iii) 105°
Steps of construction :

1. Construct ∠IDS = 90°
2. Construct ∠ADS = \(\frac{\angle \mathrm{RDS}}{2}=\frac{30^{\circ}}{2}\) = 15°
3. Now, ∠ADI = ∠IDS + ∠ADS = 90° + 15°
   ∴ ∠ADI = 105° (Q.E.D)
   

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.3

Question 1.
Construct the following angles with the help of a protractor.
i) ∠ABC = 65°
ii) ∠PQR = 136°
iii) ∠Y = 45°
iv) ∠O = 172°
Solution:
i) ∠ABC = 65°

Steps of construction :

1. Draw a ray \(\overrightarrow{\mathrm{BC}}\) of any length.
2. Place the centre point of the protractor at B and the line be aligned with \(\overrightarrow{\mathrm{BC}}\).
3. Mark a point A at 65°.
4. Join BA. ∠ABC = 65° is formed.
   Hence the required angle ∠ABC = 65° is constructed.

ii) ∠PQR = 136°
Steps of construction :

1. Draw a ray \(\overrightarrow{\mathrm{QR}}\) of any length,
2. Place the centre point of the protractor at Q and the line be aligned with \(\overrightarrow{\mathrm{QR}}\).
3. Mark a point P at 136°. ‘
4. Join PQ. ∠PQR = 136° is formed.
   Hence the required angle ∠PQR = 136° is constructed.

iii) ∠Y = 45°
Steps of construction:

1. Draw a ray \(\overrightarrow{\mathrm{YZ}}\) of any length.
2. Place the centre point of the protractor at Y
   and the line be aligned with \(\overrightarrow{\mathrm{YZ}}\).
3. Mark a point X at 45°.
4. Join YX. ∠XYZ = 45° is formed.
   Hence the required angle ∠XYZ = 45° is constructed.

iv) ∠O = 172°

Steps of construction:

1. Draw a ray \(\overrightarrow{\mathrm{OT}}\) °f any length.
2. Place the centre point of the protractor at O and the line be aligned with \(\overrightarrow{\mathrm{OT}}\)
3. Mark a point D at 172°.
4. Join OD. ∠DOT = 172° is formed.
   Hence the required angle ∠DOT = 172° is constructed.

Question 2.
Copy the following angles in your note book and find their bisectors:

Solution:

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.2 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.2

Question 1.
Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.
Solution:

Steps of construction :

1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
3. With P as centre, draw arcs below and above the line segment.
4. With the same radius and Q as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
5. Join the points X and Y. So, the line I is the perpendicular bisector of PQ.
   Hence l is required perpendicular bisector of PQ which meets at A.

Question 2.
Ravi made a line segment of length 8.6 cm. He constructed a bisector of AB on C. Find the length of AC and BC.

Solution:

As it is a bisector, it divides the line segment into two equal parts.
Each equal part is half of AB (8.6 cm) = \(\frac{\mathrm{AB}}{2}=\frac{8.6}{2}\) = 4.3 cm
∴ AC = BC = 4.3 cm

Question 3.
Using ruler and compasses, draw AB = 6.4 cm. Locate its mid point by geometric construction.
Solution:

1. Draw a line segment \(\overline{\mathrm{PQ}}\) = 6.4 cm
2. Set the compasses as radius with more than half of the length of \(\overline{\mathrm{PQ}}\).
3. With A as centre, draw arcs below and above the line segment.
4. With the same radius and B as centre draw two arcs above and below the line segment to cut the previous arcs. Name the intersecting points of arcs as X and Y.
5. Join the points X and Y. So, the line l is the perpendicular bisector of AB.
   Hence l is the required perpendicular bisector of AB which meets at M.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.1

Question 1.
Construct a line segment of length 6.9 cms using ruler and compass.

Solution:
Steps of construction:

1. Draw a line l. Mark a point A on the line l.
2. Place the metal pointer of the compass on the zero mark of the ruler. Open the compass, so that pencil point touches the 6.9 cm mark on the ruler.
3. Place the pointer at A on the line l and draw an arc to cut the line. Mark the point where the arc cuts the line as B
4. On the line l, we got the line segment AB of required length.
   

Question 2.
Construct a line segment of length 4.3 cms using ruler.
Solution:

Steps of construction:

1. Place the ruler on paper and hold it firmly.
2. Mark a point with a sharp edged pencil against 0 cm mark on the ruler. Name the point as P.
3. Mark another point against 3 small divisions just after the 4 cm mark. Name this point as Q.
4. Join points P and Q along the edge of the ruler.
   Therefore, PQ is the required line segment of length 4.3 cm.

Question 3.
Construct a circle with centre M and radius 4 cm.
Solution:

Steps of construction:

1. Mark a point M on the paper.
2. By using compasses take 4 cm as the radius on with the scale.
3. Place the metal point on M and draw a circle from M.
   Hence required circle is constructed with center M and radius 4 cm.

Question 4.
Draw any circle and mark three points A, B and C such that
(i) A is on the circle
(ii) B is in the interior of the circle
(iii) C is in
Solution:

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

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Unit Exercise

Question 1.
Give examples for each shape in the table.

Solution:

Question 2.
Look at the adjacent figure and answer the following questions,
(i) What is the name of the triangle?
(ii) Write all sides, angles and vertices of the triangle.
Solution:
i) Name of the given triangle is ΔPQR.
ii) The sides of ΔPQR are \(\overline{\mathrm{PQ}}, \overline{\mathrm{QR}}\) and \(\overline{\mathrm{PR}}\) .
The angles of ΔPQR are ∠P, ∠Q and ∠R.
The vertices of ΔPQR are P, Q and R.

Question 3.
Look at the adjacent figure and answer the following questions.
(i) Write the name of this polygon.
(ii) Write the pairs of adjacent sides and adjacent angles.
(iii) Write all vertices, pairs of opposite sides and pairs of opposite angles.
Solution:
i) Name of the given polygon is quadrilateral EFGH.
ii) EH and FG are pair of adjacent sides of EF.
EF’ and GH are pair of adjacent sides of FG.
FG and EH are pair of adjacent sides of GH.
EF and HG are pair of adjacent sides of EH.
Adjacent angles of E are ∠H and ∠F.
Adjacent angles of F are ∠E and ∠G.
Adjacent angles of G are ∠F and ∠H.
Adjacent angles of H are ∠G and ∠E

iii)Vertices of the quadrilateral EFGH are E, F, G and H.
Opposite side of EF is GH.
Opposite side of FG is EH.
Pairs of opposite angles are ∠E, ∠G and ∠F, ∠H.

Question 4.
Say true or false.
(i) We can locate only one centre in a circle. [ ]
(ii) All chords are called diameters. [ ]
(iii) A square pyramid has squares as its faces. [ ]
Solution:
i) True
ii) False
iii) False

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.4 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.4

Question 1.
Write the shapes of the following.
(i) A brick (ii) A road roller (iii) Foot ball (iv) Joker cap
Solution:
i) Shape of a brick is cuboid.
ii) Shape of a road roller is cylinder.
iii) Shape of a foot ball is shpere.
iv) Shape of a joker cap is cone.

Question 2.
Fill in the blanks.
(i) The shape of heap of grain
(ii) The shape of a dice
(iii) The shape of a bubble
(iv) The shape of a candle
Solution:
i) Cone ii) Cube iii) Sphere

Question 3.
Match the following.

Solution:

Question 4.
Fill in the table.

Verify Euler’s Formula for the data in the table.

We know that, the F + V = E + 2is Euler’s formula.
In the above table given shapes satisfy the Euler’s formula.

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 9 2D-3D Shapes Ex 9.3 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 9th Lesson 2D-3D Shapes Ex 9.3

Question 1.
Look at the given quadrilateral and answer the following questions.
i) What are the sides of the given quadrilateral?
ii) What is the opposite side of \(\overline{\mathrm{AB}}\) ?
iii) What is the opposite vertex of B?
iv) What is the opposite angle of ∠C ?
v) How many pairs of adjacent angles are there? What are they?
vi) How many pairs of opposite angles are there? What are they?

Solution:
In the given quadrilateral ABCD,
i) It has four sides. They are \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CD}} \text { and } \overline{\mathrm{AD}}\)
ii) Opposite side of \(\overline{\mathrm{AB}}\) is \(\overline{\mathrm{DC}}\)
iii) Opposite vertex to B is D.
iv) Opposite angle of ∠C is ∠BAD (or) ∠A.
v) It has four pair of adjacent angles are there. They are :
Adjacent angles of ∠A are ∠B and ∠D.
Adjacent angles of ∠B are ∠A and ∠C.
Adjacent angles of ∠C are ∠B and ∠D.
Adjacent angles of ∠D are ∠A and ∠C.
vi) It has two pairs of opposite angles are there. They are ∠A and ∠C; ∠B and ∠D.

Question 2.
Find the number of lines of symmetry in the following.

Solution:
i) Given adjacent figure is a square.
Number of lines of symmetry to a square are 4

ii) Given adjacent figure is a circle.
Number of lines of symmetry to a circle are infinite (many).

iii) Given adjacent figure is a triangle.
Number of lines of symmetry to the triangle are three.
(If it is an equilateral triangle).