Category: Class 7

SCERT AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Review Exercise

Find the missing value in the following tables and fill by using given hints:

(i)

Answer:

(ii)

Answer:

(iii)

Answer:

(iv)

Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures Ex 11.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Ex 11.4

Question 1.
The radius of a circular shaped park is 40 m. A path of width 7 m. is layed around outside the park . Find the area of circular path.
Answer:
Given the radius of a park r = 40 m
Area of inner circle = πr²
= \(\frac{22}{7}\) × 40 × 40
= \(\frac{35200}{7}\)
Area of outer circle = πr²
= \(\frac{22}{7}\) × 47 × 47
= \(\frac{48,598}{7}\) = 6942.57 sq.m
Area of the path = Outer circle area – Inner circle area
= 6942.57 – 5028.57
∴ Area of circular path = 1914 sq.m

Question 2.
Bhuvanesh builds a circular lawn of radius 28 m. in front of his house. A path of 7 m. width is laid around outside the lawn. Find the area of the circular path.
Answer:
Radius of circular lawn r = 28 m
Area of inner circle = πr²

= 2464 sq.m
A path laid outside the lawn.
Width of the path = 7 m.
Radius of the outer circle
= Inner radius + width of path
= 28 + 7 = 35 m
Area of the outer circle = πr²

= 3850 sq.m.

Question 3.
A water fountain is in circular shaped whose radius is 12 m. Its inner portion 5 m. arranged for fountain remaining part cemented. Find the area of that cemented part. Find the cost for cementing, if the rate is ₹ 150 per Sq. m.
Answer:
Radius of water fountain r = 12 m
Area of the outer circle = πr²
= \(\frac{22}{7}\) × 12 × 12
= \(\frac{3168}{7}\)
= 452.57 sq.m
Path laid inner side.
So, width of the path = 7 m
Radius of inner circle = Outer radius – width of the path
= 12 – 7 = 5m
Area of the inner circle = πr²
= \(\frac{22}{7}\) × 5 × 5
= 78.57 sq.m.
Width of the path = 5 m
Area of the path = Outer area – Inner area = 452.57 – 78.57.
Area of the circular path = 374 sq.m
Cost of cementing per sq.m = ₹ 150
Cost of cementing per 374 sq.m = 374 × 150
Total cost of cementing = ₹ 56,100

Question 4.
The radius of circular shaped cricket ground is 55 m. A lobby of 5 m width has been constructed around the ground for spectators. Find the area of lobby. Find the cost of construction of the lobby for seating arrangement, if the rate of construction is ₹ 1500 per Sq. m.
Answer:
Radius of the cricket ground r = 55 m
Area of the cricket ground = πr²
= \(\frac{22}{7}\) × 55 × 55
= \(\frac{66550}{7}\)
= 9507.14 sq.m
Lobby in constructed around the ground.
Radius of the inner circle = Outer radius – width of the path
= 55 – 5 = 50m
Area of the inner circle = πr²
Area of the path = Outer area – Inner area
= 9507.14 – 7857.14 = 1650 sq.m
Cost of constructing seating arrangement per sq.m = ₹ 1500
Cost of seating arrangement per 1650 sq.m = 1650 × 1500 = ₹ 24,75,000

SCERT AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures Ex 11.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Ex 11.3

Question 1.
The diameter of round table upper surface in Science lab is 70 cm. Find the area.

Answer:
Given diameter of table d = 70 cm
Radius of table r = \(\frac{\mathrm{d}}{2}\) = \(\frac{70}{2}\) = 35 cm
Area of the table = πr²

∴ Area of the round table = 3,850 sq.cm.

Question 2.
Radius of the circular shaped wall painting is 14 cm. Calculate the area of wall painting.

Answer:
Given radius of the wall painting
r = 14 cm
Area of the circle = πr²
= \(\frac{22}{7}\) × 14²

= 616
∴ Area of the wall painting = 616 sq.cm

Question 3.
If the area of circular shaped dart board is 1386 sq.cm. Find its radius and diameter.

Answer:
Given the area of circular shaped dart board = 1386 sq.cm
Area of the circle = πr² = 1386
⇒ \(\frac{22}{7}\) × r² = 1386
⇒ 22 × r² = 1386 × 7

⇒ r² = 441 = 21²
⇒ r² = 21²
∴ Radius of dart board r = 21 cm
Diameter of dart board d 2r
= 2 × 21
= 42 cm

Question 4.
Circumference of the circular shaped clock is 44 cm. Find the radius and surface area of the clock.

Answer:
Given the circumference of clock = 44 cm
⇒ 2πr = 44 cm
⇒ 2 × \(\frac{22}{7}\) × r = 44
⇒ 44 × r = 44 × 7

∴ Radius of the clock r = 7 cm
Surface area of the clock = πr²
= \(\frac{22}{7}\) × 7²

∴ Surface area of the clock = 154 sq.cm

Question 5.
The circumference of a circular shaped lawn in a park is 352 m. Find the area of the circular shaped lawn. If cost of grass per sq. m is ₹ 30, then find total cost of laying grass on lawn.

Answer:
Given the circumference of lawn = 352 m
⇒ 2πr = 352 m
⇒ 2 × \(\frac{22}{7}\) × r = 352
⇒ 44r = 352 × 7 7

∴ Radius of the lawn r = 56 m
Area of the lawn = πr²
= \(\frac{22}{7}\) × 56²

Area of the lawn = 9856 sq.m
Cost of grass per sq.m = ₹ 30
Cost of grass per 9856 sq.m = 9856 × 30
∴ Total cost of laying grass on lawn = ₹ 2,95,680

SCERT AP 7th Class Maths Solutions Pdf Chapter 11 Area of Plane Figures Ex 11.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Ex 11.2

Question 1.
The length and breadth of the tablet are 16 cm, 8 cm respectively. It has a inner side black border of 1 cm width around the screen. Find the area of the black border.
Answer:
Given length of the tablet l = 16 cm
breadth b = 8 cm
Area of the tablet ABCD
= l × b
= 16 × 8
= 128 sq.cm

Width of border = 1 cm
Border running around the screen.
So, length of screen
l = 16 – 2 × width
= 16 – 2 × 1
= 16 – 2 = 14 cm
Breadth of screen
b = 8 – 2 × 1
= 8 – 2 = 6 cm
Area of the screen
= l × b
= 14 × 6
= 84 sq.cm
Area of the border = Area of the tablet – Area of screen
= 128 – 84
= 44 sq.cm
∴ Area of black border = 44 sq.cm.

Question 2.
Revanth has a rectangular lawn of length 45 m and breadth 20 m in his garden. He wants to do flooring 5 m along out-side the lawn for path. Find out the area of path. Find the cost of flooring at the rate ₹ 100 per sq.m.
Answer:
Given inner length of lawn l = 45 m
breadth b = 20 m
Inner area of lawn = l × b
= 45 × 20
= 900 sq.m
Width of the path = 5 m
Path is outside the lawn.
So, outer length of lawn
L = l + 2 × w
= 45 + 2 × 5
= 45 + 10 = 55 m
Breadth B = b + 2w
= 20 + 2 × 5
= 20 + 10 = 30 m
Outer area of lawn = l × b
= 55 × 30
= 1650 sq.m
Area of the path
= outer area – inner area
= 1650 – 900
= 750 sq.m.
Cost of flooring per sq.m = ₹ 100
∴ Cost of flooring per 750 sq.m
= 750 × 100 = ₹ 75,000/-

Question 3.
The surface of water pool is in the shape of a square whose side is 450 cm. Its exterior 20 cm width and the edge part is cemented along the side of the square. Find the area of that cemented part. Find cost for cementing if the rate is ₹ 15 per sq.cm.
Answer:
Given inner side of the pool (s) = 450 cm
Inner area of waterptool ABCD = s × s
= 450 × 450
= 202500 sq.cm
Width of the edge = 20 cm

Here path is running outside the pool.
So, outer side of the pool
(s) = s + 2 × w
= 450 + 2 × 20
= 450 + 40 = 490 cm
Outer area of waterpool EFGH
= s × s
= 490 × 490
= 2,40,100 sq.cm
Area of the edge
= Outer area – Inner area
= 240100 – 202500
Area of the path = 37,600 sq.cm
Cost of cementing per sq.cm = ₹ 15
∴ Cost of cementing per 37600 sq.cm
= 37,600 × 15
= ₹ 5,64,000

Question 4.
Two pathways parallel to the length and breadth have been constructed in the centre of the rectangular park whose length is 120 m and breadth is 90 m. If the width of each pathway is 15 m, find out expenditure of flooring on the pathway at the rate ₹ 80 per sq.m,

Answer:
In the figure ABCD is a rectangular park.
Length of ABCD = 120 m
Breadth of ABCD = 90 m
Width of path = 15 m
Area of EFGH = length × width
= 120 × 15
= 1800 sq.m
Area of MNOP = breadth × width
= 90 × 15
= 1350 sq.m
Area of common path = width × width
= 15 × 15
= 225 sq.m

Area of IJKL = 225 sq.m is included in both EFGH and MNOP paths. So, we subtract once.
Area of the path = Area of EFGH + Area of MNOP – Area of IJKL .
= 1800 + 1350 – 225
= 2925 sq.m
Cost of flooring per sq.m = ₹ 80
Cost of flooring per 2925 sq.m. = 2925 × 80
∴ Expenditure for path flooring = ₹ 2,34,000.

Question 5.
A photo frame of length 28 cm. and breadth 11 cm. has done decoration of 3 cm. along inside shown in the figure. Find the total area of decoration. If the cost of decoration is ₹ 2 per Sq.cm., find total cost of decoration.
Answer:
Given the length of frame ABCD = 28 cm
Breadth = 11 cm
Area of frame ABCD = l × b
= 28 × 11
= 308 sq.cm
Width of the decoration part = 3 cm
Decoration is tying inside the frame.
So, length of EFGH = l – 2 × width
= 28 – 2 × 3
= 28 – 6 = 22 cm
Breadth = b – 2 × width
= 11 – 2 × 3
= 11 – 6 = 5 cm

Area of frame EFGH = l × b
= 22 × 5
= 110 sq.cm
Area of the decoration part
= Area of ABCD – Area of EFGH
= 308 – 110
= 198 sq.cm
Cost of decoration per sq.cm = ₹ 2
Cost of decoration per 198 sq.cm = 198 × 2
∴ Total cost of decoration = ₹ 396.

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry InText Questions

[Page No. 99]

Look at the following picture and speak about what you have noticed regarding symmetry in the picture.

Question 1.
What do you observe in this picture?
Ducks have symmetry.
Wall paintings have symmetry.
Butterfly has symmetry.
Clock tower and gaint wheels have symmetry.

Question 2.
Can you name the different shapes in picture?
Answer:
Hexagon, Circle, Rectangle, etc.

Question 3.
Which figures are looking attractive and beautiful? Why?
Answer:
Flooring, Gate arch and wall painting are beautiful as they have lines of symmetry.

Question 4.
Which figures are in symmetry?
Answer:
Ducks, wall paintings, gaint wheel and the clock tower have symmetry.

Question 5.
Can you draw line(s) of symmetry for them?
Answer:
Yes.

[Page No. 100]

Look at the following figures. If we fold them exactly to their equal halves, one half of each figure exactly coincides with other half.

Question 1.
What do we call such figures?
Answer:
Symmetric figures or shapes.

Question 2.
What do we call the line along which we fold the figures so that, one half coincides with the other?
Answer:
Line of symmetry.

Let’s Explore [Page No.102]

Question 1.
Find relation between the number of sides of regular polygon and its lines of symmetry.
Answer:

From the above table, we conclude that the number of sides of a regular polygon is equal to its number of lines of symmetry.

Question 2.
How many lines of symmetry can be drawn to a circle?
Answer:
Number of lines of symmetry can be drawn to a circle are infinite.

Let’s Think [Page No. 103]

Question 1.
Draw any three shapes basing on below sentences :
(i) no line of symmetry
Answer:

(ii) one line of symmetry
Answer:

(iii) two lines of symmetry
Answer:

(iv) three lines of symmetry
Answer:

Check Your Progress [Page No. 105]

Find whether the following letters of the English alphabet have rotational symmetry or not. If yes, find the point of rotational symmetry (approximately), and also order of rotational symmetry.

Answer:

Check Your Progress [Page No. 107]

Look at the following pictures and write their angle of symmetry and order of symmetry.

Answer:
(i) Spinner: Angle of symmetry = 120°
Order of symmetry = 3

(ii) Gaint wheel : Angle of symmetry = 60°
Order of symmetry = 6

[Page No. 108]

Tesselations: The majority of the objects what we use in daily life have atleast one type of symmetry. Most of the machine made products are symmetric.

Observe these patterns :

(i) Where do you see these?
Answer:
We see these patterns in floor designs and fabric painting etc.

(ii) How these patterns are formed? Are they symmetric as whole?
Does the basic figure which is used to form the tessellations symmetric?
Answer:
You can observe that only some patterns have symmetry as a whole as in fig(i) and (ii), through the basic figures/unit figures are symmetric. Observe the fig(iii). You may notice that the two shapes can observe as ‘Basic or unit figures’ which is square or hexagon used to draw tessellation.

Usually, these patterns are formed by arranging congruent figures side by side in all the directions to spread upon an area without any overlaps or gaps. This is called tessellation. This enhances the beauty of the diagrams.

Let’s Explore [Page No. 109]

Fill the following with different colours to get beautiful tessellation patterns.

Answer:
Not needed.
Student should colour it on his own.

Let’s Explore [Page No. 109]

Construct a square having side 3cm. Draw all possible lines of symmetry. (Steps of construction is not necessary)
Answer:

1. Let ▢ABCD be the square with side 3 cm.
2. AC, BD, LM and PQ are lines of symmetry.

Examples:

Question 1.
If an equilateral triangle ABC is rotated through 360° about the point of rotation ‘P’ (i.e., point of intersection of angle bisectors), it attains the original form three times upon rotation through 120°, 240° and 360° as shown below.
Answer:

Clearly, the above figure possesses a rotational symmetry of order 3.

Question 2.
While rotating the following figure through 360° about the point ‘O’ (mid-point of BC), it attains the original form twice (upon rotation through 180° and 360°) as shown below.
Answer:

Clearly, the above given figure possesses a rotational symmetiy of order 2.

Question 3.
Check whether the letter ‘S’ has point symmetry.

Answer:
Yes, it has a point symmetry.
Because, in the given figure, we have
i) Every part of the letter has a matching part are at the same distance from the central point.
ii) The part of the alphabet and its matching part are in the opposite direction.

[Page No- 111]

The image of an object as seen in a mirror is known as mirror image./In mirror image of an object, right side of the object appears at left side and vice versa. There are some objects whose mirror images are identical to the objects. For example, 11 capital English alphabets whose mirror images remain unchanged. They are A, H, I, M, O, T, U, V, W, X and Y.

Mirror Images of English Alphabets and some numbers :

Practice Questions [Page No. 112]

Choose the correct mirror image of the given words.

Question 1.
LATERAL

Answer:
(b)

Question 2.
QUANTITATIVE

Answer:
(d)

Question 3.
JUDGEMENT

Answer:
(c)

Question 4.
EMANATE

Answer:
(b)

Question 5.
KALINGA261B

Answer:
(d)

Question 6.
COLONIAL

Answer:
(d)

Question 7.
BR4AQ16HI

Answer:
(a)

Question 8.
R4E3N2U

Answer:
(c)

Question 9.
DL3N469F

Answer:
(b)

Question 40.
MIRROR

Answer:
(d)

Water images [Page No. 112]

The reflection of an object as seen in water is known , as water image. The upper part of the object seen downward and vice-versa. There a re some objects whose water images are identical to the objects. For example, 9 cap ital English alphabets given below, whose water images remain unchanged. They ar e B, C, D, E, H, I, K, O and X.

Practice Questions [Page No. 113]

Choose the correct water image of the given words.

Question 1.
KICK

Answer:
(d)

Question 2.
UPKAR

Answer:
(a)

Question 3.
KID

Answer:
(b)

Question 4.
SUBHAM

Answer:
(c)

Question 5.
CHIDE

Answer:
(d)

Question 6.
HIKE

Answer:
(a)

Question 7.
COT5E

Answer:
(a)

Question 8.
ab45CD67

Answer:
(b)

Question 9.
abc

Answer:
(a)

Question 10.
01234

Answer:
(a)

Mirror image of the clock [Page No. 114]

1. A clock is a device provided with three hands an hour hand, minute hand and second hand.
Hours hand also known as short hand and minutes hand is also known as long hand.

2. The face of the clock is divided into 12 equal divisions.
The angular space between any two consecutive divisions has further divided into five more divisions.

Let us observe the following images of a clocks and their mirror images :
Case I: To find the time in the mirror image subtract the actual time from 11 hours 60 minutes.

Example 1:
If actual time in clock is 9 hours 30 minutes then, what is’the time shown in mirror?
Answer:
11 hrs 60 min – 09 hrs 30 min – 2 hrs 30 min

Example 2 :
If actual time in clock is 7 hours, then what is the time shown in mirror?
Answer:
11 hrs 60 min – 07 hrs 00 min = 4 hrs 60 min = 5 hrs (1 hour — 60 minutes)

Case II : If the time in clock is between 12 hrs and 01 hr then to find the time in the mirror image subtract the actual time from 23 hr 60 min.
Example :
If actual time in clock is 12 hours T5 minutes, then what is the time shown in mirror?
Answer:
23 hrs 60 min — 12 hrs 15 min =11 hrs 45 min

Practice Questions [Page No. 114]

Choose the correct mirror image of the given words.

Question 1.
If actual time in dock is 06 hours 10 minutes, then what is the time shown in mirror?
(a) 3 hr:s 50 min
(b) 4 hrs 50 min
(c) 5 hrs 50 min
(d) 6 hrs 50 min
Answer:
(c) 5 hrs 50 min

Question 2.
If actual time in clock is 03 hours 54 minutes, then what is the time shown in mirror?
(a) 8 hrs 06 min
(b) 9 hrs 06 min
(c) 8 hrs 54 min
(d) 9 hrs 54 min
Answer:
(a) 8 hrs 06 min

Question 3.
If actual time in clock is 08 hours 26 minutes, then what is the time shown in mirror?
(a) 6 hrs 34 min
(b) 3 hrs 34 min
(c) 1 hr 34 min
(d) 3 hrs 36 min
Answer:
(b) 3 hrs 34 min

Question 4.
If actual! time in clock is 4 hours, then what is the time shown in mirror?
(a) 7 hrs
(b) 7 hrs 30 min
(c) 8 hrs
(d) 8 hrs 30 min
Answer:
(c) 8 hrs

Question 5.
If actual time in dock is 10’o clock, then what is the time shown in mirror?
(a) 2 hrs
(b) 3 hrs
(c) 4 hrs
(d) 5 hrs
Answer:
(a) 2 hrs

Question 6.
If the time shown in mirror is 10 hours 05 minutes, then what is the actual time in clock?
(a) 1 hrs 55 min
(b) 1 hrs 35 min
(c) 1 hrs 25 min
(d) 12 hrs 15 min
Answer:
(a) 1 hrs 55 min

Question 7.
If the time shown in mirror is 02 hours 47 minutes, then what is the actual time in clock?
(a) 6 hrs 13 min
(b) 7 hrs 13 min
(c) 8 hrs 13 min
(d) 9 hrs 13 min
Answer:
(d) 9 hrs 13 min

Question 8.
If actual time in clock is 11 hours 45 minutes, then what is the time shown in mirror?
(a) 1 hrs 15 min
(b) 3 hrs 15 min
(c) 6 hrs 15 min
(d) 12 hrs 15 min
Answer:
(d) 12 hrs 15 min

Question 9.
If actual time in clock is 12 hours 45 minutes, then what is the time shown in mirror?
(a) 9 hrs 15 min
(b) 10 hrs 15 min
(c) 11 hrs 15 min
(d) 12 hrs 15 min
Answer:
(c) 11 hrs 15 min

Question 10.
If actual time in clock is 12 hours 12 minutes, then what is the time shown in mirror?
(a) 11 hrs 42 min
(b) 11 hrs 48 min
(c) 10 hrs 48 min
(d) 12 hrs 42 min
Answer:
(b) 11 hrs 48 min

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Unit Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Unit Exercise

Question 1.
Fill in the blanks.
(i) Afigure has _______ symmetry if there is a line about which the figure may be folded so that the two parts of the figure coincide.
Answer:
Line

(ii) A regular pentagon has _______ line of symmetry.
Answer:
Five

(iii) A figure has _______ symmetry if after a rotation through a certain angle. The figure looks exactly the same.
Answer:
Rotational

(iv) A _______ triangle has no lines of symmetry.
Answer:
Scalene

(v) Each regular polygon has as many lines of symmetry as it has _______
Answer:
Number of sides

(vi) The concept of line symmetry is closely related to _______ reflection.
Answer:
Mirror

(vii) The quadrilateral that has four lines of symmetry and order-four rotational symmetry is a _______
Answer:
Square

(viii) The angle of rotational symmetry for letter S is _______
Answer:
180°

(ix) A line segment is symmetrical about its _______
Answer:
Perpendicular bisector

(x) Station turns an object about a fixed point. The fixed point is called _______
Answer:
Point of rotation

(xi) Each of the letters H, N, S and Z has a rotational symmetry of order _______
Answer:
2

(xii) The line of symmetry of an isosceles triangle is it’s _______ from the vertex having the equal sides.
Answer:
Median/altitude/angular bisector

Question 2.
Cut the capital letters of English and paste them in your note book. Draw possible number of lines of symmetry for each of the letter.
(i) How many letters have no line of symmetry? What are thpy?
Answer:
10. They are F, G, J, L, N, P, Q, R, S, Z.

(ii) How many letters have one line of symmetry? What are they?
Answer:
12. They are A, B, C, D, E, K, M, T, U, V, W, Y.

(iii) How many letters have two lines of symmetry? What are they?
Answer:
3. They are H, I, X.

(iv) How many letters have more than two lines of symmetry? What are they?
Answer:
1. That is 0.

(v) Which of them have rotational symmetry? What are they?
Answer:
4. They are H, I, 0, X.

(vi) Which of them have point symmetry? What are they?
Answer:
7. They are H, I, N, 0, S, X, Z.

Question 3.
Draw some natural objects which have at least one line of symmetry.
Answer:

Question 4.
Draw three tessellations and expose the basic shapes used on your tessellation.
Answer:

Question 5.
Construct a line segment of length 7cm. Draw possible line(s) of symmetry.
Answer:

1. Let AB be a line segment of length 7 cm.
2. Let M be the mid point of AB.
3. Now any line through M is a line of symmetry.
4. So, the line segment \(\widehat{\mathrm{AB}}\) has infinite lines.

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Ex 12.4 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Ex 12.4

Question 1.
Observe the following pattern and complete it.

Answer:

Question 2.
Draw any two tessellations and identify the basic shape.
Answer:

Note: Can be drawn with any regular polygon.

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Ex 12.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Ex 12.3

Question 1.
Draw lines of symmetry for the following figures. Identify which of them have point symmetry. Is there any relation between lines of symmetry and point symmetry?

Answer:

(i) Rectangle has point of symmetry.
Number of lines of symmetry = 2
Order of symmetry = 2
Lines of symmetry point or symmetry are same.

(ii) Square has a point of symmetry.
Number of lines of symmetry = 4
Order of symmetry = 4
Lines of symmetry, point of symmetry are same.

(iii) Triangle has no point of symmetry.

(iv) Regular hexagon has point of syrninetìy.
Number of lines of symmetry = 6
Order of symmetry = 6
Lines of symmetry point of symmetry are same.

(v) Pentagon has no point of symnetry

Question 2.
Identify which of the English alphabet have point symmetry in the following:

Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Ex 12.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Ex 12.2

Question 1.
Which of the following shapes have line symmetry? Which have rotational symmetry?

Answer:
(a) Has no line of symmetry and rotational symmetry.
(b) Has line of symmetry and no rotational symmetry.
(c) Has line of symmetry and rotational symmetry.
(d) Has line of symmetry and rotational symmetry.

Question 2.
Determine the order of rotational symmetry in the figures given below.

Answer:
(i) The order of rotatiorlal symmetry of the figure is 4.
(ii) The order of rotational symmetry of the figure is 6.
(iii) The order of rotational symmetry of the figure is 3.
(iv) The order of rotational symmetry of the figure is 4.

Question 3.
Draw the 2 such figures which have line symmetry and rotational symmetry of order more than 1.
Answer:
(a) Equilateral triangle has 3 lines of symmetry and rotational symmetry of order (3).

(b) Rectangle has 2 lines of symmetry and rotational symmetry of order (2).

Question 4.
Name the quadrilaterals which have both line symmetry and rotational symmetry of order more than 1.
Answer:
Rectangle, square and rhombus.

Question 5.
After rotating by 60° about its axis, a figure looks exactly the same as its original position. What other angles will this happen for the same figure?
Answer:

Question 6.
Fill in the blanks.

Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Ex 12.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Ex 12.1

Question 1.
Draw the possible line(s) of symmetry for the following figures, as shown in Figure-(i).

Answer:

Question 2.
Observe the figures with punched holes and draw the axes of symmetry.

Answer:

Question 3.
Mark the other dot to become the following dotted line in the picture as line symmetry.

Answer:

Question 4.
State the number of lines ofsyrnrnetry for the following figures and draw them.
(i) An equilateral triangle
Answer:

Number of lines of symmetry for an equilateral triangle is 3.

(ii) An isosceles triangle
Answer:

Number of lines of symmetry for an isosceles triangle is 1.
(It is the altitude on the unequal side)

(iii) A scalene triangle
Answer:

We cannot draw the line of symmetry for scalene triangle.

Question 5.
Construct an equilateral triangle with a length of side 4cm and draw all possible lines of symmetry (No need to write the steps of constructions).
Answer:

∆ADI is an equilateral triangle.
Number of lines of symmetry for an equilateral triangle is 3.

Question 6.
Construct the triangle with the base 4.5cm and base angle 45° each. Draw all possible lines of symmetry (No need to write the steps of constructions).
Answer:

∆SRI is an isosceles triangle.
Number of lines of symmetry for an isosceles triangle is 1.

SCERT AP 7th Class Maths Solutions Pdf Chapter 12 Symmetry Review Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Symmetry Review Exercise

Question 1.
In the following figures, the mirror line (i.e., the line of symmetry) is given as a dotted line.
Complete each figire performing reflection in the dotted (mirror) line. (You might perhaps place a mirror along the dotted line and look into the mirror for the image).

Answer:

Question 2.
Which of the following figures are having line symmetry? Write YES or NO in the given below the figure boxes.

Answer:

Question 3.
Draw the other half of each symmetrical shape for the following pictures.

Answer:

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles InText Questions

[Page No. 67]

Look at the picture and answer the following questions

Question 1.
Name some things from your daily life that look like triangles.
Answer:
Samosa, chapathi, window elevations, house tops, bridge trusses, floor tessellations.

Question 2.
What are the types of triangles can see in the picture?
Answer:
Right triangles,
Equilateral triangles.

Question 3.
Do you think all triangles shown are similar with their properties? What are they?
Answer:
Yes. All right triangles are similar.

All triangles are similar because they have
(a) right angle
(b) equal side and
(c) equal hypotenuse.

Check Your Progress [Page No. 69]

Question 1.
Construct an equilateral triangle ∆XYZ with XY = 4 cm.
Answer:
Given sides of an equilateral triangle ∆XYZ is XY = 4 cm.
In equilateral triangle all sides ,gresequal in length
∴ XY = YZ = ZX = 4 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment with XY = 4 cm,
3. Draw an arc with centre X and radius 4 cm.
4. Draw another arc with centre Y and radius 4 cm to intersecting the previous arc at Z.
5. Join XZ andYZ.

Hence, required ∆XYZ is constructed with the given measurements.

Question 2.
Construct an isosceles triangle ∆PQR With PQ = PR = 3 cm, QR = 5 cm.
Answer:
Given sides of an isosceles triangle ∆PQR are PQ = PR = 3 cm and QR = 5 cm
In isosceles triangle two sides are equal.
∴ PQ = PR = 3 cm

Steps of construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment QR = 5 cm.
3. Draw an arc with centre Q and radius 3 cm.
4. Draw another arc with centre R and same radius (3 cm) to intersect the previous arc at P.
5. Join PQ and PR.

Hence, required APQR is constructed with the given measurements.

Let’s Think [Page No. 69]

Can you construct ∆ABC with AB = 4 cm, BC = 5 cm and CA =10 cm? Why? Justify your answer.
Answer:
Given sides of a ∆ABC are AB = 4 cm, BC = 5 cm and CA =10 cm.
In any triangle sum of any two sides is always greater than the third side.
AB + BC = 4 cm + 5 cm = 9 cm <10 cm.
Sum of AB + BC < AC
So, with the given measurements construction of ∆ABC is not possible.

Check Your Progress [Page No. 72]

Question 1.
Construct ∆MAT with measurements MA = 5.5 cm, MT = 4 cm and ∠M=70°.
Answer:
Given measurements of ∆MAT are MA = 5.5 cm, MT = 4 cm and ∠M = 70°.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with the given measurements.
2. Draw a line segment with MA = 5.5 cm.
3. Draw a ray \(\overrightarrow{\mathrm{MX}}\) such that ∠AMX = 70°
4. Draw an arc with centre M and radius 4 cm, to intersect the \(\overrightarrow{\mathrm{MX}}\) at point T.
5. Join AT.

Hence, required ∆MAT is constructed with the given measurements.

Let’s Explore [Page No. 72]

Question 1.
Construct the triangle with the measurements of AB = 7cm, ∠B = 60° and ∠C = 70°.
Answer:
Given measurements of ∆ABC are AB = 7 cm, ∠B = 60° and ∠C = 70°.
In ∆ABC, ∠A + ∠B + ∠C = 180°
∠A + 60° + 70° = 180° .
∠A + 130° = 180°
∠A = 180° – 130° = 50°

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AB = 7 cm.
3. Draw a ray BY such that ABY = 60°.
4. Draw another ray AX such that BAX = 50°.
5. Name the intersecting of rays AX and BY is C.

Hence, required ∆ABC is constructed with the given measurements.

Check Your Progress [Page No. 73]

Question 1.
Construct ∆ABC with the measurements ∠A = 90°, ∠C = 50° and AC = 8 cm.
Answer:
Given measurements of ∆ABC are ∠A = 90°, ∠C = 50° and AC = 8 cm.

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AC = 8 cm.
3. Draw a ray AX such that ∠CAX = 90°.
4. Draw another ray CY such that ∠ACY = 50°.
5. Name the intersecting point of AX and CY as B.

Thus, required ∆ABC is constructed with the given measurements.

Let’s Think [Page No. 73]

Construct a triangle with angles 100°, 95° and a side of length of your choice. Can1 you construct the triangle?
Answer:
No. It is not possible. We know that the sum of the angles of a triangle is 180°.
But, given two angles are 100° and 95°. Their sum is greater than 180°.
So, we cannot construct the triangle.

Examples:

Question 1.
Construct ∆ABC with slides AB = 6 cm, BC = 4 cm and AC = 5 cm.
Answer:

Question 2.
Construct ∆EFD with the measurements ∠F = 70°, EF = 4 cm and FD = 5 cm.
Answer:

Question 3.
Construct ∆ABC with AB = 6 cm, BC = 7 cm, ∠BAC = 80°.
Answer:

• Step-1: Draw a rough sketch of ∆ABC and label it with the given measurements.
• Step-2: Draw a line segment AB of length 6 cm.
• Step-3: Draw a ray AX such that ∠BAX = 80°.
• Step-4: Draw an arc with center B and radius 7 cm to Intersect the ray AX at C.
• Step-5: Join B and C to get the required ∆ABC.

Question 4.
Construct a right angled triangle ∆XYZ, XY = 4 cm, XZ = 6.5 cm and ∠Y = 90°.
Answer:

• Step-1: Draw a rough sketch of ∆XYZ and label it with given measurements.
• Step-2: Draw a line segment with XY = 4 cm.
• Step-3: Draw a ray YP such that ∠XYP = 90°.
• Step-4: Draw an arc with center X and radius 6.5 cm to intersect the Ray YP at Z.
• Step-5: Join X and Z to get the required ∆XYZ.

Question 5.
Construct ∆PEN with measurements PE. = 7 cm, ∠PEN = 25° and ∠EPN = 60°
Answer:

• Step-1: Draw a rough sketch of the triangle and label it with given measurements.
• Step-2: Draw a line segment with PE = 7 cm,
• Step-3: Draw a ray PX such that ∠EPX = 60°.
• Step-4: Draw another ray EY such that ∠PEY = 25°.
• Step-5: Rays PX and EY intersect each other at N.

Hence, we get the required ∆PEN.
(If necessary, extend the rays to form a triangle.)

Question 6.
Construct ∆MPC the triangle with the measurements MP = 4 cm, ∠P = 45° arid ∠C = 80°.
Answer:
Now we have to find out the third angle.
We know that the sum of three angles in a triangle is 1800.
So, ∠M + ∠P + ∠C = 180°
⇒ ∠M + 450 + 80° = 180°
⇒ ∠M + 125° = 180°
⇒ ∠M = 180 – 125 = 55°

• Step-1: Draw a rough sketch of the triangle and label it with given measurements.
• Step-2: Draw a line segment with MP = 4 cm.
• Step-3: Draw a ray MX such that ∠PMX = 55°.
• Step-4: Draw another ray PY such that ∠MPY = 45°.
• Step-5: Rays MX and PY intersect each other at C. (If necessary extend the rays to ‘ form a triangle.)

Hence, we get the required ∆MPC.
Verification: By using the protractor and check whether ∠C is equal to 80° or not.

Practice Questions [Page No. 78]

Find the number of triangles in the given figures.

Question 1.

(a) 8
(b) 9
(c) 10
(d) 12
Answer:
(b) 9

Explanation:

Question 2.

(a) 60
(b) 65
(c) 84
(d) 90
Answer:
(a) 60

Explanation:

Number of rows = 4
Sum of numbers in each row = 1 + 2 + 3 + 4 + 5 = 15
Number of triangles in the picture = 4 × 15 = 60

Question 3.

(a) 12
(b) 13
(c) 14
(d) 15
Answer:
(b) 13

Explanation:
Formula to count number of triangles = 4n + 1
Here n = number of embedded triangles in outer triangle
In the figure n = 3
So, number of triangles = 4(3) + 1 = 12 + 1 = 13

Question 4.

(a) 16
(b) 13
(c) 9
(d) 7
Answer:
(a) 16

Question 5.

(a) 21
(b) 23
(c) 25
(d) 29
Answer:
(d) 29

Explanation:

Question 6.

(a) 10
(b) 19
(c) 21
(d) 23
Answer:
(c) 21

Explanation:

Question 7.

Answer:
(a) 5
(b) 6
(c) 8
(d) 10
Answer:
(c) 8

Question 8.

(a) 9
(b) 10
(c) 11
(d) 12
Answer:
(a) 9

Explanation:

Question 9.

(a) 19
(b) 20
(c) 16
(d) 14
Answer:
(a) 19

Question 10.

(a) 56
(b) 48
(c) 32
(d) 60
Answer:
(b) 48

Explanation:
Formula to count number of triangles = \(\frac{\mathrm{n}(\mathrm{n}+2)(2 \mathrm{n}+1)}{8}\)
Where n = number of thangles formed in a side
Here in the given figure n = 5
Number of triangles = \(\frac{5(5+2)(2 \times 5+1)}{8}=\frac{5 \times 7 \times 11}{8}\) = 48.1
So, total number of triangles = 48.