Mahesh

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.
The 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 figure 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 11 Area of Plane Figures InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures InText Questions

Check Your Progress [Page No: 84]

Question 1.
Fill the missing values in the following table.

Answer:

Question 2.
Anu has 4 right angled triangles with same size. Using those triangles, she makes a star like toy given below. Calculate the area of this toy star.

Answer:
Given base of the triangle b = 5 cm
Height of the triangle h = 12 cm
Area of the triangle = \(\frac{1}{2}\) × b × h

= 30 sq.cm
Area of star = Area of 4 triangles
∴ Area of the toy star = 4 × 30
= 120 sq.cm

Let’s Explore [Page No. 85]

Question 1.
The areas of triangular field ABC and rectangular field EFGH are equal. The length and breadth of EFGH are 15 m., 10 m. respectively. The base of ∆ABC 25 m. then find it’s height.

Answer:
Given length of rectangle l = 15 m
breadth of rectangle b = 10 m
Area of rectangle = l × b
= 15 × 10
= 150 sq.m.
Given base of the triangle b = 25 m
height of the triangle h = ? m
∴ Area of the triangle = \(\frac{1}{2}\) × b × h
⇒ \(\frac{1}{2}\) × 25 × h = 150
⇒ 25h = 150 × 2

∴ Height of triangle h = 12 m.

Question 2.
All the triangles in the given figure are on the base AB =12 cm. Find the height of each of the triangles corresponding to the base AB, by counting the grids and find the area of each triangle. What do you observe?

Answer:
Given in the figure,
Base of ∆ APB = 12 cm
Height = 8 cm
Area of ∆ APB = \(\frac{1}{2}\) × b × h

= 48 sq.cm
Base of ∆ AQB = 12 cm
Height = 8 cm
Area of ∆ AQB

= 48 sq.cm
Base of ∆ ARB = 12 cm
Height = 8 cm
Area of ∆ ARB

= 48 sq.cm
Base of ∆ ASB = 12 cm
Height = 8 cm
Area of ∆ ASB

= 48 sq.cm
By observing area of each triangle is same.

Let’s Explore [Page No: 88]

Question 1.
5 cm width white tiles arranged between square shaped blue tiles along sides as shown in the figure. If the side of the total arrangement is 150 cm, find the area of the arranged white tiles.

Answer:
In the given figure side of total arrangement ABCD = 150 cm.
Width of white tile = 5 cm

Area of white tile path EFGH = 150 × 5 = 750 sq.cm
Area of white tile path MNOP = 150 × 5 = 750 sq.cm
Area of common path IJKL = 5 × 5 = 25 sq.cm
Area of total white tile path
= Area of EFGH + Area of MNOP – Area of IJKL
= 750 + 750 – 25
∴ The area of the arranged white tiles
= 1475 sq.cm.

Question 2.
2 m wide path is present the outer side of the square grass land of side 80 m. Find the area of path and total expenditure of the path flooring with bricks, if the cost of flooring with bricks per sq.m is ₹ 200.

Answer:
Given side of the grass land = 80 m
Width of the path = 2 m
Path laid out side the grass land.
So, outer side of grassland
= side + 2 × width
= 80 + 2 × 2 = 84 m

ABCD = 80 × 80 – 6400 sq.m
Area of the grass land with path
EFGH = 84 × 84 = 7056 sq.m
Area of the path = Area of EFGH – Area of ABCD
= 7056 – 6400 = 656 sq.m.
Cost of flooring per sq.m = ₹ 200
∴ Cost of flooring per 656 sq.m
= 656 × 200 = ₹ 1,31.200

Let’s Do Activity [Page No: 88]

Prepare two coloured rectangles one is red colour of length 25 cm, breadth 20 cm. Another one is green of length 20 cm, breadth 15 cm and place smaller rectangle middle to the bigger rectangle, so, that 2.5 cm red colour formed outside to green colour rectangle. Find the area of red colour path.

Answer:
Area of red rectangle = 25 × 20 = 500 cm²
Area of green rectangle = 20 × 15 = 300 cm²
∴ Area of the coloured path = (Area of red rectangle) – (Area of green rectangle)
= 500 – 300
= 200 cm²

[Page No: 89]

In the part of ‘Nadu-Nedu’ programme, Head master decided to arrange cir¬cular shape flower bed with radius 7 m in the premises of his school. How many flower plants needed if it takes 4 plants per sq. m. ?
Answer:
Here, length of the rectangle (l)
= Half of the perimeter of circle
= πr
breadth of the rectangle (b)
= radius of circle
= r
We know that the area of the rectangle = l × b = πr × r (∵ l = πr, b = r)
= πr² – Area of the circle.
So, the area of the circle A = πr²
Now we solve above problem,
where ‘r’ = 7m
Area of flower bed = πr²
= \(\frac{22}{7}\) × 7 × 7
= 22 × 7
= 154 sq.m
Number of plants per sq.m = 4
Number of plants for 154 sq.m
= 154 × 4 = 616 plants.
So, Headmaster needs 616 plants for the flower bed.

Check Your Progress [Page No. 91]

Question 1.
The circumference of the circle shaped rangoli sheet is 88 cm. Find the radius of the circle and the area of the circle.
Answer:
Given circumference of the rangoli
2πr = 88 cm
⇒ 2 × \(\frac{22}{7}\) × r = 88
⇒ 44 r = 88 × 7

⇒ r = 14 cm
∴ Radius of rangoli r = 14 cm
Area of the rangoli = πr²
= \(\frac{22}{7}\) × 14²

∴ Area of the circle = 616 sq.cm

Question 2.
Calculate the areas of circles shown in the figure.
(i)

Answer:
From the figure radius r = 7 cm
Area of the circle = πr²
= \(\frac{22}{7}\) × 7²

= 154 sq.cm

(ii)

Answer:
From the figure diameter of circle d = 28 cm
Radius of circle r = \(\frac{\mathrm{d}}{2}\) = \(\frac{28}{2}\) = 14 cm.
Area of the circle = πr²
= \(\frac{22}{7}\) × 14²

∴ Area of the circle = 616 sq.cm

(iii)

Answer:
From the figure radius of circle (r) = 21 cm
Area of the circle = πr²
= \(\frac{22}{7}\) × 21²

∴ Area of the circle = 1386 sq.cm

Let’s Explore [Page No: 93]

Radius of circular shaped grass land is 11 m. A goat is tied with a rope of length 4 m at the centre, then find the area of grass land that the goat cannot graze.
Answer:
Given the radius of grass land = 11 m
Area of grass land = πr²
= \(\frac{22}{7}\) × 11 × 11
= \(\frac{2662}{7}\)
∴ Area of grass land = 380.29 sq.m.
Radius of the small circle = length of rope to which goat tied = 4m
Area of land that the goat can graze = πr²
= \(\frac{22}{7}\) × 4 × 4
= \(\frac{352}{7}\) = 50.29 sq.m
∴ Area of land that the goat cannot graze
= Area of grass land – Area of grazed
= 380.29 – 50.29
= 330 sq.m.

Reasoning Corner (Non-Verbal) [Page No: 96 ]

Question 1.
Embedded Figures :
The problem figure (X) is given, answer figures as (a), (b), (c) & (d) given besides. The problem figures as a hidden figure of answer figure and one should identify that figure.

Answer:

Examples

Question 1.
Find the area of the given triangles.

Answer:
(i) In ∆PQR, Base (QR) = 6 cm,
Height (PS) = 4 cm

Area of ∆PQR = \(\frac{1}{2}\) x base x height
= \(\frac{1}{2}\) × QR × PS
= \(\frac{1}{2}\) × 6 cm × 4 cm
= 12 sq.cm.

(ii) In ∆LMN, Base (MN) = 3 cm,
Height (LO) = 2 cm
Area of ∆LMN = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × MN × LO
= \(\frac{1}{2}\) × 3 cm × 2 cm
= 3 sq.cm

Question 2.
The area of the ∆ XYZ is 12 sq.cm and the height XL is 3 cm, then find base YZ.
Answer:

In ∆ XYZ, Base = YZ,
Height (XL) = 3 cm,
Area of the XYZ = 12 sq.cm.
Area of the ∆XYZ = \(\frac{1}{2}\) × base × height
= \(\frac{1}{2}\) × YZ × XL
⇒ 12 = \(\frac{1}{2}\) × YZ × 3
⇒ YZ = 12 × \(\frac{2}{3}\)
So, YZ = 8 cm

Question 3.
In ∆ ABC, AC = 8 cm, BC = 4 cm and AE = 5 cm.
Find (i) the area of the AABC (ii) BD.
Answer:

(i) In ∆ ABC, Base (BC) = 4 cm,
Height (AE) = 5 cm.
Area of the ∆ ABC = \(\frac{1}{2}\) × base × height
= = \(\frac{1}{2}\) × 4 × 5
10 sq.cm.

(ii) In ABAC, Base (AC) = 8 cm,
height (BD) = ?,
Area of ∆ BAC = 10 sq.cm
Area of the ∆ BAC = \(\frac{1}{2}\) × base × height
i. e. 10 = \(\frac{1}{2}\) × 8 × BD
BD = 10 × \(\frac{2}{8}\) = \(\frac{10}{4}\) = \(\frac{5}{2}\) = 2.5
So, height (BD) = 2.5 cm

Question 4.
Calculate the area of the given right-angled triangle with sides having right angle are 6 cm, 6 cm.
Answer:

Method – 1: Sides having right angles are 6 cm, 6 cm.
Sides forming the right angle a = 6 cm, b = 6 cm
Area of a right-angled triangle
= \(\frac{1}{2}\) × Product of sides forming the right angle
= \(\frac{1}{2}\) × a × b
= \(\frac{1}{2}\) × 6 × 6
= 6 × 3 = 18 sq.cm.

(or)

Method – 2: Observe the grid carefully, the right angled triangle covering half of the area of Square.
Area of a right-angled triangle
= \(\frac{1}{2}\) × Area of square
= \(\frac{1}{2}\) × 6 × 6 = 18 sq.cm.

Question 5.
Find the area of the triangle shaped lawn whose base and heights are 12m.,7m. respectively. Find the total cost of laying lawn, if cost of grass is ₹ 300 per Sq. m.

Answer:
Base of the triangle shaped lawn = 12 m.
Height = 7 m.
Area of triangle shaped lawn 1
= \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × 12 × 7
= 6 × 7 = 42 Sq.m
Cost of grass for laying in lawn per 1 Sq.m = ₹ 300
Cost of grass for laying in lawn for 42 Sq.m = ₹ 300 × 42 = ₹ 12,600

Question 6.
The length and breadth of a rectangular field is 65 m, 30 m respectively. A path of width 2.5 m is made around the park outside. Find the area of the path.

Answer:
In the fig, ABCD is a rectangular field and shaded area shows 2.5m. wide path. EFGH is a rectangle (field with path)
Length of ABCD (AB) = 65m.,
Breadth of ABCD (AD) = 30m.
Width of path = 2.5 m.
Area of path = Area of outer rectangular- field EFGH – Area of inner rectangular field ABCD
Length of rectangle EFGH = EF
= Length of field (AB) + (2 × Width of path)
= 65m + (2 × 2.5 m)
= 65m + 5m = 70m
Breadth of rectangle EFGH = EH
= Breadth of field (AD) + (2 × Width of path)
= 30 m + (2 × 2.5m)
= 30 m + 5m = 35m
Area of outer rectangle
EFGH = length × breadth
70m × 35m = 2450 Sq.m
Area of inner rectangle
ABCD = length × breadth
= 65 m × 30 m
= 1950 Sq.m
Area of path = Area of outer rectangle EFGH – Area of inner rectangle ABCD
= 2450 Sq.m – 1950 Sq.m
= 500 Sq.m.

Question 7.
A square shaped swimming pool of side 70m. It has 5m width path is present the outer side of the boundary. Find the area of this path. Find the expenditure of covering that path with tiles at the rate of ₹ 150 per sq.m.
Answer:
WXYZ shows a square shaped swimming pool of side 70m. 5m wide path to the outer side of the swimming pool.,
Area of the path = Area of swimming pool PQRS with path – Area of swimming pool WXYZ
PS = Side of swimming pool (WZ) + (2 × breadth of path)
= 70m + (2 × 5m)
= 70m + 10m = 80m

Area of swimming pool with path
PQRS = (Side)² = (80 m)²
= 6400 Sq.m.
Area of swimming pool
WXYZ = (Side)² = (70 m)²
= 4900 Sq.m.
Area of path = Area of swimming pool PQRS with path – Area of swimming pool WXYZ
= 6400 – 4900 = 1500 Sq.m.
If cost of covering tiles per 1 Sq.m.
= ₹ 150
Cost of covering tiles per 1500 Sq.m.
= ₹ 150 × 1500
= ₹ 2,25,000

Question 8.
The length of rectangular grass land is 55 m and breadth is 45 m in the centre of the grass land two paths of 3 m wide one parallel to the length and another parallel to breadth are situated in such a way that they intersect each other. Find the area of the path.
Answer:
In fig. ABCD is rectangular grass land.
Length of ABCD = 55 m
Breadth of ABCD = 45 m
Width of path = 3 m
Area of path EFGH = Length × Width
= 55 × 3
= 165 Sq.m

Area of path MNOP
= Breadth × Width
= 45 × 3 = 135 Sq.m.
Area of common path IJKL (situated on both paths)
= Width × Width = 3 m × 3 m = 9 Sq.m.
Area of square IJKL i.e., 9 Sq.m is included in both the paths. So we subtract one time.
Area of total path = Area of path EFGH + Area of path MNOP – Area of IJKL
= (165 Sq.m. + 135 Sq.m. – 9 Sq.m.)
= (300 9) Sq.m.
= 291 Sq.m.

Question 9.
Find the area of the rangoli of it’s radius is 21 cm.
(Use π = \(\frac{22}{7}\))

Answer:
Radius of rangoli (r) = 21 cm

Area of circular shaped rangoli = πr²
= \(\frac{22}{7}\) × 21 cm × 21 cm
= 1386 sq. cm
∴ Area of the circle = 1386 sq. cm

Question 10.
Find the surface area of a circular shaped pool whose diameter is 28 m
(Use π = \(\frac{22}{7}\)).

Answer:
The diameter of circular shaped pool (d) = 28 m
Radius (r) = \(\frac{28}{2}\) m = 14 m
Area of a circular shaped pool = πr²
= \(\frac{22}{7}\) × (14m)²
= \(\frac{22}{7}\) × 14 m × 14 m
= 22 × 2 × 14 sq.m.
= 616 sq.m.

Question 11.
In a circular shaped park inner portion is given for kids to play and outer portion is given for walking to elders. If outer radius is 35 m, width of walking track is 14 m, then find the area of walking path.

Answer:
Outer radius of park (R) = 35 m,
Width of walking track = 14 m.
Radius of playground (r) = R – w = 35 – 14 = 21m.
Area of walking track = Area of the park – Area of playing ground
= πR² – πr²
= \(\frac{22}{7}\) × 35² – \(\frac{22}{7}\) × 21²
= \(\frac{22}{7}\) (35² – 21²)
= \(\frac{22}{7}\) (1225 – 441)
= \(\frac{22}{7}\) × 784
= 22 × 112 = 2464 sq.m

Question 12.
A water fountain is in circular shaped whose radius is 10 m. Its inner portion of radius 3m is arranged for fountain and remaining part is cemented. Find the area of that cemented part and total cost of cementing if cost of cementing is ₹ 200 per sq.m.
Answer:
Radius for total water fountain (R) = 10m
Radius of fountain arranged portion (r) = 3 m
Area of cemented part = Area of total water fountain – Area of fountain arranged portion
= πR² – πr²

= \(\frac{22}{7}\) × (10)² – \(\frac{22}{7}\) × (3)²
= \(\frac{22}{7}\) [(10)² – (3)²)
= \(\frac{22}{7}\) (100 – 9) sq.m.
= \(\frac{22}{7}\) × 91 sq.m.
= 22 × 13
= 286 sq.m
Given the cost of cementing per sq.m = ₹ 200
∴ Total cost of cementing = 286 × 200 = ₹ 57,200

Practice Questions [Page No: 96]

In each question below, you are given a figure (X) followed by four figures (a), (b), (c) and (d) such that (X) is embedded in one of them. Trace out the correct alternative.

Question 1.

Answer:
a

Question 2.

Answer:
c

Question 3.

Answer:
b

Question 4.

Answer:
a

Question 5.

Answer:
c

Question 6.

Answer:
a

Question 7.

Answer:
b

Question 8.

Answer:
b

Question 9.

Answer:
d

Question 10.

Answer:
b

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

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

Question 1.

Answer:

Question 2.
Write the formulae of the following:
(i) Area of the rectangular path ……………….. .
Answer:
Area of the outer rectangle – Area of the inner rectangle.

(ii) Area of the square path ……………….. .
Answer:
Area of the outer square – Area of the – inner square.

(iii)
Area of the circular path ……………….. .
Answer:
Area of the outer circle – Area of the inner circle.
π(R + r)(R – r)

Question 3.
Find the area of a triangle if its base is 18 cm, height is 13 cm.

Answer:
Given base of the triangle b = 18 cm
Height of the triangle h = 13 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

∴ Area of the triangle = 117 sq.cm.

Question 4.
In a park a rectangular path given outside for walking around the grassland of length 28 m., and breadth 20m. If the width of the walking path is 2m., find the area of walking path.

Answer:
Given inner length of the park l = 28 m
Breadth b = 20 m

Area of the inner rectangle = l × b
= 28 × 20
= 560 sq.m
Path laid outside around the grassland.
Width of the path = 2 m
So, outer length of rectangle
l = Inner length + 2 × Width
= 28 + 2 × 2 = 28 + 4 = 32 m
Breadth b = 20 + 2 × 2
= 20 + 4 = 24m
Area of the outer rectangle = l × b = 32 × 24 = 768 sq.m
∴ Area of the walking path = outer area – inner area
= 768 – 560 = 208 sq.m.

Question 5.
The elevation of building have square – shaped window of a side 150 cm. Around this window tiles arranged with a width of 70 cm. Find the area of tiles and total cost of tiles arranged, if cost per sq.cm. is ₹ 5.
Answer:
Side of square window = 150 cm
Area of square window = 150 × 150 A
= 22500 cm²
Area of square formed by arranging tiles around the window
= (150 + 70)² = 220² = 48400 cm²
Area of tiles laid = (Area of outer square) – Area of inner square
= 48400 – 22500 = 25900
∴ Total cost by laying tiles around the window
@ ₹5 per cm² is = 259 × 5 = ₹ 1295.00

Question 6.
Two cross roads, each of width 4 m, run at right angles through the centre of a rectangular park of length 60 m and breadth 40 m and parallel to its sides. Find the area of the roads. Also find the cost of constructing the roads at the rate of ₹ 100 sq.m.

Answer:
Length of the rectangular park = 60 m
Breadth of the rectangular park = 40 m
Length of the path along length of the park = 60 m
Width of the path along length of the park = 40 m
Area of the path along length of the park = 60 × 4 = 240 m²
Similarly,
Length of the path along breadth of the park = 40 m
Width = 4m
∴ Area of the path along its breadth = 40 × 4 = 160 m²
Hence, area of the paths = Area of the path along length + Area of the path along breadth – Area of the intersecting square
= 240 + 160 – 4 × 4
= 400 – 16 = 384
∴ Cost of construction at the rate of
₹ 100 per sq.m = 384 × 100 = ₹ 38400

Question 7.
Find the area of the circular-shaped photo frame whose radius is 28 cm. If the cost of decoration is ₹ 3 per sq. cm., find the total cost of decoration.
Answer:
Radius of the circle = 28 cm
Area = πr²
= \(\frac{22}{7}\) × 28 × 28 = 2464 cm²
Circumference of the photo frame 2πr
= 2 × \(\frac{22}{7}\) × 28 = 176 cm
∴ Cost of decorative piece at the rate of ₹ 3 per cm is 176 × 3 = ₹ 528

Question 8.
Find the path area of circular-shaped grassland of radius 42 m where the width of the path is 7 m. around and outside the circle. Find the area of path and total cost of flooring, if the cost of flooring is ₹ 150 per sq.m.

Answer:
Radius of circular floor = 42 m
Width around it = 7 m
Radius of outer circle = 42 + 7 = 49 m
Area of outside path = Area of outer circle – Area of inner cirlce
Area of inner circle = πr² = \(\frac{22}{7}\) × 42 × 42
Area of the outer circle = πR² = \(\frac{22}{7}\) × 49 × 49
Area of floor = \(\frac{22}{7}\) (49 × 49 – 42× 42)
= \(\frac{22}{7}\) (2401 – 1764)
= \(\frac{22}{7}\) × 637 = 2002 m²
∴ Cost of flooring at the rate of ₹ 150 per sq.m is 2002 × 150 = ₹ 300300

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 played around outside the park. Find the area of the 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 the round table upper surface in the 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 the cost of grass per sq. m is ₹ 30, then find the total cost of laying grass on the 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 11.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 11th Lesson Area of Plane Figures Ex 11.1

Question 1.
Calculate the area of the following triangles given below:
(a)

Answer:
From the figure, base = 6 cm
Height = 3 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 9 sq.cm

(b)

Answer:
From the figure, base = 4.2 cm
Height = 3.2 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 6.72 sq.cm

(a)

Answer:
From the figure, base = 3 cm
Height = 4 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 6 sq.cm

(a)

Answer:
From the figure, base = 5 cm
Height = 2 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

= 5 sq.cm

Question 2.
Find the area of triangle with base 3.8 cm and height 4.6 cm.
Answer:
Given base of the triangle b = 3.8 cm
Height of the triangle h = 4.6 cm
Area of the triangle = \(\frac{1}{2}\) ∙ b ∙ h

∴ Area of the triangle = 8.74 sq.cm.

Question 3.
The surface area of a triangular shape window is 24 sq.m, and height is 6m., then find the base of the triangle, if the cost of glass fitting per sqm is ₹ 250, then find total cost of glass fitting for window.

Answer:
Let,
base of the triangular window = b m
Height of the triangular window h= 6m
Area of triangular window = \(\frac{1}{2}\) ∙ b ∙ h

= 24 sq.cm
⇒ 3b = 24
⇒ b = \(\frac{24}{3}\)
= 8
∴ Base of the triangular window = 8m
Cost of glass fitting per sq.m = ₹ 250
Cost of glass fitting per 24 sq.m = 250 × 24
∴ Total cost of glass fitting for window = ₹ 6000/-

Question 4.
A traffic signal plate in the shape of triangle is having base 20 cm, height 15 cm. Find the area of the triangle. If the cost of painting is ₹ 2 per Sq.cm, then find the total cost for painting the signal board on one side.

Answer:
Given the base of the signal plate b = 20 cm
Height of the signal plate h = 15 cm
Area of the signal plate = \(\frac{1}{2}\) ∙ b ∙ h

= 150 sq.cm
∴ Area of the signal plate = 150 sq.cm
Cost of painting per sq.cm = ₹ 2
Cost of painting for 150 sq.cm = 150 × 2
∴ Total cost for painting the signal board = ₹ 300

Question 5.
find the area of triangular shaped wall painting whose base is 24m, height is 38m. Find the area of the triangle, if the cost of painting is ₹ 50 per Sq. m. and also find the total cost of painting.

Answer:
Given base of the triangular wall painting = 24 m Height = 38 m
Area of wall painting = \(\frac{1}{2}\) ∙ b ∙ h

= 456 sq.m.
Cost of painting per 1 sq.m = ₹ 50
Cost of painting for 456 sq.m = 456 × 50
∴ Total cost for painting per 456 sq.m = ₹ 22,800

Question 6.
The area of triangle-shaped elevation of a house is 195 Sq.m. Its base is 26m. Find the height of elevation. Find the total cost of cementing, if the cost of cementing per Sq.m, is ₹ 250.

Answer:
Given, base of triangle shaped elevation of house b = 26 m height h = ?
Area of the triangular elevation = 195 sq.m
= \(\frac{1}{2}\) ∙ b ∙ h

⇒ 13 h = 195

Height of triangular elevation = 15 m
Cost of cementing per sq.m = ₹ 250
Cost of cementing per 195 sq.m
= 195 × 250
∴ Total cost of cementing of triangular elevation = ₹ 48,750

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 the angles 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, the total number of triangles = 48.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Ex 10.3 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Ex 10.3

Question 1.
Construct ∆DEF with measurements ∠D = 60°, ∠F = 50° and DF = 4 cm.
Answer:
Given measurements of ∆DEF are ∠D = 60°, ∠F = 50° and DF = 4 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with DF = 4 cm.
3. Draw a ray DX such that ∠FDX = 60°.
4. Draw another ray FY such that ∠DFY = 50° which intersects the previous ray at E. Thus, required ∆DEF is constructed with the given measurements.

Question 2.
Construct the triangle with the measurements XY = 7.2 cm, ∠Y = 30° and ∠Z = 100°.
Answer:
Given measurements of ∆XYZ are XY = 7.2 cm, ∠Y = 30° and ∠Z = 100°.
We know stun of angles of triangle is 180°
So, ∠X + ∠Y + ∠Z = 180°
⇒ ∠X + 30° + 100° = 180°
⇒ ∠X= 180°-130°
∴ ∠X = 50°

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with XY = 7.2 cm.
3. Draw a ray \(\overrightarrow{\mathrm{XP}}\) such that ∠YXP = 50°
4. Draw another ray \(\overrightarrow{\mathrm{YQ}}\) such that ∠XYQ = 30° which intersects the previous ray at Z. Thus, required ∆XYZ is constructed with the given measurements.

Question 3.
Construct ∆PQR with the measurements ∠P = ∠Q = 60° and PQ = 7 cm.
Answer:
Given measurements of ∆PQR are ∠P = ∠Q = 60° and PQ = 7 cm.

Steps of Construction:

1. Draw a rough sketch of the triangle and label it with given measurements.
2. Draw a line segment with PQ = 7 cm.
3. Draw a ray PX such that ∠QPX — 60°
4. Draw another ray QY such that ∠PQY = 60° which intersects the previous ray at R. Thus, required ∆PQR is constructed with the given measurements.

SCERT AP 7th Class Maths Solutions Pdf Chapter 10 Construction of Triangles Ex 10.2 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Ex 10.2

Question 1.
Construct ∆ABC with measurements AB = 4.5 cm, BC = 6 cm and ∠B=75°
Answer:
Given measurements of ∆ABC are AB = 4.5 cm, BC = 6 cm and ∠B = 75°.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with BC = 6 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠CBX = 15°.
4. Draw an arc with centre B and radius 4.5 cm, to intersect \(\overrightarrow{\mathrm{BX}}\) at point A.
   Thus, required ∆ABC is constructed with the given measurements.

Question 2.
Construct an isosceles triangle with measurements DE = 7 cm, EF = 7 cm and ∠E = 60°
Answer:
Given measurements of ∆DEF are DE = 7 cm, EF = 7 cm and ∠E = 60°.

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with DE = 7 cm.
3. Draw a ray \(\overrightarrow{\mathrm{EX}}\) such that ∠DEX = 60°.
4. Draw an arc with centre E and radius 7 cm to intersect \(\overrightarrow{\mathrm{EX}}\) at point F.
5. Join DF.
   Hence, required ∆DEF is constructed with the given measurements.

Question 3.
Draw a triangle with,measurements ∠B = 50°, AB = 3 cm and AC = 4 cm.
Answer:
Given measurements of a triangle are ∠B = 50°, AB = 3 cm and AC = 4cm..

Steps of Construction:

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AB = 3 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠ABX = 50°.
4. Draw an arc with centre A and radius 4 cm to intersect \(\overrightarrow{\mathrm{BX}}\) at point C.
5. Join AC.
   Hence, required ∆ABC is constructed with the given measurements.

Question 4.
Construct a right angled triangle in which, XY = 5 cm, XZ = 6 cm and right tingle at X.
Answer:
Given measurements of ∆XYZ are XY = 5 cm, XZ = 6 cm and ZX = 90° (∠YXZ = 90°)

Steps of Construction:

1. Draw a rough sketch of triangle arid label it with given measurements.
2. Draw a line segment with XY = 5 cm.
3. Draw a ray \(\overrightarrow{\mathrm{XP}}\) such that ∠YXP = 90°.
4. Draw an arc with centre X and radius 6 cm to intersect XP at point Z.
5. Join YZ.
   Hence, required triangle ∆XYZ is constructed with the given measurements.

Question 5.
Construct a right angled with measurements ABC, ∠B = 90°, AB = 8 cm and AC = 10 cm.
Answer:
Given measurements of ∆ABC are ∠B = 90°, AB = 8 cm and AC = 10 cm.

Steps of Construction :

1. Draw a rough sketch of triangle and label it with given measurements.
2. Draw a line segment with AB = 8 cm.
3. Draw a ray \(\overrightarrow{\mathrm{BX}}\) such that ∠ABX = 90°.
4. Draw an arc with centre A and radius 10 cm to intersect \(\overrightarrow{\mathrm{BX}}\) at point C.
5. Join AC.
   Thus, the required ∆ABC is constructed with the given measurements.

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

AP State Syllabus 7th Class Maths Solutions 10th Lesson Construction of Triangles Review Exercise

Question 1.
Draw the angles 70° and 110° by using a protractor,
(i) ∠AOB = 70°
Answer:
∠AOB = 70°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\)
2. At ‘O’ erect a ray OB such that ∠AOB = 70°

(ii) ∠AOB = 110°
Answer:
∠AOB = 110°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. At ‘O’ erect a ray \(\overrightarrow{\mathrm{OB}}\) such that ∠AOB = 110°

Question 2.
Construct the angles 60° and 120° by using ruler and compass.
(i) ∠AOB = 60°
Answer:
∠AOB = 60°

1. Draw a ray \(\overrightarrow{\mathrm{OA}}\).
2. Draw an arc with any radius (convenient) taking ‘O’ as a centre and cutting \(\overrightarrow{\mathrm{OA}}\) at X.
3. Draw an arc with the same radius – taking X as a centre and meeting the previous arc at Y.
4. Join OY and produce it to B.
5. ∠AOB is formed.
6. ∠AOB = 60°

(ii) ∠ADI = 120°
Answer:
∠ADI = 120°

1. Draw a ray \(\overrightarrow{\mathrm{DA}}\).
2. Draw an arc with any radius taking ‘D’ as centre intersecting \(\overrightarrow{\mathrm{DA}}\) at X.
3. Now draw two successive arcs from X intersecting the previous arc at Y and Z.
4. Join D and Z and produce DZ to I.
5. Now the angle formed ∠ADI is equal to 120°.

Question 3.
Draw a line segment PQ = 4.5 cm and construct its perpendicular bisector by using ruler and compass.
Answer:

∠PMY = ∠QMY = 90°
So, PQ ⊥ XY

1. Draw a line segment PQ of length 4.5 cm.
2. Draw two arcs with centre P and radius > \(\frac{1}{2}\)PQ on both sides of PQ.
3. Repeat step 2 with centre ‘Q’ intersecting the previous arc at X and Y.
4. Join X and Y and produce it on either sides to form \(\overrightarrow{\mathrm{XY}}\).
5. XY is the perpendicular bisector of PQ.

Question 4.
Construct ∠DEF = 60° and its angular bisector by using ruler and compass.
Answer:

∠DEF = 60°
∠DEK = ∠FEK = \(\frac{\angle \mathrm{DEF}}{2}=\frac{60^{\circ}}{2}\) = 30°

1. Construct ∠DEF = 60°
2. Draw an arc with a convenient radius taking E as centre intersecting ED at X and EF at Y.
3. Draw two arcs with the same radius taking X and Y as centres intersecting at Z.
4. Join E, Z and produce it to K.
5. ∠DEK = ∠KEF = 30°.

Question 5.
Construct an angle of 90° without using a protractor.
Answer:
∠SRI = 90°

1. Draw a ray \(\overrightarrow{R S}\).
2. Draw an arc of convenient radius taking R as a centre and intersecting \(\overrightarrow{R S}\) at X.
3. With the same radius draw two successive arcs from X intersecting the previous arc at Y & Z.
4. Draw bisector to \(\widehat{\mathrm{YZ}}\) intersecting at I.
5. Join R, I.

SCERT AP 7th Class Maths Solutions Pdf Chapter 9 Algebraic Expressions InText Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 9th Lesson Algebraic Expressions InText Questions

Check your progress [Page No. 49]

Question 1.
How many number of terms are there in each of the following expressions ?
(i) 5x² + 3y + 7
Answer:
Given expression is 5x² + 3y + 7 Number of terms = 3

(ii) 5x²y + 3
Answer:
Given expression is 5x²y + 3
Number of terms = 2

(iii) 3x²y
Answer:
Given expression is 3x²y
Number of terms = 1

(iv) 5x – 7
Answer:
Given expression is 5x – 7
Number of terms = 2

(v) 7x³ – 2x
Answer:
Given expression is 7x³ – 2x
Number of terms = 2

Question 2.
Write numeric and algebraic terms in ‘ the above expressions.
(i) 5x² + 3y + 7
Answer:
Given expression is 5x² + 3y + 7
Numerical terms = 7
Algebraic terms = 5x², 3y

(ii) 5x²y + 3
Answer:
Given expression is 5x²y + 3
Numerical terms = 3
Algebraic terms = 5x²y

(iii) 3x²y
Answer:
Given expression is 3x²y
Numerical terms = No
Algebraic terms = 3x²y

(iv) 5x – 7
Answer:
Given expression is 5x – 7
Numerical terms = – 7
Algebraic terms = 5x

(v) 7x³ – 2x
Answer:
Given expression is 7x³ – 2x
Numerical terms = No
Algebraic terms = 7x³, – 2x

Question 3.
Write the terms in the following expressions.
– 3x + 4, 2x – 3y, \(\frac{4}{3}\)a² + \(\frac{5}{2}\)b,
1.2ab + 5.1b – 3.2a
Answer:

Let’s Explore (Page No: 50)

Question 1.
Identify the terms which contain m² and write the coefficients of m².
(i) mn² + m²n
Answer:
Given expression is mn² + m²n

(ii) 7m² – 5m – 3
Answer:
Given expression is 7m² – 5m – 3

(iii) 11 – 5m² + n + 8 mn
Answer:
Given expression is 11 – 5m² + n + 8 mn

Check Your Progress [Page No. 52]

Question 1.
Write like terms from the following :
– xy², – 4yx, 8x, 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x
Answer:
Given – xy², – 4yx, 8x, 2xy², 7y, – 11x², – 100x, – 11yx, 20x²y, – 6x², y, 2xy, 3x

• – xy², 2xy² are like terms because they contain same algebraic factor xy².
• – 4yx, – 11yx, 2xy are like terms because they contain same algebraic factor xy.
• 8x, – 100x, 3x are like terms because they contain same algebraic factor x.
• 7y, y are like terms because they contain same algebraic factor y.
• – 11x², – 6x². are like terms because they contain same algebraic factor x².

Question 2.
Write 3 like terms for
(i) 3x²y
Answer:
Like terms of 3x²y are – 8x²y, \(\frac{5}{3}\) x²y, 2.9 x²y

(ii) – ab²c
Answer:
Like terms of – ab²c are 3ab²c, 5.8 ab²c, \(\frac{-1}{4}\) ab²c.

Let’s Explore [page No: 52]

Question 1.
Jasmin says that 3xyz is a trinomial. Is she right ? Give reason.
Answer:
Given expression 3xyz.
In this expression only one term is there. So, it is a monomial only. But, not trinomial.
So, Jasmin is wrong.

Question 2.
Give two examples each for Monomial and Binomial algebraic expression.
Answer:

Check Your progress [Page No. 54]

Question 1.
Add the following like terms.
(i) 12ab, 9ab, ab
Answer:
Sum of 12ab, 9ab, ab
= 12ab + 9ab + ab
= (12 + 9 + 1) ab
= 22 ab.

(ii) 10x², – 3x², 5x²
Answer:
Sum of 10x², – 3x², 5x²
= 10x² + (- 3x²) + 5x²
= 10x² – 3x² + 5x²
= (10 – 3 + 5) x²
= 12x²

(iii) – y², 5y², 8y², – 14y²
Answer:
Sum of – y², 5y², 8y², – 14y²
= – y² + 5y² + 8y² + (- 14y²)
= – 1y² + 5y² + 8y² – 14y²
= (- 1 + 5 + 8 – 14) y²
= – 2y²

(iv) 10mn, 6mn, – 2mn, – 7mn
Answer:
Sum of 10mn, 6mn, – 2mn, – 7mn
= 10mn + 6mn + (- 2mn) + (- 7mn)
= 10mn + 6mn – 2mn – 7mn
= (10 + 6 – 2 – 7) mn
= 7mn

Let’s Think [Page No: 54]

Reshma simplified the expression
4p + 6p + p like this.
4p + 6p + p = 10p. Is she correct ?
Answer:
4p + 6p + p = (4 + 6 + 1)p
= 11p ≠ 10p
So, Reshma simplified is wrong.

Check Your Progress [Page No: 54]

Question 1.
Write the standard form of the following expressions :
(i) – 5l + 2l² + 4
Answer:
2l² – 5l + A

(ii) 4b² + 5 – 3b
Answer:
4b² – 3b + 5

(iii) z – y – x
Answer:
– x – y + z.

Check Your Progress [Page No. 56]

Question 1.
Add the following expressions in both Row and Column methods:
(i) x – 2y, 3x + 4y
Answer:

(ii) 4m² – 7n² + 5mn, 3n² + 5m² – 2mn
Answer:

(iii) 3a – 4b, 5c – 7a + 2b
Answer:

Let’s Explore [Page No. 56]

Think of at least two situations in each of which you need to form two algebraic expressions and add them.

(i) Aditya and Madhu went to a store. Aditya bought 6 books and Madhu bought 2 books. All the books are same cost. How much money did both spend ₹
Answer:
Let the cost of each book ₹a.
Number of books Aditya bought = 6
Cost of 6 books = 6 × a = ₹ 6a
Number of books Madhu bought = 2
Cost of 2 books = 2 × a = ₹ 2a
Money spend on books bought by Aditya and Madhu
= 6a + 2a
= (6 + 2)a
= ₹ 8a

(ii) Sreeja and Sreekari went to icecream parlour. Sreeja bought Vineela icecreams 2 and Sreekari bought Butter Scotch 3. Cost of icecreams are different. How much money they gave to the shop keeper ₹
Answer:
Let the cost of Vineela is ₹x and cost of Butter Scotch is ₹y.
Number of Vineela icecreams Sreeja bought = 2
Cost of 2 iceqreams = 2 × x = ₹ 2x
Number of Butter Scotch icecreams Sreekari bought = 3
Cost of 3 icecreams = 3 × y = ₹ 3y
Money given to shopkeeper = 2x + 3y
= ₹(2x + 3y)

Check Your Progress [Page No. 57]

Question 1.
Subtract the first term from second term :
(i) 2xy, 7xy
Answer:
7xy – 2xy
= (7 – 2) xy
= 5xy

(ii) 4a², 10a²
Answer:
10a² – 4a² = (10 – 4)a² = 6a²

(iii) 15p, 3p
Answer:
3p – 15p = (3 – 15)p = – 12p

(iv) 6m²n, – 20m²n
Answer:
– 20 m²n – 6m²n = (- 20 – 6) m²n
= – 26m²n

(v) a²b², – a²b²
Answer:
– a²b² – a²b² = (- 1 – 1)a²b² = – 2a²b²

Let’s Explore [Page No. 58]

Add and subtract the following expressions in both Horizontal and Vertical method x – 4y + z, 6z – 2x + 2y
Answer:
Addition

Subtraction:

Let’s Explore [Page No. 60]

Question 1.
Write an expression whose value is – 15 when x = – 5.
Answer:
Given x = – 5 and value = – 15
Value = – 15 ,
= 3 × – 5 ,
= – 3 × x (∵ x = – 5)
∴ Expression = 3x

Question 2.
Write an expression whose value is 15 when x = 2.
Answer:
Given x = 2 and value = 15
Value = 15
= \(\frac{30}{2}\)
= \(\frac{1}{2}\) × 15 × 2
= \(\frac{1}{2}\) × 15 × x (∵ x = 2)
∴ Expression = \(\frac{15 x}{2}\)

Lets Think [Page No. 60]

While finding the value of an algebraic expression 5x when x = – 2, two stu¬dents solved as follows:

Can you guess who has done it correctly? Justify!
Answer:
Given expression is 5x when x = – 2
5x = 5(- 2) = – 10
Chaitanya is correct.
Here we have to multiply 5 and,- 2.
But Reeta subtracted.
So, Reeta is wrong.

Examples

Question 1.
How many number of terms are there in each of the following expressions ?
(i) a + b
Answer:
a + b ……….. 2 terms

(ii) 3t²
Answer:
3t² ……….. 1 term

(iii) 9p³ + 10q – 15
Answer:
9p³ + 10q – 15 ……….. 3 terms

(iv) \(\frac{5 \mathrm{~m}}{3 \mathrm{n}}\)
Answer:
\(\frac{5 \mathrm{~m}}{3 \mathrm{n}}\) ……….. 1 term

(v) 4x + 5y – 3z – 1
Answer:
4x + 5y – 3z – 1 ……….. 4 terms

Question 2.
In the following expressions, write the number of terms and identify numeri¬cal and algebraic expressions in them.
(i) 8p
Answer:
8p = 1 term – Algebraic expression

(ii) 5c + s – 7
Answer:
5c + s – 7 = 3 terms – Algebraic expression

(iii) – 6
Answer:
– 6 = 1 term – Numerical expression

(iv) (2 + 1) – 6
Answer:
(2 + 1) – 6 = 2 terms – Numerical expression

(v) 9t + 15
Answer:
9t + 15 = 2 terms – Algebraic expression

Question 3.
Write the coefficients of
(i) p in 8pq
Answer:
8pq = p(8q)
so, coefficient of p in 8pq is 8q.

(ii) x in \(\frac{x y}{3}\)
Answer:
\(\frac{x y}{3}\) = x\(\left(\frac{\mathrm{y}}{3}\right)\)
so, coefficient of x in \(\frac{x y}{3}\) is \(\left(\frac{\mathrm{y}}{3}\right)\)

(iii) abc in (- abc)
Answer:
(- abc) = – (abc)
so, coefficient of abc is – 1.

Question 4.
Identify like terms among the following and group them:
10ab, 7a, 8b, – a²b², – 7ba, – 105b, 9b²a², – 5a², 90a.
Answer:
(7a, 90a) are like terms because they
contain same algebraic factor ‘a’.
(10ab, – 7ba) are like terms as they have same algebraic factor ’ab’.
(8b, -105b) are like terms because they contain same algebraic factor ‘b’.
(- a²b², 9b²a²) are like terms because they contain same algebraic factor ‘a²b²‘.

Question 5.
State with reasons, classify the following expressions into monomials, binomials, trinomials.
a + 4b, 3x²y, px² + qx + 2, qz², x² + 2y, 7xyz, 7x² + 9y³ – 10z⁴, 3l² – m², x, – abc.
Answer:

Question 6.
Find the sum of the following like terms :
(i) 3a, 9a
Answer:
Sum of 3a, 9a = 3a + 9a
= (3 + 9)a = 12a

(ii) 5p²q, 2p²q
Answer:
Sum of 5p²q, 2p²q = 5p²q + 2p²q
= (5 – 2) p²q = 7p²q

(iii) 6m, – 15m, 2m
Answer:
Sum of 6m, – 15m, 2m
= 6m + (- 15m) + 2m
= 6m – 15m + 2m
= (6 – 15 + 2)m = – 7m

Question 7.
Write the perimeter of the given figure.

Answer:
The perimeter of the figure
= 10 + 4 + x + 3+ y
= x + y + (10 + 4 + 3)
= x + y + 17

Question 8.
Simplify:
6a² + 3ab + 5b² – 2ab – b² + 2a² + 4ab + 2b² – a²
Answer:
6a² + 3ab + 5b² – 2ab – b² + 2a² + 4ab + 2b² – a²
= (6a² + 2a² – a²) + (3ab – 2ab + 4ab) + (5b² – b² + 2b²)
= [(6 + 2 – 1) a²] + [(3 – 2 + 4)ab] + [(5 – 1 + 2)b²]
= 7a² + 5ab + 6b²

Question 9.
Add 2x² – 3x + 5 and 9 + 6x² in vertical method.

Answer:

Question 10.
Find the additive inverse of the follow-ing expressions:
(i) 35
Answer:
Additive inverse of 35 = – 35

(ii) – 5a
Answer:
Additive inverse of – 5a = – (-5a) = 5a

(iii) 3p – 7
Answer:
Additive inverse of 3p – 7 = – (3p – 7) = – 3p + 7

(iv) 6x² – 4x + 5
Answer:
Additive inverse of 6x² – 4x + 5
= (6x² – 4x + 5) = – 6x² + 4x – 5

Question 11.
Subtract 2p² – 3 from 9p² – 8.
Answer:
9p² – 8 – (2p² – 3) = 9p² – 8 – 2p² + 3
= (9 – 2) p² – 8 + 3
= 7p² – 5

Question 12.
Subtract 3a + 4b – 2c from 6a – 2b + 3c in row method.
Answer:
Let A = 6a – 2b + 3c, B = 3a + 4b – 2c
Subtracting 3a + 4b – 2c from 6a – 2b + 3c is equal to adding additive inverse of 3a + 4b – 2c to 6a – 2b + 3c
i.e., A – B = A + (-B)
additive inverse of (3a + 4b – 2c) = – (3a + 4b – 2c) = – 3a – 4b + 2c
A – B = A + (-B)
= 6a – 2b + 3c + (- 3a – 4b + 2c)
= 6a – 2b + 3c – 3a – 4b +2c = (6 – 3)a – (2 + 4)b + (3 + 2)c
Thus, the required answer = 3a – 6b + 5c

Question 13.
Subtract 3m³ + 4 from 6m³ + 4m² + 7m – 3 in stepwise method.
Answer:
Let us solve this in stepwise.
Step 1: 6m³ + 4m² + 7m – 3 – (3m³ + 4)
Step 2: 6m³ + 4m² + 7m – 3 – 3m³ – 4
Step 3: 6m³ – 3m³ + 4m² + 7m – 3 – 4 (rearranging like terms)
Step 4: (6 – 3)m³ + 4m² + 7m – 7 (distributive law)
Thus, the required answer = 3m³ + 4m² + 7m – 7

Question 14.
Subtract 4m² – 7n² + 5mn from 3n² + 5m² – 2mn.
(For easy understanding, same colours are taken to like terms)
Answer:

Question 15.
Find the value of the following expressions, when x = 3.
(i) x + 6
Answer:
When x = 3, the value of x + 6
(substituting 3 in the place of x) is
(3) + 6 = 9

(ii) 8x – 1
Answer:
When x = 3,
the value of 8x – 1 = 8(3) – 1
= 24 – 1 = 23

(iii) 14 – 5x
Answer:
When x = 3,
the value of 14 – 5x = 14 – 5(3)
= 14 – 15 = – 1

Practice Questions [Page No: 65]

Question 1.
In a certain code BOARD: CNBQE, how ANGLE will be written in that code?
(a) BMHKF
(b) CNIJE
(c) BLGIF
(d) CMIKF
Answer:
(a) BMHKF

Explaination:
Each letter in a word is shifted to 1 position is alphabetical order and went 1 position is backward like this.

ANGLE = BMHKF

Question 2.
In a certain code, MOBILE: 56, how PHONE will be written in that code?
(a) 52
(b) 54
(c) 56
(d) 58
Answer:
(d) 58

Explaination:
Taking sum of positions of letters in for ward direction.

So, PHONE = 16 + 8 + 15 + 14 + 5 = 58

Question 3.
If BEAN: ABNE, then NEWS ?
(a) WSNE
(b) WSEN
(c) WNSE
(d) WNES
Answer:
(c) WNSE

Explaination:
Follow the same to decode the given word.

Question 4.
If ROSE : 6821, CHAIR : 73456, PREACH : 961473, then SEARCH ?
(a) 241673
(b) 214673
(c) 216473
(d) 216743
Answer:
(b) 214673

Explaination:
By comparing each letter with the symbol, by writing one below the other.

Question 5.
If COMPUTER: RFUVQNPC, then MEDICINE?
(a) EDJOJMEF
(b) EOJDJEFM
(c) EOJJDFEM
(d) EDJJOFME
Answer:
(b) EOJDJEFM

Explaination:
In the coded form the first & last letters have been interchanged while the rem¬aining letters are coded by taking their immediate next letters in the reverse order.

Question 6.
If LAKE = 7@$5, WALK = %@7$, then WAKE = ?
(a) @%75
(b) %@$5
(c) %5@7
(d) %@57
Answer:
(b) %@$5

Explaination:
By comparing each letter with symbol, by writing One below the other.

Question 7.
If MANY = OCPA, then LOOK = ?
(a) NQQM
(b) MQQN
(c) QMQN
(d) QNQM
Answer:
(a) NQQM

Explaination:
Each letter in a word is shifted to 2 position forward in alphabetical order.

∴ LOOK = NQQM

Question 8.
If SOME = PUB, then BODY ?‘
(a) LABY
(b)YBAL
(c) YLAV
(d) ABLY
Answer:
(c) YLAV

Explaination:
Bach letter in a word is shifted to 3 position backward in alphabetical order.

So, BODY = YLAV

Question 9.
If ARC = CVI, then RAY?
(a) TEU
(b) TEE
(c) TED
(d) TEF
Answer:
(b) TEE

Explaination:
Each letter in a word is forwarded alphabetical order as follows.

So, RAY = TEE

Question 10.
If MEAN = KGYI, then MODE = ?
(a) QBGK
(b) KBQC
(c) KGBQ
(d) kQBG
Answer:
(d) kQBG

Explaination:
Each letter in the word is shifted to 2 position backward and 2 position forward as following.

So, MODE = KQBG

Question 11.
If FIND = DNIF, then DONE ?
(a) ENOD
(b) ENDO
(c) NEOD
(d) ONED
Answer:
(a) ENOD

Explaination:
By reversing the word from left to right.
So, DONE = ENOD .

Question 12.
If BASE = SBEA, then AREA = ?
(a) AARE
(b) EAAR
(c) EARA
(d) REAA
Answer:
(b) EAAR

Explaination:
Follow the same to decode the given word.

Question 13.
If LESS = 55, then MORE = ?
(a)54
(b)50
(c) 51
(d) 52
Answer:
(c) 51

Explaination:
Taking sum of positions of letters in forward direction.

Question 14.
If BACK = 17, then CELL =?
(a) 33
(b) 30
(c) 31
(d) 32
Answer:
(d) 32

Explaination:
Taking sum of positions of letters in for ward direction.

3 + 5 + 12 + 12 = 32

Question 15.
If BIG = 63, then SMALL =?
(a) 76
(b) 78
(c) 74
(d) 72
Answer:
(b) 78

Explaination:
Taking sum of positions of letters in reverse direction.

8 + 14 + 26 + 15 + 15 = 78