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AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures Ex 9.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures Exercise 9.2

Question 1.
A rectangular acrylic sheet is 36 cm by 25 cm. From it, 56 circular buttons, each of diameter 3.5 cm have been cut out. Find the area of the remaining sheet.
Solution:
The dimensions of a rectangular acrylic sheet = 36 cm × 25 cm.
∴ Its area = l × b
= 36 × 25 = 900 sq.cm
The diameter of a circular button = 3.5 cm
radius = r = \(\frac{d}{2}=\frac{3.5}{2}\) = 1.75 cm.
∴ Area of each button = πr²
= \(\frac{22}{7}\) × (1.75)²
= \(\frac{22}{7}\) × 1.75 × 1.75
= 9.625 sq.cm.
The area of 56 such buttons = 56 × 9.625 = 539 sq.cm.
∴ The area of remaining sheet
= Area of a rectangular sheet – Area of 56 buttons
= 900 – 539
= 361 sq.cm

Question 2.
Find the area of a circle inscribed in a square of side 28 cm.
[Hint. Diameter of the circle is equal to the side of the square]

Solution:
Diameter of a circle = d = side of a square = 28 cm.
∴ d = 28 cm
⇒ r = \(\frac{d}{2}=\frac{28}{2}\) = 14 cm
∴ Area of a circle = \(\frac{22}{7}\) × 14 × 14
= 22 × 28
= 616 sq.cm.

Question 3.
Find the area of the shaded region in each of the following figures.
[Hint:d + \(\frac{d}{2}\) + d/2 = 42]
d = 21
∴ side of the square 21 cm
i)

Solution:

∴ Side of a square, d = 21 cm.
Radius of a semicircle = r = \(\frac{\mathrm{d}}{2}=\frac{21}{2}\) = 10.5 cm
∴ Area of a semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 1.5 × 10.5
= 173.250 sq.cm.

∴ Area of 4 shaded semi-circles
= 4 × Area of each semi-circle
= 4 × 173.25 = 693 sq.cm.

Diameter of a big circle = 21 m

Diameter of a semicircle = 10.5 m.
Radius = \(\frac{10.5}{2}\) = 5.25 m.
∴ Area of semi-circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 0.75 × 5.25
= 43.3125 sq.m.
Area of two semicircles
= 2 × 43.3125 = 86.6250 sq.m.
∴ Area of shaded region
= Area of big circle – Area of 2 semicircles
= 346.5 – 86.6250
= 259.8750 sq.m.

Question 4.
The adjacent figure consists of four small semi-circles of equal radii and two big semi-circles of equal radii (each 42 cm). Find the area of the shaded region
Solution:
Diameter of a bigger semi circle d = 42 cm
Radius, r = \(\frac { 42 }{ 2 }\) = 21 cm
Area of a bigger semi circle = \(\frac{\pi r^{2}}{2}\)

= 11 × 3 × 21
= 693 sq.cm

∴ Area of the shaded region
= Area of two bigger semi circles – Area of two smaller semi circles + Area of two smaller semi circles
= Area of two bigger semi circles
= 2 × Area of bigger semi circles
= 2 × 693 = 1386 sq.cm.

Question 5.
The adjacent figure consists of four half circles and two quarter circles.
If OA=OB=OC=OD= 14cm. Find the area of the shaded region.

Solution:

OA = OB = OC = OD = 14CM
Radius of \(\frac { 1 }{ 4 }\) of circle BXD, r = 14cm

= 154 sq.cm.
∴ Area of the quarter circle AYC
= 154 sq.cm
.’. Area of the shaded region = Area of quarter circle BXD – Area of semi circle OPB + Area of semi circle OQD + Area of quarter circle AYC – Area of semi circle ARO + Area of semi circle OSC
= Area of quarter circle BXD + Area of quarter circle AYC
= 154 + 154
= 308 sq.cm.

Question 6.
In adjacent figure A, B, C and D are centres of equal circles which touch externally is pairs and ABCD is a square of side 7 cm. Find the area of the shaded region.

Solution:

Side of a square,
AB = BC = CD = DA = 7 cm
Area of a square = s × s
= 7 × 7 = 49 sq.cm.
From the above figure,
Radius of a circle, r = \(\frac{\text { Side of a square }}{2}\)
= \(\frac{7}{2}\) = 3.5
Areas of 4 equal sectors are equal.
Angle of a sector APQ, x = 90°
Radius, r = 3.5 cm
Area of sector APQ

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.
Total area of 4 equal sectors = 4 × 9.625 = 38.5 sq.cm.
∴ Area of shaded region = Area of square ABCD – Area of 4 equal sectors
= 49 – 38.5
= 10.5 sq.cm.

Question 7.
The area of an equilateral triangle is 49√ cm². Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in the figure. Find the area of the portion in the triangle not included in the circles.

Solution:
From the given figure,
Δ ABC is an equilateral triangle.
∴ Area of an equilateral triangle
= 49√3 sq.em (given)
3 equal circles (whose radii are equal) are touch externally.
Each angle of a sector in the circle = 60°,
Radius r = 7 cm.
∴ The areas of 3 sectors are equal.
∴ Area of a sector APQ

∴ The area of 3 sectors
= 3 × \(\frac { 77 }{ 3 }\) = 77 sq. cm
∴ The area of the portion in the triangle not included in the circles
= area of ΔABC — area of 3 equal sectors
=49√3 – 77
=49 × 1.7321 – 77 (∵√3 = 1.7321)
= 84.8729 – 77
= 7.8729 sq.cm.

Question 8.
(i) Four equal circles, each of radius ‘a’ touch one another. Find the area between them.
(ii) Four equal circles are described about the four corners of a square so that each circle
touches two of the others. Find the area of the space enclosed between the circumferences
of the circles, each side of the square measuring 14 cm.
Solution:

The side of a square ABCD
AB = BC = CD = DA = a + a = 2a units.
∴ Area of a square = s × s
= 2a × 2a = 4a2 sq.units.
4 equal sectors are there.
In each sector,
The angle of a sector APQ, x = 90°
Radius, r = a units
Area of each sector APQ

∴ Area of 4 equal sectors
= 4 × \(\frac{\pi \mathrm{a}^{2}}{4}\) = πa² sq. units
= 2a × 2a = 4a2 sq.units. 4 equal sectors are there.
∴ The area between the circles
= Area of square sectors
= 4a² – πa²
= a² (4 – π) sq. units
= a² (4 – \(\frac { 22 }{ 7 }\) ) = \(\frac{6 a^{2}}{7}\) sq. units

Question 9.
From a piece of cardbord, in the shape ofa trapezium ABCD, and AB||CD and ∠BCD = 90°,= quarter circle is removed. Given AB = BC = 3.5 cm and DE =2 cm. Calcualte the area of the remaining piece of the cardboard. (Take π to be \(\frac{22}{7}\) )

Solution:
ABCD is a trapezium
AB || CD and ZBCD = 90°
AB = BC = 3.5 cm; DE = 2 cm.
Area of a trapezium ABCD :
Length of parallel sides, AB = 3.5 cm
CD = DE + EC
= 2 + 3.5 = 5.5 cm.

Distance between parallel sides,
BC = 3.5 cm.
∴ Area of a trapezium
= \(\frac { 1 }{ 2 }\) h (a + b)
= \(\frac { 1 }{ 2 }\) × BC (AB + CD)
= \(\frac { 1 }{ 2 }\) × 3.5 (3.5 + 5.5)
= \(\frac { 1 }{ 2 }\) × 3.5 × 9
= 3.5 × 4.5
= 15.75 sq.cm.

Area of quarter circle EBC :
Radius, r = 3.5 cm
∴ Area of quarter circle

= 5.5 × 0.5 × 3.5
= 9.625 sq.cm.

Area of the remaining piece of card board = Area of a trapezium – \(\frac { 1 }{ 2 }\) th of area of a circle
= 15.75 – 9.625
= 6.125 sq.cm.

Question 10.
A horse is placed for grazing inside a rectangular field 70 m by 52 m and is tethered to one corner by a rope 21 m long. How much area can it graze?

Solution:
From the above figure,
The sector represents the grazing area of a horse.
The angle of a sector OPQ, x = 90°
Radius of a sector OPQ, r = 21 m .
∴ Area of a sector


= 346.5 sq.m.
∴ The area of grazing = 346.5 sq.m.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 9 Area of Plane Figures Ex 9.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 9th Lesson Area of Plane Figures Exercise 9.1

Question 1.
Divide the given shapes as instructed.

(i)

Solution:
Rectangle ABCD
Rectangle CEFG
Rectangle FHIJ

(ii)

Solution:
Rectangle ABCD
Rectangle EFGH
Rectangle CHIJ

(iii)

Solution:
Trapezium ABEF Trapezium BCDE

(iv)

Solution:
Triangle ABC
Triangle DEF B
Rectangle ACDF

(v)

Solution:
Triangle BCD
Triangle BDE

Question 2.
Find the area enclosed by each of the following figures

Solution:
i) Area of a pentagon ABCDE = Area of square ACDE + area of Δ ABC
= AE × ED + \(\frac{1}{2}\) AC × BF
= 4 × 4 + \(\frac{1}{2}\) × 4 × 2
[∵ AC = ED = 4 cm;
BF = BD-CD = 6-4 = 2 cm]
= 16 + 4 = 20 sq.cm.

ii)

Area of a figure ABCDEF
= Area of square ABCF + area of trapezium CDEF
= AB × BC + \(\frac{1}{2}\) CG × (ED + CF)
= 18 × 18 + \(\frac{1}{2}\) × 8 × (7+ 18)
[∵h = CG = 8 cm, FC = AB = 18 cm]
= 324 + 4 × 25
= 324 + 100
= 424 cm²

iii)

Area of a figure ABCDEF
= Area of rectangle ABCD + area of trapezium ADEF
= AB × BC + \(\frac{1}{2}\) × GA × (EF + AD)
= 20 × 15 + \(\frac{1}{2}\) × 8 × (6 + 15)
[∵ GA = GB-AB = 28-20 = 8cm;
AD = BC – 15 cm]
= 300 + 4 × 21
= 300 + 84
= 384 sq.cm

Question 3.
Calculate the area of a quadrilateral ABCD when length of the diagonal AC = 10 cm and the lengths of perpendiculars from B and D on AC be 5 cm and 6 cm respectively.
Solution:
From the figire given below

h₁ = 6 cm,
h₂ = 5 cm,
d = 10 cm.
Area of a quadrilateral ABCD
= \(\frac{1}{2}\) d(h₁+h₂) . .
= \(\frac{1}{2}\) AC × (DE + BF)
= \(\frac{1}{2}\) × 10(6+5)
= 5 × 11 = 55sq.cm.

Question 4.
Diagram of the adjacent picture frame has outer dimensions 28 cm × 24 cm and inner dimensions 20 cm × 16 cm. Find the area of shaded part of frame, if width of each section is E the same.

Solution:
The outer dimensions of a picture frame = 28 cm × 24 cm

Length of the outer edge = 28 cm.
Breadth of the outer edge = 24 cm.
Dimensions of inner picture frame = 20 cm × 16 cm
Length of inner edge = 20 cm
Breadth of inner edge = 16 cm.
Area of AABC = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × AC × BC
= \(\frac{1}{2}\) × 4 × 4 = 8 cm²
Area of rectangle CDEB = l × b
= CD × DE
= 20 × 4
= 80 cm²

Area of ADEF = \(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × DF × DE
= \(\frac{1}{2}\) × 4 = 8sq.cm
∴ The area of the shaded region
= ar ΔABC + ar ▭ CDEB + ar ADEF
= 8 + 80 + 8
= 96 sq.cm.

Question 5.
Find the area of each of the following fields. All dimensions are in metres
Solution:

(i) Area of a trapezium ABCH
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (30 + 40)
= 40 (70)
= 2800 sq.m.

Area of Δ HCD =\(\frac{1}{2}\) × HC × HD
= \(\frac{1}{2}\) × 40 × 80
= 1600 sq.m.

Area of Δ EID = \(\frac{1}{2}\) × El × ID
= \(\frac{1}{2}\) × 60 × 40
= 1200 sq.m.

Area of trapezium FGIE
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 70 × (50 + 60)
= 35 (110)
= 3850 sq.m.

Area of Δ FGA = \(\frac{1}{2}\) × FG × GA
= \(\frac{1}{2}\) × 25 × 50
= 1250 sq.m.

∴ Area of the field

= 2800 + 1600 + 1200 + 3850 + 1250
= 10700 sq.m.

ii) Area of ΔABK = \(\frac{1}{2}\) × KB × KA
= \(\frac{1}{2}\) x 30 × 50
= 750 sq.m.

Area of trapezium KBCI
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 60 (30 + 40)
= 30 (70)
= 2100 sq.m.

Area of trapezium ICDE
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (40 + 50)
= 40 (90)
= 3600 sq.m.

Area of Δ FHE = \(\frac{1}{2}\) × FH × HE
=\(\frac{1}{2}\) × 20 × 40
= 400 sq.m.

Area of trapezium GJHF
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 80 × (40 + 20)
= 40 (60)
= 2400 sq.m.

∴ Area of ΔGJA = \(\frac{1}{2}\) × GJ × JA
= \(\frac{1}{2}\) × 40 × 70
= 1400 sq.m.

∴ Area of the field

= 750 + 2100 + 3600 + 400 + 2400 + 1400
= 10650 sq.m.

Question 6.
The ratio of the length of the parallel sides of a trapezium is 5:3 and the distance between them is 16cm. If the area of the trapezium is 960 cm², find the length of the parallel sides.
Solution:
The ratio of the length of parallel sides of a trapezium = 5:3.
Let the parallel sides be 5x, 3x say.
∴ a = 5x cm, b = 3x cm
The distance between the parallel sides
(h) = 16 cm.
∴ Area of a trapezium
= \(\frac{1}{2}\) × h(a + b)
= \(\frac{1}{2}\) × 16 x (5x + 3x)
= 8(8x)
= 64 x

According to the problem,
Area of the trapezium = 960 sq.cm.
∴ 64x = 960
x = \(\frac{960}{64}\) = 15
∴ The length of parallel sides
a = 5x = 5 × 15 = 75 cm
b = 3x = 3 × 15 = 45 cm

Question 7.
The floor of a building consists of around 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of flooring if each tile costs rupees 20 per m².
Solution:

The floor of a building is paved with 3000 tiles.
The shape of each tile is a rhombus.
Diagonals of a tile are
d₁ = 45 cm, d₂ = 30 cm.
Area of each tile = \(\frac{1}{2}\) × d₁ d₂
= \(\frac{1}{2}\) × 45 × 30
= 675 sq.cm.
∴ Area of floor
= 3000 × 675
= 2025000 sq.cm
= \(\frac{2025000}{10000}\) sq m
(∵ 1 sq.m. = 10000 sq.cm) 2025
= \(\frac{2025}{10}\) sq.m.
= 202.5 sq.m.
The cost of tile per square meter = ₹ 20
∴ The cost of flooring = ₹ 202.5 × 20
= ₹ 4050

Question 8.
There is a pentagonal shaped parts as shown in figure. For finding its area Jyothi and Rashida divided it in two different ways. Find the area in both ways and what do you
observe?

Solution:

Area of pentagon ABCDE

= \(\frac{1}{2}\) × AF × (AB + CF) + \(\frac{1}{2}\) FE x (ED + CF)
= \(\frac{1}{2}\) × 7.5 × (15+30) + \(\frac{1}{2}\) x 7.5 x (15+30)
= \(\frac{1}{2}\) × 7.5 × 45 + \(\frac{1}{2}\) x 7.5 x 45
= 2 × (\(\frac{1}{2}\) × 7.5 × 45)
= 7.5 × 45 = 337.50 sq.cm.

Rashida’s Method :

Area of square ABDE = side × side
= AE × ED = 15 × 15 = 225 sq.cm.

Area of ABDC =\(\frac{1}{2}\) × b × h
= \(\frac{1}{2}\) × BD × CF
= \(\frac{1}{2}\) × 15 × 15
( ∵ CF = 30 – 15 = 15 )
= \(\frac{225}{2}\) = 112.50 sq.cm
∴ Area of pentagon ABCDE
= ar □ ABDE + ar ΔBDC
= 225 + 112.50 = 337.50 sq.cm
The area of a polygon can’t changed if its procedures are changed.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.2

Question 1.
Cut the bold type English alphabets (capital) and paste in your note book. Draw possible number of lines of symmetry for each of the letter.
(i) How many letters have no linear symmetry?
(ii) How many letters have one line of symmetry?
(iii) How many letters have two lines of symmetry?
(iv) How many letters have more than two lines of symmetry?
(v) Which of them have rotational symmetry?
(vi) Which of then have point symmetry?
Solution:

i) letters don’t have line of symmetry
→ F, G, J, L, N, P, Q, R, S, Z.
ii) Liters which are having 1 line of symmetry
→ B, C, D, E, H, I, K, M, 0, T, U, V,
iii) Letirs which are having 2 lines of symmetry
→ H, I, O, X.
iv) Letters which are having more than two lines of symmetry O, X.
v) Letters which are having rotational symmetry
→ B, D, E, H, I, M, 0, S, T, W, X, Z.
vi) Letters which are having point symmetry
→ 0, X, M, W, H, 1, E, D.

Question 2.
Draw lines of symmetry for the following figures. Identify which of them have point symmetry. Is there any implication between lines of symmetry and point symmetry?

Solution:

The given figures has point of symmetry. Which are having line of symmetry are also having point of symmetry.

Question 3.
Name some natural objects with faces which have atleast one line of symmetry.
Solution:
1) Moon (Full)
2) Face of a human being
3) Orange fruit
4) Sunflower
5) Butterfly

Question 4.
Draw three tessellations and name the basic shapes used on your tessellation.
Solution:
I may notice that the basic shapes used to draw tessellations are pentagon, rectangle, squares, equilateral triangles, hexagons etc. Most tessellations can’t be formed with these shapes.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 8 Exploring Geometrical Figures Ex 8.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 8th Lesson Exploring Geometrical Figures Exercise 8.1

Question 1.
Name five pairs of congruent objects, you use daily.
Solution:
Solution:
i) Pair of ear rings.
ii) Wheels of a cycle.
iii) Equal lengths of scales.
iv) Pair of bangles.
v) Mathematics textbooks of same class.

Question 2.
(a) Draw two congruent figures. Are they similar? Explain
(b) Take two similar shapes. If you slide rotate or flip one of them, does the similarity remain?
Solution:
a) ΔABC ≅ ΔPQR

From ΔABC & ΔPQR
AB = PQ
AC = PR
BC = QR
∠A = ∠P
∠B = ∠Q
∠C = ∠R
∴ Congruent triangles are similar to each other.
But similar triangles are not congruent.

b)

From fig. ΔXYZ and ΔSTU are similar triangles. If these triangles are rotated even they are similar to each other.

Question 3.
If Δ ABC ≅ Δ NMO, name the congruent sides and angles.
Solution:

From the congruent triangles
ΔABC ≅ ΔNMO
AB = NM
BC = MO
AC = NO
∠A = ∠N
∠B =∠M
∠C = ∠O

Question 4.
State whether the following statements are true. Explain with reason.
(i) Two squares of side 3 cm each and one of them rotated through 45° are congruent.
(ii) Any two right triangles with hypotenuse 5 cm, are congruent.
(iii) Any two circles of radii 4 cm each are congruent.
(iv) Two equilateral triangles of side 4 cm each but labeled as A ABC and A LHN arc not congruent.
(v) Mirror image of a polygon is congruent to the original.
Solution:
i) True. If the square is rotated 45° then it is remaining congruent.
ii) True. If the hypotenuse of two right triangles are equal, then the two triangles are congruent to each other, [∵ Their corresponding sides and angles will also be equal]

(iii) True
Since they are having equal radii (r₁ = r₂ = 4 cm )
= 4 cm).
∴ The given circles are congruent.

iv) False.
If two sides of one triangle are equal to the corresponding two sides of another triangle, then the third sides are in the proportion.
∴ ΔABC ≅ ΔLHN

But given tXABC,’L\LHN. Then it is false.

v) True.
Always the mirror images of a polygon is congruent to the original.

Question 5.
Draw a polygon on a square dot sheet. Also draw congruent figures in different directions and mirror image of it.
Solution:

Question 6.
Using a square dot sheet or a graph sheet draw a rectangle and construct a similar figure.
Find the perimeter and areas of both and compare their ratios with the ratio of their
corresponding sides.
Solution:

Question 7.
7 pillars are used to hold a slant iron gudder as shown in the figure. If the distance between every two pillars is 1 m and height of the last piller is 10.5 m. Find the height of pillar.

Solution:

The height of the 1st pillar
h₁ = \(\frac{1}{7}\) x 10.5 = 1.5m.

The height of 2nd pillar
h₂ = \(\frac{2}{7}\) x 10.5 = 3m.

The height of 3rd pillar
h₃ = \(\frac{3}{7}\) x 10.5 = 4.5 m.

The height of 4th pillar
h₄ = \(\frac{4}{7}\) x 10.5 = 6 m.

The height of 5th pillar
h₅ = \(\frac{5}{7}\) x 10.5 = 7.5 m.

The height of 6th pillar
h₆ =\(\frac{6}{7}\) x 10.5 = 9 m.

Question 8.
Standing at 5 m apart from a vertical pole of height 3 m, Sudha observed a building at the back of the piller that tip of the pillar is in line with the top of the building. If the distance
between pillar and building is 10 m, estimate the height of the building. [Here height of
Sudha is neglected]
Solution:
ΔAOAB ~ ΔOCD
The corresponding sides of similar triangles are in a same proportion^?

⇒ h = 3 x 3 = 9
∴ h = height of a building = 9 mts.

Question 9.
Draw a quadrilateral of any measurements. Construct a dilation of scale factor 3. Measure their corresponding sides and verify whether they are similar.
Solution:

Observe the following dilation ABCD, it is a square drawn on a graph sheet.
All vertices A, B, C, D are joined from the sign ’O’ and produced to 3 times the length upto A’, B’, C’ and D’ respec¬tively. Then A’, B’, C’, D’ are joined to form a rectangle which 3 times has enlarged sides of ABCD. Here, 0 is called centre of dilation and \(\frac{\mathrm{OA}^{\prime}}{\mathrm{OA}}=\frac{3}{1}\) = 3 is called scale factor.
∴ □ ABCD ~ □ A’B’C’D’
[ ∵ their corresponding sides are equal]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.3

Question 1.
The following table gives the distribution of45 students across the different levels of intelligent Quotient. Draw the histogram for the data.

Solution:

Steps of construction:
1. Calculate the difference between mid values of two consecutive classes
∴ h = 75 – 65 = 10
∴ Class interval (C.I) = 10

2. Select such a right scale
on X-axis 1 cm = 10 units
on Y-axis 1 cm 1 student

3. Construct a histogram with C.Is as width and frequencies as lengths.

Question 2.
Construct a histogram for the marks obtained by 600 students in the VII class annual
examinations.

Solution:
Classes will be prepared on class marks.
Step – 1: Take the difference between the mid values of two consecutive classes.
∴ h = 400 – 360 = 40

Step – 2: Let the lower and upper boundaries be taken as x – \(\frac{h}{2}\) , x + \(\frac{h}{2}\)
∴ x – \(\frac{h}{2}\) = 360 – \(\frac{40}{2}\) = 360 – 20 = 340
x + \(\frac{h}{2}\) = 360 + \(\frac{40}{2}\) = 360 + 20 = 380

Step – 3 : Select the scale
on X-axis 1 cm = 1 C.I (mid values)
on Y-axis 1 cm = 20 students

Step – 4 : Take C.I’s as width, frequencies as lengths.
Then construct the histogram.

Class Marks	Class Interval	Frequency
360	340- 380	100
400	380- 420	125
440	420-460	140
480	460-500	95
520	500-540	80
560	540-580	60

Scale : On Y – axis no. of students = 20, On X – axis take marks of students.

Question 3.
Weekly wages of 250 workers in a factory are given in the following table. Construct the histogram and frequency polygon on the same graph for the data given.
using the histogram. (Use separate graph sheets)

Solution:

Class Interval (Weekly wages)	Frequency (No. of workers)	Mid values
500-550	30	525
550-600	42	575
600-650	50	625
650-700	55	675
700-750	45	725
750 -800	28	775
	N = 250

Steps of constructIon:

1. C.I. = Difference of two consecutive mid values = h = 575 – 525 = 50
2. Scale : On X-axis 1 cm = ₹ 50
   On Y-axis 1 cm = lo members
3. Take on X-axis width of C.I, on Y-axis frequncies.
4. Keep points A, B, C, D, E, F, G, H on the mid points of rectangles.
5. The area of a histogram is equal to the area of a polygon.

Question 4.
Ages of 60 teachers in primary schools of a Mandai are given in the following frequency distribution table. Construct the Frequency polygon and frequency curve for the data without using the histogram. (Use separate graph sheets)

Solution:
Construction of frequency polygon:

1. Class interval = The difference between two mid values of classes
   = 30 – 26 = 4
2. Scale : On X-axis take the age of teachers
   On Y-axis take number of teachers.
3. Scale: On X-axis 1 cm = 4units
   On Y- axis 1 cm = 2 units
4. Take the widths of classes on X – axis. Frequencies on Y – axis.
5. The points are formed on the graph sheet are joined by a scale then the required frequency polygon and if the points are joined by hand frequency curve will be formed.

Question 5.
Construct class intervals and frequencies for the foliowing distribution tabie. Also draw the ogive curves for the same.

Solution:
Steps of constructIon:
Step – 1: If the given frequency distribution is in inclusive form, then convert it into an
exclusive form.

Step – 2: Calculate the less than cumulative frequency.

Step – 3: Mark the upper boundaries of the class intervals along X-axis and their corresponding cumulative frequencies along Y-axis.

Select the scale:
X – axis 1 cm = 1 class interval
Y – axis 1 cm = 10 students

Step – 4: Also, plot the lower boundary of the first class (upper boundary of the class previous to first class) interval with cumulative frequency 0.

Step – 5: Join these points by a free hand curve to obtain the required ogive.
Similarly we can construct ‘greater than cumulative frequency curve by taking greater than cumulative on Y – axis and corresponding ‘lower boundaries’ on the
X-axis.

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.2

Question 1.
Given below are the ages of 45 people in a colony.
Construct grouped frequency distribution for the given data with 6 class intervals.

Solution:
Number of classes = 6
Range = Maximum value – Minimum value = 63 – 5 = 58

Question 2.
Number of students in 30 class rooms in a school are given below. Construct a frequency
distribution table for the data with a exclusive class interval of 4 (students).

Solution:
Class Interval (C.I.) = 4
Range = Maximum value – Minimum value = 40 – 15 = 25
No. of classes = \(\frac{\text { Range }}{\mathrm{C} . \mathrm{I}}=\frac{25}{4}\) = 6.25 *(approx ‘6’)

Question 3.
Class intervals in a grouped frequency distribution are given as 4 – 11, 12 – 19, 20 – 27, 28 – 35, 36 – 43. Write the next two class intervals. (1) What is the length of each class
interval? (ii) Write the class boundaries of all classes, (iii) What are the class marks of
each class?
Solution:
The given class intervals are 4-11,12 -19, 20 – 27, 28 – 35, 36 – 43
The next two class intervals are 44 – 51, 52 – 59
i) The length of each class interval is 8
ii) Boundaries of classes :
iii) Class = \(\frac{3.5+11.5}{2}=\frac{15}{2}\) = 7.5

Class Interval	Boundaries of classes	Class Marks
4-11	3.5-11.5	7.5
12-19	11.5-19.5	15.5
20- 27	19.5- 27.5	23.5
28-35	27.5-35.5	31.5
36-43	35.5-43.5	39.5
44-51	43.5-51.5	47.5
52-59	51.5-59.5	55.5

Question 4.
In the following grouped frequency distribution table class marks are given.

(i) Construct class intervals of the data. (Exclusive class intervals)
(ii) Construct less than cumulative frequencies and
(iii) Construct greater than cumulative frequencies.
Solution:

To determine the lower boundary of first class :
The difference between two marks of consecutive classes = h = 22-10 = 12
Let the marks of each class be ‘x’ then the boundaries of the classes be
Lower boundary of first class = x – \(\frac{\mathrm{h}}{2}\) = 10 – \(\frac{12}{2}\) = 10 – 6 = 4
Upper boundary = 10 + \(\frac{12}{2}\) = 10 + 6 = 16
We can determine the remaining classes in the same way.

Question 5.
The marks obtained by 35 students in a test in statistics (out of 50) are as below.

Construct a frequency distribution table with equal class intervals, one of them being 10-20(20 is not included).
Solution:
The frequency distribution table is

Class Interval	Frequency
0-10	2
10-20	10
20-30	4
30-40	9
40-50	10

C.I. = 10 (from 10 – 20)
Range = 48 – 1 = 47
No. of classes = = 4.7 = 5 (approx)

Question 6.
Construct the class boundaries of the following frequency distribution table. Also construct less than cumulative and greater than cumulative frequency tables.

Solution:

Question 7.
Cumulative frequency table is given below. Which type of cumulative frequency is given. Try to build the frequencies of respective class intervals.

Solution:
In the given table frequencies are increasing from top to bottom. So, it is less than cumulative frequency distribution

Less than C.F.	Frequency
3	3
8	(8 – 3) 5
19	(19 – 8) 11
25	(25 – 19)6
30	(30 – 25) 5

∴ Thee required less than C.F. distribution table is

Question 8.
Number of readers in a library are given below. Write the frequency of respective classes.
Also write the less than cumulative fequency table.

Solution:

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 7 Frequency Distribution Tables and Graphs Ex 7.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 7th Lesson Frequency Distribution Tables and Graphs Exercise 7.1

Question 1.
Find the arithmetic mean of the sales per day in a fair price shop in a week.
₹ 10000, ₹ 10250, ₹ 10790, ₹ 9865, ₹ 15350, ₹ 10110
Solution:

Question 2.
Find the mean ofthe data; 10.25, 9, 4.75, 8, 2.65, 12, 2.35
Solution:

Question 3.
Mean of eight observations is 25. If one observation 11 is excluded, find the mean of the remaining.
Solution:
Given that
The mean of 8 observations = 25
The sum of 8 observations = 8 × 25
= 200
If the observation 11 is subtracted from the sum
⇒ 200 – 11 = 189
∴ The mean of remaining 7 observations = \(\frac { 189 }{ 7 }\) = 27

Question 4.
Arithmetic mean of nine observations is calculated as 38. But in doing so, an observation 27 is mistaken for 72. Find the actual mean of the data.
Solution:
The mean of 9 observations = 38
The sum of 9 observations = 38 × 9
= 342
If the observation 27 is mistaken for 72
then correct observation = 72 – 27
= 45
Sum of correct observations = 342 + 45
= 387
∴ Correct Mean = \(\frac { 387 }{ 9 }\) = 43

Question 5.
Five years ago mean age of a family was 25 years. What is the present mean age of the family ?
Solution:
When all the observations of the data are increased or decreased by a certain number, the mean also increases or decreases by the same number.
Five years ago mean age of a family = 25 years
Present mean age of the family = 25 + 5 = 30 y

Question 6.
Two years ago the mean age of 40 people was 11 years. Now a person left the group and the mean age is changed to 12 years. Find the age of the person who left the group.
Solution:
The present mean age of 40 people = 11 + 2 = 13 years
∴ The total age of 40 people = 40 × 13
= 520 years
If 1 person is left from the group of 40 people then remaining will be = 40 – 1 = 39.
The mean age of 39 people = 12 years
The total age of 39 people = 39 × 12
= 468 years
Age of the person who left from the group = 520 – 468 = 52 years

Question 7.
Find the sum of deviations of all observations of the data 5,8, 10, 15,22 from their mean.
Solution:
A.M. of the observations 5,8,10,15,22 is


∴ Sum of deviations = 0

Question 8.
If sum of the 20 deviations from the mean is 100, then find the mean deviation.
Solution:

Question 9.
Marks of 12 students inaunittestare givenas4,21, 13, 17,5,9, 10,20, 19, 12,20, 14.
Assume a mean and calculate the arithmetic mean of the data. Assume another number as
mean and calculate the arithmetic mean again. Do you get the same result? Comment.
Solution:
Given observations are 4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14
Estimated mean = 10 say
A.M. Estimated Mean + Average of deviations

= 10 + \(\frac { 44 }{ 12 }\) = 10 + 3.67 = 13.67
∴ A.M = \(\bar{x}\) = 13.67
Another Estimated Mean = 12 then
\(\bar{x}\) = A.M. = Estimated Mean + Average of deviations

= 12 + \(\frac { 20 }{ 12 }\) = 12 + 1.67 = 13.67
\(\bar{x}\) = 13.67
∴ In all cases \(\bar{x}\) = 13.67

Question 10.
Arithmetic mean of marks (out of 25) scored by 10 students was 15. One of the student, named Karishma enquired the other 9 students and find the deviations from her marks are noted as – 8, – 6, – 3, – 1, 0, 2, 3,4, 6. Find Karishma’s marks.
Solution:
Average marks of 10 students = 15
Total marks of 10 students = 10 × 15
= 150
Marks obtained by Karishma = x say
Deviations of 9 students =
-8, -6-3,-1,0, 2, 3,4,6
∴ x-8 + x-6 + x-3 + x-l + x + x + 2 + x + 3 + x + 4 + x + 6 + x = 150
⇒ 10x- 3 = 150
⇒ 10x = 150 + 3 = 153
x = \(\frac { 153 }{ 10 }\) = 15.3

(Or)

The given observations are
-8,-6,-3,-1,0, 2, 3, 4,6
Let the number of marks of Karisima =
Her deviation from x’ are – 8 + x, – 6 + x, – 3 + x, – 1 + x, x + 0, 2 + x, 3 + x, 4 + x, 6 + x

[ ∵ Mean = 15 is given]
10x – 3 = 15 × 10
10x =150 + 3
10x= 153
x = \(\frac{153}{10}\)
x = 15.3
∴ Number of marks obtained by Karishma = 15.3

Question 11.
The sum of deviations of ‘n’ observations from 25 is 25 and sum of deviations of the same ‘n’ observations from 35 is – 25. Find the mean of the observations.
Solution:
From the value ’25’ the sum of deviations of a data = Σx₁ = 25
∴ Mean of the observation = 25 + \(\frac{\Sigma \mathrm{x}_{1}}{\mathrm{n}}\)
= \(25+\frac{25}{n}=\frac{25 n+25}{n}\)
From the value ’35’ the sum of deviations of a data = Σx₁ = -25
∴ Mean of the observation = 35 + \(\frac{\Sigma \mathrm{x}_{1}}{\mathrm{n}}\)
= \(35+\frac{-25}{n}=\frac{35 n+25}{n}\)
Mean of the given data

(Or)

Let x₁, x₂, x₃,…… x_n are the number of
deviations from 25
∴ Number of observations are
x₁ – 25, x₂ – 25, …………. x_n – 25
∴ Sum of the observations = 25
∴ x₁ – 25 + x₂ – 25 +…………. + x_n – 25 = 25
⇒ x₁+ x₂ + + x_n = 25 + 25n …………(1)
= 25 (n + 1)
Deviations from 35 are
x₁ – 35, x₂ – 35y …………., x_n – 35
∴ Sum of the deviations = – 25
∴ (x₁ – 35) + (x₂ – 35) + … (x_n – 35) = – 25
⇒x₁ + x₁ +………………….. + x_n = – 25 + 35 x n …………….(2)
From (1) and (2),
– 25 + 35n = 25 + 25n
⇒ 10n = 50
n = 5
∴ Mean of the observations

Question 12.
Find the median of the data ; 3.3, 3.5, 3.1, 3.7, 3.2, 3.8
Solution:
The ascending order of 3.3, 3.5, 3.1, 3.7, 3.2, 3.8 is 3.1, 3.2, 3.3, 3.5, 3.7, 3.8
∴ n = 6 (is an even)
Median = Mean of \(\frac{n}{2},\left(\frac{n}{2}+1\right)\) terms
= \(\frac{6}{2}, ([latex]\frac{6}{2}\) + 1)[/latex]
= Mean of 3, 4 terms
= \(\frac{3.3+3.5}{2}\)
= \(\frac{6.8}{2}\) = 3.4

Question 13.
The median of the following observations, arranged in ascending order is 15. 10, 12, 14, x – 3, x, x + 2, 25.Then find x.
Solution:
Given observations are
10, 12, 14, x – 3,x ; x + 2, 25
⇒ n = 7 (is an odd)
∴ Median = \(\frac{n+1}{2}=\frac{7+1}{2}\) = 4th term
= x – 3
⇒ x-3 = 15
⇒ x = 15 + 3
∴ x = 18

Question 14.
Find the mode of 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10
Solution:
The given observations are
10, 12, 11, 10, 15, 20, 19,21,11,9, 10
∴ The most frequently occurring observation is 10.
∴ Mode = 10

Question 15.
Mode of certain scores is x. 1f each score is decreased by 3, then find the mode of the new series.
Solution:
The mode of some observations is ‘x’. If ‘3’ is subtracted from every observa¬tion then the mode = x – 3

Question 16.
Find the mode of all digits used in Titing the natural numbers from 1 to 100.
Solution:
Natural numbers from 1 to 100 are 1,2,3, …………….99, 100
∴ It has no mode.
[ ∵ Every number appears only once]

Question 17.
Observations of a raw data are 5, 28, 15, 10, 15, 8, 24. Add four more numbers so that mean and median of the data remain the same, but mode increases by 1.
Solution:
The given observations are 5, 28, 15, 10, 15, 8, 24

Ascending order of the observations 5, 8, 10, 15, 15, 24, 28
Median = \(\frac{n+1}{2}=\frac{7+1}{2}\) = 4 th term
= 15 [n = 7 is an odd]
∴ Mode = 15
If 4 observations are inserted for the given data then mode increases by 1.
Let x₁ = x₂ = x₃ = x say .
∴ The mean
= \(\frac{x+x+x+x_{4}+10}{11}\) = 15
⇒ 3x + x₄ = 165 – 105 = 60
∴ 3x + x₄ = 60
If mode will increases by 1 then
x= 15 +1 = 16
∴ 3x + x₄ = 60
⇒ 3 x 16 + x₄ =60
⇒ x₄ = 60 – 48 = 12
∴ The required 4 observations are 12, 16, 16, 16

Question 18.
If the mean of a set of observations x₁, x₁ , x₁₀ is 20. Find the mean of x₁ + 4, x₂ + 8, x₃ + 12 , x₁₀ + 40.
Solution:
Mean of observations x₁, x₂……………….. ,x₁₀ is

= 20 + 2 × 11 = 20 + 22 = 42

Question 19.
Six numbers from a list of nine integers are 7,8,3, 5,9 and 5. Find the largest possible value of the median of all nine numbers in this list.
Solution:
The given 6 integers are 7, 8, 3, 5, 9, 5
Ascending order = 3, 5, 5, 7, 8, 9
Let the remaining 3 observations be x₁, x₂, x₃ say
Median of 3, 5, 5, 7, 8, 9, x₁, x₂, x₃ is
= \(\frac{n+1}{2}\) [ ∵ n is an odd (9)]
= \(\frac{9+1}{2}\) = 5th term = 8
∴ The largest possible median will be 8

Question 20.
The median of a set of 9 distinct observations is 20. If each of the largest 4 observations of the set is increased by 2, find the median of the resulting set.
Solution:
Let the 9 observations be x₁, x₂………..x₉ say
Median = \(\frac{n+1}{2}\) [∵ n = 9 is an odd]
= \(\frac{n+1}{2}\) = 5th term = x₅ = 20
∴ x₅ = 20
If 2 is added to the last 4 largest numbers then x₁, x₂, x₃, x₄, x₅, x₆ + 2 x₇ + 2, x₈ + 2, x₉ +2
∴ Median = \(\frac{n+1}{2}=\frac{9+1}{2}\) = 5th term
∴ x₅ = 20

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.5 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.5

Question 1.
Find the cube root of the following numbers by prime factorization method.
(i) 343
(ii) 729
(iii) 1331
(iv) 2744
Solution:

Question 2.
Find the cube root of the following numbers through estimation’?
(i) 512
(ii) 2197
(iii) 3375
(iv) 5832
Solution:
Step 1: Start making groups of three digits starting from the unit place.
i.e, \(\overline{512}\) First group is 512

Step 2: First group i.e 512 will give us the units digit of the cube root. As 512 ends with 2, then its cube root ends with 8 (2 x 2 x 2) So the units place of the cube root will be 8.

Step 3: Now take the second group i.e. 0. Which is 0³ < 1 < 2³.
So the least number is ‘0′.
∴ Tens digit of a cube root of a number be 0.
∴ \(\sqrt[3]{512}\) = 08 = 8

(ii) 2197
Step 1: Start making groups of three digits starting from the unit place.

Step 2: First group i.e., 197 will give us the units digit of the cube root.
As 197 ends with 7, its cube root ends with 3. ‘
[∵ 3 x 3 x 3 = 27]
∴ Its units digit is 7.

Step 3: Now take the second group i.e.,2
We know that i3 < 2 < 2
∴ The least number be 1.
∴ The required number is 13.
∴ \(\sqrt[3]{2197}=\sqrt[3]{13 \times 13 \times 13}=\sqrt[\not]]{(13)^{8}}\)
= 13

(iii) 3375
Step 1: Start making groups of three digits starting from the unit place.
i.e.;

Step 2: First group is 375. Its units digit is 5.
∴ The cube root is also ends with 5.
∴ The units place of the cube root will be 5.

Step 3: Now take the second group,
i.e., 3 we know that 1³ < 3³ <2³
∴ The least number is 1.
∴ The tens digit of a cube root will be 1.
∴ The required number = 15
\(\sqrt[3]{3375}=\sqrt[3]{15 \times 15 \times 15}=\sqrt[\not]{15^{\not 3}}=15\)

(iv) 5832
Step 1: Start making groups of three digits starting from the unit place.

Step 2: The units digit of 832 is 2.
∴ The cube root of the number ends with units digit 8.
[∵ 8 x 8 x 8 = 512]

Step 3: In the second group i.e., 5 lie between 1 and 6
i.e., 1³ < 5 < 2³
∴ The tens digit of a number will bel.
∴ The required number is 18.
∴ \(\sqrt[3]{5832}=\sqrt[3]{18 \times 18 \times 18}=\sqrt[3]{(18)^{3}}\)
= 18

Question 3.
State true or false?
(i) Cube of an even number is an odd number
(ii) A perfect cube may end with two zeros
(iii) If a number ends with 5, then its cube ends with 5
(iv) Cube of a number ending with zero has three zeros at its right
(v) The cube of a single digit number may be a single digit number.
(vi) There is no perfect cube which ends with 8
(vii) The cube of a two digit number may be a three digit number.
Solution:
(i) Cube of an even number is an odd number (F)
(ii) A perfect cube may end with two zeros (F)
(iii) If a number ends with 5, then its cube ends with 5. (T)
(iv) Cube of a number ending with zero has three zeros at its right. (T)
(v) The cube of a single digit number may be a single digit number. (F)
(vi) There is no perfect cube which ends with 8(F)
(vii) The cube of a two digit number may be a three digit number. (F)

Question 4.
Find the two digit number which is a square number and also a cubic number.
Solution:
The two digited square and cubic
number is 64
∴ 64 = 8 x 8 = 8² ⇒ \(\sqrt{64}\) = 8
64 = 4 x 4 x 4 = 4³ ⇒ \(\sqrt[3]{64}\) = 4

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.4 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.4

Question 1.
Find the cubes of the following numbers
(1) 8
(ii) 16
(iii) 21
(iv) 30
Solution:

Number	Cube Of a Number
i) 8	83 =  8 × 8 × 8 = 512
ii) 16	163 = 16 × 16 × 16 = 4096
iii) 21	213 = 21 × 21 × 21 = 9261
iv) 30	303 = 30 × 30 × 30 = 27000

Question 2.
Test whether the given numbers are perfect cubes or not.
(i) 243
(ii) 516
(iii) 729
(iv) 8000
(v)2700
Solution:

Number	Cube Of a Number	Yes / No
i) 243	3 × 3 × 3 × 3 × 3 = 35	No
ii) 516	2 × 2 × 3 × 43	No
iii) 729	9 × 9 × 9 = 93	Yes
iv) 8000	20 × 20 × 20 = (20)3	Yes
v) 2700	(30) × (30) × 3	No

Question 3.
Find the smallest number by which 8788 must be multiplied to obtain a perfect cube?
Solution:
The prime factorisation of 8788
= (2 × 2) × (13 × 13 × 13)

∴ From the above product 2 ¡s left in the triplet.
∴ 2 should be multiplied with 8788 we will get a perfect cube number.

Question 4.
What smallest number should 7803 be multiplied with so that the product becomes a perfect cube?
Solution:
The prime factorisation of 7803
= (3 × 3 × 3) × (17 × 17)

∴ From the above product 17 is left in
the triplet.
∴ 17 should be multiplied to 7803 then we will get a perfect cube number.

Question 5.
Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube’?
Solution:
The prime factorisation of 8640
= (2 × 2 × 2) × (2 × 2 × 2) × 5 × (3 × 3 × 3)
= (2)³ × (2)³ × 5 × (3)³

Question 6.
Ravi made a cuboid of plasticine ofdimensions 12cm, 8cm and 3cm. How many minimum number of such cuboids will be needed to form a cube’?
Solution:
The volume of a plasticine cuboid
= l × b × h
= 12 × 8 × 3
= 288 cm³
If the minimum no. of such cuboids will be needed to form a cube then its volume be less than 288 i.e., 216 cm³
∴ s³ = 216
s = \(\sqrt[3]{216}=\sqrt[3]{6 \times 6 \times 6}=\sqrt[3]{6^{3}}\) = 6
∴ The side of the cube 6 cm

Question 7.
Find the smallest prime number dividing the sum 3¹¹ +5¹³.
Solution:
The units digit in 3¹¹ is 7

∴ The units digit in 3¹¹ is 7
The units digit in 5¹³ is 5
7 + 5 = 12 is divided by a smallest prime number 2.
∴ The smallest prime number that divide the sum 311 + 513 = 2

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.3 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.3

Question 1.
Find the square roots of the following numbers by division method.
(i) 1089
(ii) 2304
(iii) 7744
(iv) 6084
(v) 9025
Solution:
(i) 1089

(ii) 2304

(iii) 7744

(iv) 6084

(v) 9025

Question 2.
Find the square roots of the following decimal numbers.
(i) 2.56
(ii) 18.49
(iii) 68.89
(iv) 84.64
Solution:
(i) 2.56

(ii) 18.49

(iii) 68.89

(iv) 84.64

Question 3.
Find the least number that is to be subtracted from 4000 to make it perfect square
Solution:
Square root of 4000 by
Division Method:

∴ The least number 31 should be subtracted from 4000 we will get a perfect square number4
∴ 4000 – 31 = 3969
= \(\sqrt{3969}=\sqrt{63 \times 63}\) = 63

Question 4.
Find the length of the side of a square whose area is 4489 sq.cm.
Solution:
Area of a square (A) = 4489 sq.cms
A = s²
s² = 4489
s = \(\sqrt{4489}=\sqrt{67 \times 67}\) = 67cms.
∴ The side of a square (s) = 67cms.

Question 5.
A gardener wishes to plant 8289 plants in the form of a square and found that there were 8 plants left. How many plants were planted in each row?
Solution:
No. of plants are planted = 8289 If 8289 plants are planted in a square shape, 8 plants are left.
Then remaining plants = 8289 – 8 = 8281

∴ No. of plants for each row = 91
∴ 8281 plants are planted in a square shape then no. of plants are planted for each row = 91

Question 6.
Find the least perfect square with four digits.
Solution:
The smallest number of 4 digits = 1000

∴ 24 should be added tö 1000 then 1000 + 24 = 1024
∴ The smallest 4 digited perfect square number is 1024.
[∵ \(\sqrt{1024}\) = 32]

Question 7.
Find the least number which must be added to 6412 to make it a perfect square?
Solution:

∴ The least number 149 should be added to 6412 then we will get a perfect square number.
∴ 6412 + 149 = 6561
∴ \(\sqrt{6561}=\sqrt{81 \times 81}\) = 81

Question 8.
Estimate the value of the following numbers to the nearest whole number
(i) \(\sqrt{97}\)
(ii) \(\sqrt{250}\)
(iii) \(\sqrt{780}\)
Solution:
ï) \(\sqrt{97}\) , 97 lie between the perfect
square numbers 81 and 100.
∴ 81 <97< 100
9² < 97 < 10²
=9 < \(\sqrt{97}\) < 10
∴ \(\sqrt{97}\) Is nearest to 10.
[∵ 97 is nearest to 100]

(ii) \(\sqrt{250}\), 250 lie between the perfect square numbers 225 and 256.
∴ 225 < 250 < 256
15² < 250 < 16²
= 15 < \(\sqrt{250}\) <16
∴ \(\sqrt{250}\) is nearest to 16.
[ ∵ 250 is nearest to 256]

ii) \(\sqrt{780}\), 780 lie between the perfect
square numbers 729 and 784.
∴ 729 < 780 < 784
27² < 780 < 28²
= 27< \(\sqrt{780}\) <28
∴ \(\sqrt{780}\) is nearest to 28.
[ ∵ 780 is nearest to 784]

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.2 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.2

Question 1.
Find the square roots of the following numbers by Prime factorization method.
(1) 441
(ii) 784
(iii) 4096
(iv) 7056
Solution:

Question 2.
Find the smallest number by which 3645 must be multiplied to get a perfect square.
Solution:
The prime factorization of 3645
= (3 × 3) × (3 × 3) (3 × 3) × 5
∴ Deficiency of one ‘5’ is appeared in the above product.
∴ 3645 is multiplied with 5 then we will get a perfect square.

Question 3.
Find the smallest number by which 2400 is to be multiplied to get a perfect square and also find the square root of the resulting number.
Solution:
The prime factorization of 2400
=(2 × 2) × (2 × 2) × 2 × (5 × 5) × 3
∴ 2,3 are needed to form a pair
∴ 2 × 3 = 6
∴ 6 should be multiplied with 2400 then we will get a perfect square number.
∴ 2400 × 6 = 14400
∴ \(\sqrt{14400}\) = 120

Question 4.
Find the smallest number by which 7776 is to be divided to get a perfect square.
Solution:

The prime factorization of 7776
=(2 × 2) × (2 × 2) × 2 × (3 × 3) × (3 × 3) × 3
∴ 2, 3 are needed to form a pair
∴ 2 × 3 = 6
∴ 7776 should be divided by 6 then we will get a perfect square number.

Question 5.
1521 trees are planted in a garden in such a way that there are as many trees in each row as there are rows in the garden. Find the number of rows and number of trees in each row.
Solution:
Let the no. of trees planted in a garden for each row = x say.
No. of rows in the garden = x
∴ Total no. of trees in the garden = x × x = x²
According to the sum x² = 1521
x = \(\sqrt{1521}=\sqrt{39 \times 39}\) = 39
∴ No. of trees for each row = 39
No. of rows in the garden = 39

Question 6.
A school collected ₹ 2601 as fees from its students. If fee paid by each student and number students in the school were equal, how many students were there in the school?
Solution:
Let the no. of students in a school = x say
The (amount) fee paid by each student = ₹ x
Amount collected by all the students
= x × x = x²
According to the sum
∴ x² = 2601
x = \(\sqrt{2601}=\sqrt{51 \times 51}\) = 51
∴ x = 51
∴ No. of students in the school = 51

Question 7.
The product of two numbers is 1296. If one number is 16 times the other, find the two numbers?
Solution:
Given that the product of two numbers = 1296.
Let the second number = x say
Then first number = 16 × x = 16x
∴ The product of two numbers
= x × 16x= 16x²
According to the sum
16x² = 1296
⇒ x² = \(\frac { 1296 }{ 16 }\) = 81
⇒ x² = 81
⇒ x = \(\sqrt{8} \overline{1}=\sqrt{9 \times 9}\) = 9
⇒ x = 9
∴ The first number = 16x
= 16 × 9
=144
The second number = x = 9

Question 8.
7921 soldiers sat in an auditorium in such a way that there are as many soldiers in a row as there are rows in the auditorium. How many rows are there in the auditorium’?
Solution:
Let the number of soldiers sat in an auditorium for each row = x say
∴ No. of rows in an auditorium = x
∴ Total no. of soldiers = x × x = x²
According to the sum,
x² = 7921
x = \(\sqrt{7921}=\sqrt{89 \times 89}\) = 89
∴ No. of rowS in an auditorium = 89

Question 9.
The area of a square field is 5184 m². Find the area of a rectangular field, whose perimeter is equal to the perimeter of the square field and whose length is twice of its breadth.
Solution:
Area of a square field = 5184 m²
A = s² = 5184
:. s = \(\sqrt{5184}=\sqrt{72 \times 72}\) = 72
∴ s = 72
∴ Perimeter of the square field = 4 × s
= 4 × 72
= 288 m
According to the sum,
Perimeter of a rectangular field
= Perimeter of a square field = 288 m
Let the breadth of a rectangular field
= x m say
∴ Length = 2 × x = 2 × m
∴ Perimeter of the rectangular field
= 2 (1 + b)
= 2 (2x + x)
= 2 × 3x
= 6x
∴ 6x = 288 .
x = \(\frac { 288 }{ 6 }\)
x = 48
∴ Breadth of the rectangular field
= x = 48 m
Length of the rectangular field = 2x
= 2 × 48
=96m
∴ Area of the rectangular field
= l × b
= 96 × 48
= 4608 m²

AP State Syllabus AP Board 8th Class Maths Solutions Chapter 6 Square Roots and Cube Roots Ex 6.1 Textbook Questions and Answers.

AP State Syllabus 8th Class Maths Solutions 6th Lesson Square Roots and Cube Roots Exercise 6.1

Question 1.
What will be the units digit of the square of the following numbers?
(i) 39
(ii) 297
(iii) 5125
(iv) 7286
(v) 8742
Solution:

Number	Square of the units digIt	Units digit of a squared number
i) 39	92 = 9 x 9 = 81	1
ii) 297	72 = 7 x 7 = 49	9
Iii) 5125	52 = 5 x 5 = 25	5
iv) 7286	62 = 6 x 6 = 36	6
v) 8742	22 = 2 x 2 = 4	4

Question 2.
Which of the following numbers are perfect squares?
(i) 121
(ii) 136
(iii) 256
(iv) 321
(v) 600
Solution:

Number	Prime factorizatlon	Perfect square numbers Yes/No
i) 121	121 = 11 x 11 = 112	yes
ii) 136	136 = 8 x 17 = 2 x 2 x 2 x 17	No
iii) 256	256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 28 = (24)2	yes
iv) 321	321 = 3 x 107	No
v) 600	600 = 120 x 5 = 12 x 10 x 5 = 2 x 2 x 2 x 3 x 5 x 5	No

Question 3.
The following numbers are not perfect squares. Give reasons?
(i) 257
(ii) 4592
(iii) 2433
(iv) 5050
(v) 6098
Solution:
i) 257 → The units digit of the number is 7. So it is not a perfect square number.
ii) 4592 → The units digit of the number is 2. So it is not a perfect square number.
iii) 2433 → The units digit of the number is 3. So, it is not a perfect square number.
Iv) 5050 → The last two digits of the number are not two zero’s. So, it is not a perfect square number.
v) 6098 → The units digit of the number is 8. So It is not a perfect square number

Question 4.
Find whether the square of the following numbers are even or odd?
(i) 431
(ii) 2826
(iii) 8204
(iv) 17779
(v) 99998
Solution:

Number	Units digit of square of a number	Even / Odd
(i) 431	12 = 1	1 , odd
(ii) 2826	62 = 36	6, odd
(iii) 8204	42 = 16	6 , even
(iv)17779	92 = 81	1 , odd
(v) 99998	82 = 64	4 , even

Question 5.
How many numbers lie between the square of the following numbers
(i) 25; 26
(ii) 56; 57
(iii) 107;108
Solution:
The numbers lie between the square of the numbers are:
1) 25,26 → 2 x 25=50
ii) 56, 57 → 2 x 56 = 112
iii) 107, 108 → 2 x 107 = 214

Question 6.
Without adding, find the sum of the following numbers
(i) 1 + 3 + 5 + 7 + 9 =
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 =
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 =
Solution:
(i) 1 + 3 + 5 + 7 + 9 = (5)² = 5 x 5 = 25
[∵ Sum of ‘n’ consecutive odd number = n²]
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 9² = 9 x 9 = 81
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 = 13² = 13 x 13 = 169