AP Board Class 8 Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbook Solutions PDF: Download Andhra Pradesh Board STD 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Book Answers |

## Andhra Pradesh Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbooks Solutions PDF

Andhra Pradesh State Board STD 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbooks. These Andhra Pradesh State Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.## Andhra Pradesh State Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Books Solutions

Board | AP Board |

Materials | Textbook Solutions/Guide |

Format | DOC/PDF |

Class | 8th |

Subject | Maths |

Chapters | Maths Chapter 9 Area of Plane Figures Ex 9.1 |

Provider | Hsslive |

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## AP Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 8th Maths Chapter 9 Area of Plane Figures Ex 9.1 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams: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 + 12 AC × BF

= 4 × 4 + 12 × 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 + 12 CG × (ED + CF)

= 18 × 18 + 12 × 8 × (7+ 18)

[∵h = CG = 8 cm, FC = AB = 18 cm]

= 324 + 4 × 25

= 324 + 100

= 424 cm^{2}

iii)

Area of a figure ABCDEF

= Area of rectangle ABCD + area of trapezium ADEF

= AB × BC + 12 × GA × (EF + AD)

= 20 × 15 + 12 × 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

h_{1} = 6 cm,

h_{2} = 5 cm,

d = 10 cm.

Area of a quadrilateral ABCD

= 12 d(h_{1}+h_{2}) . .

= 12 AC × (DE + BF)

= 12 × 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 = 12 × b × h

= 12 × AC × BC

= 12 × 4 × 4 = 8 cm^{2}

Area of rectangle CDEB = l × b

= CD × DE

= 20 × 4

= 80 cm^{2}

Area of ADEF = 12 × b × h

= 12 × DF × DE

= 12 × 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

= 12 × h(a + b)

= 12 × 80 × (30 + 40)

= 40 (70)

= 2800 sq.m.

Area of Δ HCD =12 × HC × HD

= 12 × 40 × 80

= 1600 sq.m.

Area of Δ EID = 12 × El × ID

= 12 × 60 × 40

= 1200 sq.m.

Area of trapezium FGIE

= 12 × h(a + b)

= 12 × 70 × (50 + 60)

= 35 (110)

= 3850 sq.m.

Area of Δ FGA = 12 × FG × GA

= 12 × 25 × 50

= 1250 sq.m.

∴ Area of the field

= 2800 + 1600 + 1200 + 3850 + 1250

= 10700 sq.m.

ii) Area of ΔABK = 12 × KB × KA

= 12 x 30 × 50

= 750 sq.m.

Area of trapezium KBCI

= 12 × h(a + b)

= 12 × 60 (30 + 40)

= 30 (70)

= 2100 sq.m.

Area of trapezium ICDE

= 12 × h(a + b)

= 12 × 80 × (40 + 50)

= 40 (90)

= 3600 sq.m.

Area of Δ FHE = 12 × FH × HE

=12 × 20 × 40

= 400 sq.m.

Area of trapezium GJHF

= 12 × h(a + b)

= 12 × 80 × (40 + 20)

= 40 (60)

= 2400 sq.m.

∴ Area of ΔGJA = 12 × GJ × JA

= 12 × 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 cm^{2}, 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

= 12 × h(a + b)

= 12 × 16 x (5x + 3x)

= 8(8x)

= 64 x

According to the problem,

Area of the trapezium = 960 sq.cm.

∴ 64x = 960

x = 96064 = 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 m^{2}.

Solution:

The floor of a building is paved with 3000 tiles.

The shape of each tile is a rhombus.

Diagonals of a tile are

d_{1} = 45 cm, d_{2} = 30 cm.

Area of each tile = 12 × d_{1} d_{2}

= 12 × 45 × 30

= 675 sq.cm.

∴ Area of floor

= 3000 × 675

= 2025000 sq.cm

= 202500010000 sq m

(∵ 1 sq.m. = 10000 sq.cm) 2025

= 202510 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

= 12 × AF × (AB + CF) + 12 FE x (ED + CF)

= 12 × 7.5 × (15+30) + 12 x 7.5 x (15+30)

= 12 × 7.5 × 45 + 12 x 7.5 x 45

= 2 × (12 × 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 =12 × b × h

= 12 × BD × CF

= 12 × 15 × 15

( ∵ CF = 30 – 15 = 15 )

= 2252 = 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 Board Textbook Solutions PDF for Class 8th Maths

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- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం Ex 15.1 Textbook Solutions PDF
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- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం Ex 15.3 Textbook Solutions PDF
- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం Ex 15.4 Textbook Solutions PDF
- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం Ex 15.5 Textbook Solutions PDF
- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం Ex 15.6 Textbook Solutions PDF
- AP Board Class 8 Maths Chapter 15 సంఖ్యలతో ఆడుకుందాం InText Questions Textbook Solutions PDF

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