AP Board Class 9 Maths Chapter 11 Areas InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 11 Areas InText Questions Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

AP Board Class 9 Maths Chapter 11 Areas InText Questions Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 11 Areas InText Questions Book Answers

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Andhra Pradesh Board Class 9th Maths Chapter 11 Areas InText Questions Textbooks Solutions PDF

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Andhra Pradesh State Board Class 9th Maths Chapter 11 Areas InText Questions Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 11 Areas InText Questions
Provider	Hsslive

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AP Board Class 9th Maths Chapter 11 Areas InText Questions Textbooks Solutions with Answer PDF Download

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AP State Syllabus 9th Class Maths Solutions 11th Lesson Areas InText Questions

Activity

Question

Observe Figure I and II. Find the area of both. Are the areas equal?
Trace these figures on a sheet of paper, cut them. Cover fig. 1 with fig. II. Do they cover each other completely? Are they congruent?
Observe fig. Ill and IV. Find the areas of both. What do you notice?
Are they congruent?
Now, trace these figures on sheet of paper. Cut them let us cover fig. Ill by fig. IV by coinciding their bases (length of same side).
As shown in figure V are they covered completely?
We conclude that Figures I and II are congruent and equal in area. But figures III and IV are equal in area but they are not congruent

Think, Discuss and Write

Question 1.
If 1 cm represents 5 in, what would be an area of 6 cm² represents ?
[Page No. 247]
Solution:
1 cm² = 5 m
1 cm² = 1 cm × 1 cm = 5m × 5m = 25m²
∴ 6 cm² = 6 × 25 m² = 150 m²

Question 2.
Rajni says 1 sq. m = 100² sq. cm. Do you agree ? Explain.
Solution:
No
1 sq. m = 100 cm²

Think, Discuss and Write

Question
Which of the following figures lie on the same base and between the same parallels? In such cases, write the common base and two parallels. [Page No. 249]

Solution:
a) In figure (a) ΔPCD and □ ABCD lie on same base CD and between the same parallels AB//CD.
b) No,
c) ΔTRQ and □ PQRS lie on the same base QR and between the same par-allels PS//QR.
d) ΔAPD and □ ABCD lie on the same base AD and between the same par-allels AD//BC.
e) No.

Activity

Question
Take a graph sheet and draw two par-allelograms ABCD and PQCD on it as show in the Figure, [Page No. 250]

The parallelograms are on the same base DC and between the same parallels PB and DC. Clearly the part DCQA is common between the two parallelo-grams. So if we can show that ΔDAP and ΔCBQ have the same area then we can say ar(PQCD) = ar(ABCD)

Activity

Draw pairs of triangles one the same base or ( equal bases) and between the same parallels on the graph sheet as shown in the Figure.
Let AABC and ADBC be the two triangles lying on the same base BC and between parallels BC and FE.
Draw CE II AB and BF II CD. Parallelograms AECB and FDCB are on the same base BC and are between the same parallels BC and EF.

Thus ar (AECB) = ar (FDCB).
We can see ar (ΔABC) = 12 ar (parallelogram AECB) …………….(i)
and ar (ΔDBC) = 12 ar (parallelogram FDCB) ……………..(ii)
From (i) and (ii), we get ar (ΔABC) = ar (ΔDBC)
You can also find the areas of ΔABC and ΔDBC by the method of counting the squares in graph sheet as we have done in the earlier activity and check the areas are whether same.[Page No. 254]

Think, Discus and Write

Draw two triangles ABC and DBC on the same base and between the same parallels as shown in the figure with P as the point of intersection of AC and BD. Draw CE//BA and BF//CD such that E and F lie on line AD.
Can you show ar(ΔPAB) = ar(ΔPDQ) ?[Page No. 254]

[Hint: These triangles are not congruent but have equal areas.
Solution:
□ ABCE = 2 × ΔABC
[∵ ΔABC; □ABCE lie on the same base BC and between the same parallels BC // AE]
ΔABC = 12 × □ ABCE ……………(1)
Also □ BCDF = 2 × ΔBCD…………..
[∵ΔBCD and □ BCDE lie on the same base BC and between , the same parallels BC//DE]
ΔBCD = 12 × □ BCDF ……………… (2)
But □ABCE = □BCDF
[ ∵ □ABCE and □BCDF lie on the same base BC and between the same parallels BC//FE]
From (1) & (2); ΔABC = ΔBCD
ΔPAB + ΔPBC = ΔPBC + ΔPDC
⇒ ΔPAB = ΔPDC
Hence proved.

AP Board Textbook Solutions PDF for Class 9th Telugu

Andhra Pradesh Board Class 9th Maths Chapter 11 Areas InText Questions Textbooks for Exam Preparations

Andhra Pradesh Board Class 9th Maths Chapter 11 Areas InText Questions Textbook Solutions can be of great help in your Andhra Pradesh Board Class 9th Maths Chapter 11 Areas InText Questions exam preparation. The AP Board STD 9th Maths Chapter 11 Areas InText Questions Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 11 Areas InText Questions Books State Board syllabus with maximum efficiency.

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