RS Aggarwal Solutions for Class 9 Maths Chapter 10–Area

Class 9: Maths Chapter 10 solutions. Complete Class 9 Maths Chapter 10 Notes.

RS Aggarwal Solutions for Class 9 Maths Chapter 10–Area

RS Aggarwal 9th Maths Chapter 10, Class 9 Maths Chapter 10 solutions

Question 1.
Solution:
Given : In the figure, ABCD is a quadrilateral and
AB = CD = 5cm

Question 2.
Solution:
In ||gm ABCD,
AB = 10cm, altitude DL = 6cm
and BM is altitude on AD, and BM = 8 cm.

Question 3.
Solution:
Diagonals of rhombus are 16cm and 24 cm.
Area = 12 x product of diagonals
= 12 x 1st diagonal x 2nd diagonal
= 12 x 16 x 24
= 192 cm² Ans.

Question 4.
Solution:
Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm

Question 5.
Solution:
from the figure
(i) In ∆ BCD, ∠ DBC = 90°

Question 6.
Solution:
In the fig, ABCD is a trapezium. AB || DC
AB = 7cm, AD = BC = 5cm.
Distance between AB and DC = 4 cm.
i.e. ⊥AL = ⊥BM = 4cm.

Question 7.
Solution:
Given : In quad. ABCD. AL⊥BD and CM⊥BD.
To prove : ar(quad. ABCD)

Question 8.
Solution:
In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD

Question 9.
Solution:
Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : ar(∆ AOD) = ar(∆ BOC)

Question 10.
Solution:
Given : In the figure,
DE || BC.

Question 11.
Solution:
Given : In ∆ ABC, D and E are the points on AB and AC such that
ar( ∆ BCE) = ar( ∆ BCD)
To prove : DE || BC.
Proof : (∆ BCE) = ar(∆ BCD)
But these are on the same base BC.
Their altitudes are equal.
Hence DE || BC
Hence proved.

Question 12.
Solution:
Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.

Question 13.
Solution:
Given : In quad. ABCD.
A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.
To prove : ar( ∆ ABP) = ar(quad. ABCD).
Const. Join AC

Question 14.
Solution:
Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and
ar( ∆ ABC) = ar( ∆ DBC).

Question 15.
Solution:
Given : In ∆ ABC, AD is the median and P is a point on AD
BP and CP are joined
To prove : (i) ar(∆BDP) = ar(∆CDP)
(ii) ar( ∆ ABP) = ar( ∆ ACP)

Question 16.
Solution:
Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD
To prove : ar(∆ ABC) = ar(∆ ADC)
Proof : In ∆ ABD,

Question 17.
Solution:
In ∆ ABC,D is mid point of BC
and E is midpoint of AD and BE is joined.

Question 18.
Solution:
Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.
To prove : ar( ∆ BEC) = 12 ar( ∆ ABC)
Proof : In ∆ ABD,
E is midpoint of AD
BE is its median
ar(∆ EBD) = ar(∆ ABE)

Question 19.
Solution:
Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.
To prove that ar( ∆ BOE) = 18 ar(∆ ABC).

Question 20.
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC.
To prove : ar( ∆ AOB) = ar( ∆ AOD)
Const. Join BD which intersects AC at P
Proof : In ∆ OBD,
P is midpoint of BD
(Diagonals of ||gm bisect each other)

Question 21.
Solution:
Given : ABCD is a ||gm.
P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.
PQ, QR, RS and SP are joined.

Question 22.
Solution:
Given : In pentagon ABCDE,
EG || DA meets BA produced and
CF || DB, meets AB produced.
To prove : ar(pentagon ABCDE) = ar(∆ DGF)

Question 23.
Solution:
Given ; A ∆ ABC in which AD is the median.
To prove ; ar( ∆ ABD) = ar( ∆ ACD)
Const : Draw AE⊥BC.
Proof : Area of ∆ ABD

Question 24.
Solution:
Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.

Question 25.
Solution:
Given : In ∆ ABC,
D is a point on BC such that
BD = 12 DC
AD is joined.

Question 26.
Solution:
Given : In ∆ ABC, D is a point on BC such that
BD : DC = m : n
AD is joined.

RS Aggarwal Solutions for Class 9 Maths Chapter 10: Download PDF

RS Aggarwal Solutions for Class 9 Maths Chapter 10–Area

Download PDF: RS Aggarwal Solutions for Class 9 Maths Chapter 10–Area PDF

Chapterwise RS Aggarwal Solutions for Class 9 Maths :

• Chapter 1–Real Numbers
• Chapter 2–Polynomials
• Chapter 3–Introduction to Euclid’s Geometry
• Chapter 4–Lines and Triangles
• Chapter 5–Congruence of Triangles and Inequalities in a Triangle
• Chapter 6–Coordinate Geometry
• Chapter 7–Areas
• Chapter 8–Linear Equations in Two Variables
• Chapter 9–Quadrilaterals and Parallelograms
• Chapter 10–Area
• Chapter 11–Circle
• Chapter 12–Geometrical Constructions
• Chapter 13–Volume and Surface Area
• Chapter 14–Statistics
• Chapter 15–Probability

About RS Aggarwal Class 9 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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