Solution:<\/strong>
We know that sum of angles of a quadrilateral is 360\u00b0
Now, sum of three angles = 56\u00b0 + 115\u00b0 + 84\u00b0 = 255\u00b0
Fourth angle = 360\u00b0 \u2013 255\u00b0 = 105\u00b0 Ans.<\/p>\n\n\n\n
Question 2.<\/strong> Question 3.<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 7.<\/strong> Question 8.<\/strong> Question 9.<\/strong> Question 10.<\/strong> Question 1.<\/strong> Question 2.<\/strong> Question 3.<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 7.<\/strong> Question 8.<\/strong> Question 9.<\/strong> Question 10.<\/strong> Question 11.<\/strong> Question 12.<\/strong> Question 13.<\/strong> Question 14.<\/strong> Question 15.<\/strong> Question 16.<\/strong> Question 17.<\/strong> Question 18.<\/strong> Question 19.<\/strong> Question 20.<\/strong> Question 21.<\/strong> Question 1.<\/strong> Question 2.<\/strong> Question 3.<\/strong> Question 4.<\/strong> Question 5.<\/strong> Question 6.<\/strong> Question 8.<\/strong> Question 9.<\/strong> Question 10.<\/strong> Question 11.<\/strong> Question 12.<\/strong> But diagonals of a ||gm bisect each other PR and QS bisect each other. Question 13.<\/strong> RS Aggarwal Solutions for Class 9 Maths Chapter 9\u2013Quadrilaterals and Parallelograms<\/p>\n\n\n\n Download PDF<\/strong>: RS Aggarwal Solutions for Class 9 Maths Chapter 9\u2013Quadrilaterals and Parallelograms PDF<\/a><\/p>\n\n\n\n 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.<\/p>\n\n\n\n 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.<\/p>\n\n\n\n Why must I refer to the RS Aggarwal textbook? Why should you refer to RS Aggarwal textbook solutions on Indcareer? Does IndCareer cover RS Aggarwal Textbook solutions for Class 6-12?
Solution:<\/strong>
Sum of angles of a quadrilateral = 360\u00b0
Their ratio = 2 : 4 : 5 : 7
Let first angle = 2x
then second angle = 4x
third angle = 5x
and fourth angle = 7x
\u2234 2x + 4x + 5x + 7x = 360\u00b0
=> 18x = 360\u00b0
=> x = 360o18 = 20\u00b0
Hence, first angle = 2x = 2 x 20\u00b0 = 40\u00b0
Second angle = 4x = 4 x 20\u00b0 = 80\u00b0
Third angle = 5x = 5 x 20\u00b0 = 100\u00b0
and fourth angle = 7x = 7 x 20\u00b0 = 140\u00b0Ans.<\/p>\n\n\n\n
Solution:<\/strong>
In the trapezium ABCD
DC || AB
\u2234 \u2220 A + \u2220 D = 180\u00b0 (Co-intericr angles)
\u2234 55\u00b0+ \u2220D = 180\u00b0
\u2220D = 180\u00b0 \u2013 55\u00b0
\u2234 \u2220D = 125\u00b0
Similarly, \u2220B + \u2220C = 180\u00b0
(Co-interior angles)
=> 70\u00b0 + \u2220C = 180\u00b0
=> \u2220C = 180\u00b0 \u2013 70\u00b0
\u2220C = 110\u00b0
Hence \u2220C = 110\u00b0 and \u2220D = 125\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In the figure, ABCD is a square and \u2206 EDC is an equilateral triangles on DC. AE and BE are joined.
To Prove :<\/strong> (i) AE = BE
(ii) \u2220DAE = 15\u00b0
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q4.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q4.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In the figure,
BM \u22a5 AC, DN \u22a5 AC.
BM = DN
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q5.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In quadrilateral ABCD,
AB = AD and BC = DC
AC is joined.
To Prove :<\/strong> (i) AC bisects \u2220 A and \u2220 C
(ii) BE = DE
(iii) \u2220ABC = \u2220ADC
Const. Join BD.
Proof<\/strong> :<\/strong> (i) In \u2206 ABC and \u2206 ADB
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q6.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In square ABCD,
\u2220 PQR = 90\u00b0
PB = QC = DR
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q7.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q7.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In quadrilateral ABCD, O is any point inside it. OA, OB, OC and OD are joined.
To Prove :<\/strong> OA + OB + OC + OD > AC + BD
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q8.1.png\")
<\/figure>\n\n\n\n![\"\"](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q8.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> In quadrilateral ABCD, AC is its one diagonal.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q9.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q9.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q9.3.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given :<\/strong> A quadrilateral ABCD
To Prove :<\/strong> \u2220A + \u2220B + \u2220C + \u2220D = 360\u00b0
Const. Join AC.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q10.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9A-Q10.2.png\")
<\/figure>\n\n\n\nEx 9A<\/h4>\n\n\n\n
Solution:<\/strong>
In parallelogram ABCD.
\u2220 A = 72\u00b0
But \u2220 A = \u2220 C (opposite angle of a ||gm)
\u2234 \u2220 C = 72\u00b0
\u2234 AD || BC
\u2234 \u2220 A + \u2220 B = 180\u00b0 (co-interior angles)
=> 72\u00b0 + \u2220B = 180\u00b0
=> \u2220B = 180\u00b0 \u2013 72\u00b0
=> \u2220B = 108\u00b0
But \u2220 B = \u2220 D (opposite angles of a ||gm)
\u2234 \u2220D = 108\u00b0
Hence \u2220D = 108\u00b0, \u2220 C = 72\u00b0 and \u2220 D = 108\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
In || gm ABCD, BD is its diagonal
and \u2220DAB = 80\u00b0 and \u2220DBC = 60\u00b0
\u2234AB || DC
\u2234\u2220DAB + \u2220ADC \u2013 180\u00b0
(co-interior angles)
=> 80\u00b0 + \u2220ADC = 180\u00b0
=> \u2220 ADC = 180\u00b0 \u2013 80\u00b0
=> \u2220 ADC = 100\u00b0
But \u2220 ADB = \u2220 DBC (Alternate angles)
\u2234\u2220 ADB = 60\u00b0
But \u2220 ADB + \u2220 CDB = 100\u00b0
(\u2234\u2220 ADC = 100\u00b0)
60\u00b0 + \u2220CDB = 100\u00b0
=> \u2220CDB = 100\u00b0 \u2013 60\u00b0 = 40\u00b0
Hence \u2220CDB = 40\u00b0 and \u2220 ADB = 60\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD,
\u2220 A = 60\u00b0 Bisectors of \u2220 A and \u2220 B meet DC at P.
To Prove : (i) \u2220APB = 90\u00b0
(ii) AD = DP and PB = PC = BC
(iii) DC = 2AD
Proof: \u2234 AD || B (opposite sides of a ||gm)
\u2234\u2220 A + \u2220 B = 180\u00b0 (co-interior angles)
But AP and BP are the bisectors of \u2220A and \u2220B
\u223412\u2220A + 12 \u2220B = 90\u00b0
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q3.1.png\")
<\/figure>\n\n\n\n
=> \u2220PAB + \u2220PBA = 90\u00b0
But in \u2206 APB,
\u2220PAB + \u2220PBA + \u2220APB = 180\u00b0 (angles of a triangle)
=> 90\u00b0 + \u2220APB = 180\u00b0
=> \u2220APB = 180\u00b0 \u2013 90\u00b0 = 90\u00b0
Hence \u2220APB = 90\u00b0
(ii) \u2220 A + \u2220 D = 180\u00b0 (co-interior angles)
and \u2220 A \u2013 60\u00b0
\u2234 \u2220D = 180\u00b0 \u2013 60\u00b0 = 120\u00b0
But \u2220DAP = 12 \u2220A = 12 x 60\u00b0 = 30\u00b0
\u2234\u2220DPA = 180\u00b0 \u2013 (\u2220DAP + \u2220D)
= 180\u00b0 \u2013 (30\u00b0 + 120\u00b0)
= 180\u00b0 \u2013 150\u00b0 = 30\u00b0
\u2220DAP = \u2220DPA (each = 30\u00b0)
Hence AD = DP (sides opposite to equal angles)
In \u2206 BCP,
\u2220 C = 60\u00b0 (opposite to \u2220 A)
\u2220CBP = 12 \u2220 B = 12 x 120\u00b0 = 60\u00b0
But \u2220CPB + \u2220CBP + \u2220C = 180\u00b0
(Angles of a triangle)
=> \u2220CPB + 60\u00b0 + 60\u00b0 = 180\u00b0
=> \u2220CPB + 120\u00b0 = 180\u00b0
=> \u2220CPB = 180\u00b0 \u2013 120\u00b0 = 60\u00b0
\u2206 CBP is an equilateral triangle and BC = CP = BP
=> PB \u2013 PC = BC
(iii) DC = DP + PC
= AD + BC
(\u2234 DP = AD and PC = BC proved)
= AD + AD (\u2234 AD = BC opposite sides of a ||gm)
= 2AD
Hence DC = 2AD.
Hence proved.<\/p>\n\n\n\n
Solution:<\/strong>
In ||gm ABCD,
AC and BD are joined
\u2220BAO = 35\u00b0, \u2220 DAO = 40\u00b0
\u2220COD = 105\u00b0
\u2234 \u2220AOB = \u2220COD
(vertically opposite angles)
\u2234\u2220AOB = 105\u00b0
(i) Now in \u2206 AOB,
\u2220ABO + \u2220AOB + \u2220OAB = 180\u00b0
(angles of a triangle)
=> \u2220ABO + 105\u00b0 + 35\u00b0 = 180\u00b0
=> \u2220ABO + 140\u00b0 = 180\u00b0
=> \u2220ABO = 180\u00b0 \u2013 140\u00b0
\u2220 ABO = 40\u00b0
(ii) \u2234 AB || DC
\u2234 \u2220 ABO = \u2220 ODC (alternate angles)
\u2234 \u2220 ODC = 40\u00b0
(iii) \u2234 AD || BC
\u2234 \u2220 ACB = \u2220 DAO or \u2220 DAC
(alternate angles)
= 40\u00b0
(iv) \u2234 \u2220 A + \u2220 B = 180\u00b0 (co-interior angles)
=> (40\u00b0 + 35\u00b0) + \u2220B = 180\u00b0
=> \u2220B = 180\u00b0 \u2013 75\u00b0 = 105\u00b0
=> \u2220 CBD + \u2220ABO = 105\u00b0
=> \u2220CBD + 40\u00b0 = 105\u00b0
=> \u2220CBD = 105\u00b0 \u2013 40\u00b0 = 65\u00b0
Hence \u2220 CBD = 65\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
In ||gm ABCD
( \u2220A = (2x + 25)\u00b0 and \u2220 B = (3x \u2013 5)\u00b0
\u2234AD || BC (opposite sides of parallelogram)
\u2234\u2220 A + \u2220B = 180\u00b0 (co-interior angles)
=> 2x + 25\u00b0 + 3x \u2013 5\u00b0 = 180\u00b0
=> 5x + 20\u00b0 = 180\u00b0
=> 5x = 180\u00b0 \u2013 20\u00b0
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q5.1.png\")
<\/figure>\n\n\n\n
=> 5x = 160\u00b0 => x = 160o5 = 32\u00b0
\u2234x = 32\u00b0
Now \u2220A = 2x + 25\u00b0 = 2 x 32\u00b0 + 25\u00b0
= 64\u00b0 + 25\u00b0 = 89\u00b0
\u2220B = 3x \u2013 5 = 3 x 32\u00b0 \u2013 5\u00b0
= 96\u00b0 \u2013 5\u00b0 = 91\u00b0
\u2220 C = \u2220 A (\u2234 opposite angles of ||gm)
= 89\u00b0
Similarly \u2220B = \u2220D
\u2220D = 91\u00b0
Hence \u2220 A = 89\u00b0, \u2220 B = 91\u00b0, \u2220 C = 89\u00b0, \u2220D = 91\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
Let \u2220A and \u2220B of a ||gm ABCD are adjacent angles.
\u2220A + \u2220B = 180\u00b0
Let \u2220B = x
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q6.1.png\")
<\/figure>\n\n\n\n
Then \u2220 A = 45 x
\u2234 x + 45 x = 180\u00b0
95 x = 180\u00b0
=>180oX59 = 100\u00b0
\u2234 \u2220A = 45 x 100\u00b0 = 80\u00b0
and \u2220B = 100\u00b0
But \u2220 C = \u2220 A and \u2220 D = \u2220 B
(opposite angles of a || gm)
\u2234\u2220C = 80\u00b0, and \u2220D = 100\u00b0
Hence \u2220A = 80\u00b0, \u2220B = 100\u00b0, \u2220C = 80\u00b0 and \u2220 D = 100\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
Let the smallest angle \u2220 A and the other angle \u2220 B
Let \u2220 A = x
Then \u2220 B = 2x \u2013 30\u00b0
But \u2220 A + \u2220 B = 180\u00b0 (co-interior angles)
\u2234x + 2x \u2013 30\u00b0 = 180\u00b0
=> 3x = 180\u00b0 + 30\u00b0 = 210\u00b0
=> x = 210o3 = 70\u00b0
\u2234 \u2220 A = 70\u00b0
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q7.1.png\")
<\/figure>\n\n\n\n
and \u2220 B = 2x \u2013 30\u00b0 = 2 x 70\u00b0 \u2013 30\u00b0
= 140\u00b0 \u2013 30\u00b0 = 110\u00b0
But \u2220C = \u2220 A and \u2220D = \u2220B
(opposite angles of a ||gm)
\u2220C = 70\u00b0 and \u2220D = 110\u00b0
Hence \u2220A = 70\u00b0, \u2220B = 110\u00b0, \u2220C = 70\u00b0 and \u2220D = 110\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
In ||gm ABCD,
AB = 9.5 cm and perimeter = 30 cm
=> AB + BC + CD + DA = 30cm
=> AB + BC + AB + BC = 30 cm
( \u2234 AB = CD and BC \u2013 DA opposite sides)
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q8.1.png\")
<\/figure>\n\n\n\n
=> 2(AB + BC) = 30cm
=> AB + BC = 15cm
=> 9 5cm + BC = 15cm
\u2234BC = 15cm \u2013 9.5cm = 5.5cm
Hence AB = 9.5cm, BC = 5.5cm,
CD = 9.5cm and DA = 5.5cm Ans.<\/p>\n\n\n\n
Solution:<\/strong>
ABCD is a rhombus
AB = BC = CD = DA
(i)\u2234 AB || DC
\u2234 \u2220 B + \u2220 C = 180\u00b0 (co-interior angles)
=> 110\u00b0 + \u2220C = 180\u00b0
=> \u2220C = 180\u00b0 \u2013 110\u00b0 = 70\u00b0
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q9.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q9.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q9.3.png\")
<\/figure>\n\n\n\n\n\n
Solution:<\/strong>
In a rhombus,
Diagonals bisect each other at right angles
\u2234 AC and BC bisect each other at O at right angles.
But AC = 24 cm and BD = 18 cm
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q10.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Let ABCD he the rhombus whose diagonal are AC and BD which bisect each other at right angles at O.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q11.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q11.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
ABCD is a rectangle whose diagonals AC and BD bisect each other at O.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q12.1-1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q12.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q12.3.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
ABCD is a square. A line CX cuts AB at X and diagonal BD at O such that
\u2220 COD = 80\u00b0 and \u2220 OXA = x\u00b0
\u2234\u2220 BOX = \u2220 COD
(vertically opposite angles)
\u2234\u2220BOX = 80\u00b0
\u2234Diagonal BD bisects \u2220 B and \u2220 D
\u2234 \u2220ABO or \u2220ABD = \u2220 ADO or \u2220 ADB
\u2234 \u2220OBA or \u2220OBX = 45\u00b0
Now in \u2206 OBX,
Ext. \u2220 OXA = \u2220 BOX + \u2220 OBX
=>x\u00b0 = 80\u00b0 + 45\u00b0 = 125\u00b0 Ans.<\/p>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD, AC is joined. AL \u22a5 BD and CM \u22a5 BD
To prove :
(i) \u2206 ALD \u2245 \u2206 CMB
(ii) AL = CM
Proof : In \u2206 ALD and \u2206 BMC
AD = BC (opposite sides of ||gm)
\u2220L = \u2220M (each 90\u00b0)
\u2220 ADL = \u2220 CBM (Alternate angles)
\u2234 \u2206 ALD \u2245 \u2206 BMC. (AAS axiom)
\u2234 or A ALD \u2245 A CMB.
AL = CM (c.p.c.t.) Hence proved.<\/p>\n\n\n\n
Solution:<\/strong>
Given : In \u2206 ABCD, bisectors of \u2220A and \u2220B meet each other at P.
To prove : \u2220APB = 90\u00b0
Proof : AD || BC
\u2220A + \u2220B = .180\u00b0 (co-interior angles)
PA and PB are the bisectors of \u2220 A and \u2220B
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q15.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q15.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
In ||gm ABCD,
P and Q are the points on AD and BC respectively such that AP = 13 AD and CQ = 13 BC
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q16.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD, diagonals AC and BD bisect each other at O.
A line segment EOF is drawn, which meet AB at E and DC at F.
To -prove : OE = OF
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q17.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : ABCD is a ||gm.
AB is produced to E. Such that AB = BE
DE is joined which intersects BC in O.
To prove : ED bisects BC i.e. BO = OC
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q18.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD, E is the midpoint BC
DE is joined and produced to meet AB on producing at F.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q19.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : \u2206 ABC and lines are drawn through A, B and C parallel to respectively BC, CA and AB forming \u2206 PQR.
To prove : BC = 12 QR<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q20.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In \u2206 ABC, parallel lines are drawn through A, B and C respectively to the sides BC, CA and AB intersecting each other at P, Q and R.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9B-Q21.1.png\")
<\/figure>\n\n\n\nEx 9C<\/h4>\n\n\n\n
Solution:<\/strong>
Given : In trapezium ABCD,
AB || DC and E is the midpoint of AD.
A line EF ||AB is drawn meeting BC at F.
To prove : F is midpoint of BC
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q1.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD, E and F are the mid points of AB and CD respectively. A line segment GH is drawn which intersects AD, EF and BC at G, P and H respectively.
To prove : GP = PH
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q2.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In trapezium ABCD, AB || DC
P, Q are the midpoints of sides AD and BC respectively
DQ is joined and produced to meet AB produced at E
Join AC which intersects PQ at R.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q3.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In \u2206 ABC,
AD is the mid point of BC
DE || AB is drawn. BE is joined.
To prove : BE is the median of \u2206 ABC.
Proof : In \u2206 ABC
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q4.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In \u2206 ABC, AD and BE are the medians. DF || BE is drawn meeting AC at F.
To prove : CF = 14 BC.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q5.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In ||gm ABCD, E is mid point of DC.
EB is joined and through D, DEG || EB is drawn which meets CB produced at G and cuts AB at F.
To prove : (i)AD = 12 GC
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q6.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q6.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q6.3-1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q6.3.png\")
<\/figure>\n\n\n\n
Question 7.<\/strong>
Solution:<\/strong>
<\/p>\n\n\n\n\n\n
Given : In \u2206 ABC,
D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and FD are joined.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q7.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q7.3.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In \u2206 ABC, D, E and F are the mid points of sides BC, CA and AB respectively
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q8.1.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In rectangle ABCD, P, Q, R and S are the midpoints of its sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
To prove : PQRS is a rhombus.
<\/p>\n\n\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q9.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In rhombus ABCD, P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q10.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q10.2.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In square ABCD, P,Q,R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q11.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q11.2.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q11.3.png\")
<\/figure>\n\n\n\n
Solution:<\/strong>
Given : In quadrilateral ABCD, P, Q, R and S are the midpoints of PQ, QR, RS and SP respectively PR and QS are joined.
To prove : PR and QS bisect each other
Const. Join PQ, QR, RS and SP and AC
Proof : In \u2206 ABC,
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q12.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q12.2.png\")
<\/figure>\n\n\n\n
\u2234 PR and QS bisect each other<\/p>\n\n\n\n
Solution:<\/strong>
Given : ABCD is a quadrilateral. Whose diagonals AC and BD intersect each other at O at right angles.
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively. PQ, QR, QS and SP are joined.
<\/p>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q13.1.png\")
<\/figure>\n\n\n\n![\"RS](\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RS-Aggarwal-Class-9-Solutions-Chapter-9-Quadrilaterals-and-Parallelograms-Ex-9C-Q13.2.png\")
<\/figure>\n\n\n\nRS Aggarwal Solutions for Class 9 Maths Chapter 9: Download PDF<\/strong><\/h2>\n\n\n\n
Chapterwise RS Aggarwal Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n
About RS Aggarwal Class 9 Book<\/h2>\n\n\n\n
FAQs<\/h2>\n\n\n\n
<\/strong>RS Aggarwal is one of the most important reference books for high school grades and is recommended to every high school student. The book covers every single topic in detail. It goes in-depth and covers every single aspect of all the mathematics topics and covers both theory and problem-solving. The book is true of great help for every high school student. Solving a majority of the questions from the book can help a lot in understanding topics in detail and in a manner that is very simple to understand. Hence, as a high school student, you must definitely dwell your hands on RS Aggarwal!<\/p>\n\n\n\n
<\/strong>RS Aggarwal is a book that contains a few of the hardest questions of high school mathematics. Solving them and teaching students how to solve questions of such high difficulty is not the job of any neophyte. For solving such difficult questions and more importantly, teaching the problem-solving methodology to students, an expert teacher is mandatory!<\/p>\n\n\n\n
<\/strong>RS Aggarwal is available for grades 6 to 12 and hence our expert teachers have formulated detailed solutions for all the questions of each edition of the textbook. On our website, you’ll be able to find solutions to the RS Aggarwal textbook right from Class 6 to Class 12. You can head to the website and download these solutions for free. All the solutions are available in PDF format and are free to download!<\/p>\n\n\n\nRead More<\/h2>\n\n\n\n