NCERT Solutions for Class 9 Maths Chapter 9
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1, 9.2, 9.3 and Exercise 9.4 in English Medium prepared and updated for the academic session 2020-2021.
NCERT Textbook Solutions of Prashnavali 9.1, 9.2, 9.3 and Prashnavali 9.4 of Samantar Chaturbhujon aur Tribhujon ka kshetrfal in Hindi Medium is also available to study online or download in PDF free based on new CBSE Curriculum 2020-2021. All the textbook solutions for 2020-21 and Offline Apps are applicable for CBSE Board, Gujrat Board, Uttarakhand Board, UP Board, MP Board and all other Boards using NCERT Books 2020-2021 for their course book of the academic year 2020-2021.
NCERT Solutions for Class 9 Maths Chapter 9
Class 9: | Mathematics |
Chapter 9: | Areas of Parallelograms and Triangles |
9th Maths Chapter 9 Solutions in English and Hindi Medium
CBSE NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles all exercises are given below for new session 2020-2021. For other questions visit to Class 9 Maths main page. Join the Discussion Forum to share your knowledge and to solve your doubts.
9th Maths Exercise 9.1 Solutions
9th Maths Exercise 9.2 Solutions
9th Maths Exercise 9.3 Solutions
9th Maths Exercise 9.4 Solutions
Study Material for 2020-21
Class 9 Maths Exercise 9.1 and 9.2 Explanation in Videos
Class 9 Maths Exercise 9.3 and 9.4 Explanation in Videos
Class 9 Maths Exercise 9.4 Question 8 Solution
Class 9 Maths Exercise 9.1 and 9.2 Solutions in Video
Class 9 Maths Exercise 9.3 and 9.4 Solutions in Video
Important Notes on 9th Maths Chapter 9
1. Area of a figure is a number associated with the part of the plane enclosed by that figure.
2. Two congruent figures have equal areas but the converse need not be true.
3. A diagonal of a parallelogram divides it into two triangles of equal areas.
4. Two figures are said to be on the same base and between same parallels, if they have a common base and vertices opposite to the common base of each figure lie on a line parallel to the base.
5. Parallelograms on the same base and between same parallels are equal in area.
6. A parallelogram and a rectangle on the same base and between same parallels are equal in area.
Do you know?
1. Parallelograms on the same base and having equal areas lie between the same parallels.
2. Triangles on the same base and between the same parallels are equal in area.
3. If a parallelogram and a triangle are on the same bases and between the same parallels, them area of the triangle is half the area of parallelogram.
4. Area of a triangle is half the product of its base and the corresponding altitude.
5. Triangles with equal bases and equal areas have equal corresponding altitudes.
6. Triangles on the same base and having equal areas lie between the same parallels.
7. A median of a triangle divides it into two triangles of equal areas.
Important Questions on 9th Maths Chapter 9
यदि आधार DC है तो
ABCD का क्षेत्रफल = 1/2 × DC × AE … (1)
तथा यदि आधार AD है तो
ABCD का क्षेत्रफल = 1/2 × AD × FC … (2)
समीकरण (1) और (2) से
1/2 × DC × AE = 1/2 × AD × FC
⇒ DC × AE = AD × FC
⇒ AB × AE = AD × FC [∵ DC = AB]
⇒ 16 × 8 = AD × 10
⇒ AD = (16 × 8)/10 = 12.8
अतः, AD = 12.8 cm
Hence, ar(ABD) = ar(ACD) … (1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔEBC, ED is median. [∵ Given]
Hence, ar(EBD) = ar(ECD) … (2)
Subtracting equation (2) from (1), we get
ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)
⇒ ar(ABE) = ar(ACE)
Therefore, PO = OR and SO = OQ
In ΔPQS, PO is median. [∵ SO = OQ]
Hence, ar(PSO) = ar(PQO) … (1)
[∵ A median of a triangle divides it into two triangles of equal areas.]
Similarly, in ΔPQR, QO is median. [∵ PO = OR]
Hence, ar(PQO) = ar(QRO) … (2)
And in ΔQRS, RO is median. [∵ SO = OQ]
Hence, ar(QRO) = ar(RSO) … (3)
From the equations (1), (2) and (3), we get
ar(PSO) = ar(PQO) = ar(QRO) = ar(RSO)
Hence, in parallelogram PQRS, diagonals PR and QS divide it into four triangles in equal area.
अतः, ar(PQRS) = ar(ABRS)
[∵ एक ही आधार वाले और एक ही समांतर रेखाओं के बीच स्थित समांतर चतुर्भुज क्षेत्रफल में बराबर होते है]
त्रिभुज APQ और समांतर चतुर्भुज PQRS एक ही आधार PQ और एक ही समांतर रेखाओं PQ || SR के बीच स्थित हैं।
अतः, ar(APQ) = 1/2 ar(PQRS)
[∵ यदि एक त्रिभुज और एक समांतर चतुर्भुज एक ही आधार और एक ही समांतर रेखाओं के बीच स्थित हों, तो त्रिभुज का क्षेत्रफल समांतर चतुर्भुज के क्षेत्रफल का आधा होता है।]
अतः, किसान एक फसल को APQ में बो सकती है तथा दूसरी फसल को खेत के बचे हुए भाग ASP और ARQ में बो सकती है।
Therefore, DE || BC
[∵Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.]
Hence, ar(ABD) = ar(ABC)
[∵ Triangles on the same base (or equal bases) and between the same parallels are equal in area.]
Subtracting ar(ABO) form both the sides
ar(ABD) – ar(ABO) = ar(ABC) – ar(ABO)
⇒ ar(AOD) = ar(BOC)
∠F = 90° [∵आयत का कोण]
AD > AF [∵ समकोण त्रिभुज में कर्ण सबसे लंबी भुजा होती है]
दोनों ओर AB जोड़ने पर, AD + AB > AF + AB
2 से गुणा करने पर’ 2[AD + AB] > 2[AF + AB]
⇒ समांतर चतुर्भुज का परिमाप > आयत का परिमाप