NCERT Solutions for Class 9 Maths Chapter 9

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.1, Exercise 9.2, Exercise 9.3 and Exercise 9.4 in English Medium and Hindi Medium updated for academic session 2020-21.

Download प्रश्नावली 9.1, प्रश्नावली 9.2, प्रश्नावली 9.3 and प्रश्नावली 9.4 of समांतर चतुर्भुजों और त्रिभुजों के क्षेत्रफल in Hindi Medium to study online or download in PDF free for academic session 2020-21. Download NCERT Solutions 2020-21 and Offline Apps for CBSE Board, Gujrat board, Uttarakhand Board, UP Board, MP Board and all other Boards using NCERT Books 2020-21 for the exams of academic year 2020-2021.

NCERT Solutions for Class 9 Maths Chapter 9

Class:	9
Subject:	Maths – गणित
Chapter 9:	Areas of Parallelograms and Triangles

9th Maths Chapter 9 Solutions in English & Hindi Medium

CBSE NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles all exercises are given below for new session 2020-21. 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
- Study Material for 2020-21 – English Medium
- Study Material for 2020-21 – Hindi Medium

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

ABCD एक समांतर चतुर्भुज है, AE ⊥ DC और CF ⊥ AD है। यदि AB = 16 cm, AE = 8 cm और CF = 10 cm है, तो AD ज्ञात कीजिए।

हम जानते हैं कि समांतर चतुर्भुज का क्षेत्रफल = 1/2 × आधार × शीर्षलंब
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

E is any point on median AD of a ΔABC. Show that ar (ABE) = ar (ACE).

In ΔABC, AD is median. [∵ Given]
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)

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Diagonals of parallelogram bisect each other.
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.

PQRS और ABRS समांतर चतुर्भुज हैं तथा X भुजा BR पर स्थित कोई बिंदु है । दर्शाइए कि ar(PQRS) = ar(ABRS)

समांतर चतुर्भुज PQRS और ABRS एक ही आधार RS और एक ही समांतर रेखाओं SR || PB के बीच स्थित हैं।

अतः, ar(PQRS) = ar(ABRS)

[∵ एक ही आधार वाले और एक ही समांतर रेखाओं के बीच स्थित समांतर चतुर्भुज क्षेत्रफल में बराबर होते है]

एक किसान के पास समांतर चतुर्भुज PQRS के रूप का एक खेत था। उसने RS पर स्थित कोई बिंदु A लिया और उसे P और Q से मिला दिया। खेत कितने भागों में विभाजित हो गया है? इन भागों के आकार क्या हैं? वह किसान खेत में गेंहूँ और दालें बराबर-बराबर भागों में अलग-अलग बोना चाहती है। वह ऐसा कैसे करे?

खेत 3 भागों में विभाजित हो गया है: APS, APQ और ARQ
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।

D and E are points on sides AB and AC respectively of ΔABC such that ar (DBC) = ar (EBC). Prove that DE||BC.

ΔDBC and ΔEBC are on the same base BC and ar(DBC)=ar(EBC).
Therefore, DE || BC
[∵Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.]

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

Triangles ABD and ABC are on the same base AB and between same parallels, AB || CD.
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)

समांतर चतुर्भुज ABCD और आयत ABEF एक ही आधार पर स्थित हैं और उनके क्षेत्रफल बराबर हैं। दर्शाइए कि समांतर चतुर्भुज का परिमाप आयत के परिमाप से अधिक है।

AFD में,
∠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