NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles

NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 9 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 9 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 9 Mathematics are an important part of exams for Class 9 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 9 Mathematics and also download more latest study material for all subjects. Chapter 9 Areas of Parallelograms and Triangles is an important topic in Class 9, please refer to answers provided below to help you score better in exams

Chapter 9 Areas of Parallelograms and Triangles Class 9 Mathematics NCERT Solutions

Class 9 Mathematics students should refer to the following NCERT questions with answers for Chapter 9 Areas of Parallelograms and Triangles in Class 9. These NCERT Solutions with answers for Class 9 Mathematics will come in exams and help you to score good marks

Chapter 9 Areas of Parallelograms and Triangles NCERT Solutions Class 9 Mathematics

 

Exercise 9.1

Q.1) Which of the following figures lie on the same base and between the same parallels? In such a case, write the common base and the two parallels.

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Sol.1) (i) Trapezium ABCD and π›₯𝑃𝐷𝐢 lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium 𝑆𝑀𝑁𝑅 lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram 𝑃𝑄𝑅𝑆 and π›₯𝑅𝑇𝑄 lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and π›₯𝑃𝑄𝑅 do not lie on the same base but between the same parallel lines BC and AD.
(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and between the same parallel lines AD and BQ.
(vi) Parallelogram 𝑃𝑄𝑅𝑆 and parallelogram ABCD do not lie on the same base SR but between the same parallel lines SR and PQ.

Exercise 9.2

Q.1) In the figure, ABCD is a parallelogram, 𝐴𝐸 βŠ₯ 𝐷𝐢 and 𝐢𝐹 βŠ₯ 𝐴𝐷. If 𝐴𝐡 = 16 π‘π‘š, 𝐴𝐸 = 8 π‘π‘š and 𝐢𝐹 = 10 π‘π‘š, find 𝐴𝐷.
Sol.1) Area of parallelogram ABCD
= 𝐴𝐡 Γ— 𝐴𝐸
= 16 Γ— 8 π‘π‘š2 = 128 π‘π‘š2
Also, area of parallelogram ABCD
= 𝐴𝐷 Γ— 𝐹𝐢 = (𝐴𝐷 Γ— 10)π‘π‘š2
∴ 𝐴𝐷 Γ— 10 = 128
β‡’ 𝐴𝐷 = 128/10 = 12.8 π‘π‘š

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Q.2) If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that π‘Žπ‘Ÿ (𝐸𝐹𝐺𝐻) = 1/2 π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷).
Sol.2) Given : A parallelogram ABCD Β· E, F, G, H are mid-points of sides AB, BC, CD, DA respectively.
To Prove : π‘Žπ‘Ÿ (𝐸𝐹𝐺𝐻) = 1/2 π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷)
Construction : Join AC and HF.

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Proof : In Ξ”ABC, E is the mid-point of AB.
F is the mid-point of BC.
β‡’ 𝐸𝐹 is parallel to AC and 𝐸𝐹 = 1/2
𝐴𝐢 ... (i)
Similarly, in Δ𝐴𝐷𝐢, we can show that
𝐻𝐺 || 𝐴𝐢 and 𝐻𝐺 = 1/2
𝐴𝐢 ... (ii)
From (i) and (ii)
𝐸𝐹 || 𝐻𝐺 and 𝐸𝐹 = 𝐻𝐺
∴ 𝐸𝐹𝐺𝐻 is a parallelogram. [One pour of opposite sides is equal and parallel]
In quadrilateral ABFH, we have
𝐻𝐴 = 𝐹𝐡 π‘Žπ‘›π‘‘ 𝐻𝐴 || 𝐹𝐡 [𝐴𝐷 = 𝐡𝐢 β‡’ 1/2
𝐴𝐷 = 1/2
𝐡𝐢 β‡’ 𝐻𝐴 = 𝐹𝐡]
∴ 𝐴𝐡𝐹𝐻 is a parallelogram. [One pair of opposite sides is equal and parallel]
Now, triangle HEF and parallelogram 𝐻𝐴𝐡𝐹 are on the same base HF and between the same parallels HF and AB.
∴ Area of Δ𝐻𝐸𝐹 = Β½ area of HABF ... (iii)
Similarly, area of Δ𝐻𝐺𝐹 = Β½ area of HFCD ... (iv)
Adding (iii) and (iv),
Area of Δ𝐻𝐸𝐹 + area of Δ𝐻𝐺𝐹 = 1/2
(area of HABF + area of HFCD)
β‡’ π‘Žπ‘Ÿ (𝐸𝐹𝐺𝐻) = 1/2
π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷) Proved.

Q.3) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram π΄π΅πΆπ·. Show that π‘Žπ‘Ÿ (𝐴𝑃𝐡) = π‘Žπ‘Ÿ (𝐡𝑄𝐢).
Sol.3) Given : A parallelogram ABCD. P and Q are any points on DC and AD respectively.
To prove : π‘Žπ‘Ÿ (𝐴𝑃𝐡) = π‘Žπ‘Ÿ (𝐡𝑄𝐢)
Construction : Draw 𝑃𝑆 || 𝐴𝐷 and 𝑄𝑅 || 𝐴𝐡.
Proof : In parallelogram 𝐴𝐡𝑅𝑄, 𝐡𝑄 is the diagonal.
∴ area of Δ𝐡𝑄𝑅 = 1/2 area of 𝐴𝐡𝑅𝑄 ... (i)
In parallelogram CDQR, CQ is a diagonal.
∴ area of Δ𝑅𝑄𝐢 = 1/2 area of 𝐢𝐷𝑄𝑅 ... (ii)
Adding (i) and (ii), we have
area of Δ𝐡𝑄𝑅 + area of Δ𝑅𝑄𝐢
= 1/2 [area of 𝐴𝐡𝑅𝑄 + area of CDQR]
β‡’ area of Δ𝐡𝑄𝐢 = 1/2 area of 𝐴𝐡𝐢𝐷 ... (iii)
Again, in parallelogram 𝐷𝑃𝑆𝐴, 𝐴𝑃 is a diagonal.
∴ area of Δ𝐴𝑆𝑃 = 1/2 area of 𝐷𝑃𝑆𝐴 ... (iv)
In parallelogram BCPS, PB is a diagonal.
∴ area of Δ𝐡𝑃𝑆 = 1/2 area of 𝐡𝐢𝑃𝑆 ... (v)
Adding (iv) and (v)
area of Δ𝐴𝑆𝑃 + area of Δ𝐡𝑃𝑆 = 1/2 (area of DPSA + area of BCPS)
β‡’ area of Δ𝐴𝑃𝐡 = 1/2 (area of ABCD) ... (vi)
From (iii) and (vi), we have
area of Δ𝐴𝑃𝐡 = area of Δ𝐡𝑄𝐢. Proved.

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Q.4) In the figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) π‘Žπ‘Ÿ (𝐴𝑃𝐡) + π‘Žπ‘Ÿ (𝑃𝐢𝐷) = 1/2 π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷)
(ii) π‘Žπ‘Ÿ (𝐴𝑃𝐷) + π‘Žπ‘Ÿ (𝑃𝐡𝐢) = π‘Žπ‘Ÿ(𝐴𝑃𝐡) + π‘Žπ‘Ÿ (𝑃𝐢𝐷)

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Sol.4) Given : A parallelogram ABCD. P is a point inside it.
To prove : (i) π‘Žπ‘Ÿ (𝐴𝑃𝐡) + π‘Žπ‘Ÿ(𝑃𝐢𝐷) = 1/2 π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷)
(ii) π‘Žπ‘Ÿ (𝐴𝑃𝐷) + π‘Žπ‘Ÿ (𝑃𝐡𝐢) = π‘Žπ‘Ÿ (𝐴𝑃𝐡) + π‘Žπ‘Ÿ (𝑃𝐢𝐷)
Construction : Draw EF through P parallel to AB, and GH through P parallel to AD.
Proof : In parallelogram FPGA, AP is a diagonal,
∴ area of Δ𝐴𝑃𝐺 = area of Δ𝐴𝑃𝐹 ... (i)
In parallelogram BGPE, PB is a diagonal,
∴ area of Δ𝐡𝑃𝐺 = area of Δ𝐸𝑃𝐡 ... (ii)
In parallelogram DHPF, DP is a diagonal,
∴ area of Δ𝐷𝑃𝐻 = area of Ξ”DPF ... (iii)
In parallelogram HCEP, CP is a diagonal,
∴ area of Δ𝐢𝑃𝐻 = area of Δ𝐢𝑃𝐸 ... (iv)
Adding (i), (ii), (iii) and (iv)
area of Δ𝐴𝑃𝐺 + area of Δ𝐡𝑃𝐺 + area of Δ𝐷𝑃𝐻 + area of Δ𝐢𝑃𝐻
= area of Δ𝐴𝑃𝐹 + area of Δ𝐸𝑃𝐡 + area of Δ𝐷𝑃𝐹 + area Δ𝐢𝑃𝐸
β‡’ [area of Δ𝐴𝑃𝐺 + area of Δ𝐡𝑃𝐺] + [area of Δ𝐷𝑃𝐻 + area of Δ𝐢𝑃𝐻]
= [area of Δ𝐴𝑃𝐹 + area of Δ𝐷𝑃𝐹] + [area of Ξ”EPB + area of Δ𝐢𝑃𝐸]
β‡’ area of Δ𝐴𝑃𝐡 + area of Δ𝐢𝑃𝐷 = area of Δ𝐴𝑃𝐷 + area of Δ𝐡𝑃𝐢 ... (v)
But area of parallelogram ABCD
= area of Δ𝐴𝑃𝐡 + area of Δ𝐢𝑃𝐷 + area of Δ𝐴𝑃𝐷 + area of Δ𝐡𝑃𝐢 ... (vi)
From (v) and (vi)
area of Δ𝐴𝑃𝐡 + area of Δ𝑃𝐢𝐷 = 1/2 area of ABCD
or, π‘Žπ‘Ÿ (𝐴𝑃𝐡) + π‘Žπ‘Ÿ (𝑃𝐢𝐷) = 1/2 π‘Žπ‘Ÿ (𝐴𝐡𝐢𝐷)                    Proved.
(ii) From (v),
β‡’ π‘Žπ‘Ÿ (𝐴𝑃𝐷) + π‘Žπ‘Ÿ (𝑃𝐡𝐢) = π‘Žπ‘Ÿ (𝐴𝑃𝐡) + π‘Žπ‘Ÿ (𝐢𝑃𝐷)            Proved.

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Q.5) In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ (𝐴𝐡𝑅𝑆)
(ii) π‘Žπ‘Ÿ (𝐴𝑋𝑆) = 1/2 π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆)
Sol.5) Given : PQRS and ABRS are parallelograms and X is any point on side BR.

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To prove : (i) π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ (𝐴𝐡𝑅𝑆)

(ii) π‘Žπ‘Ÿ (𝐴𝑋𝑆) = 1/2 π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆)
Proof : (i) In Δ𝐴𝑆𝑃 and BRQ, we have
βˆ π‘†π‘ƒπ΄ = βˆ π‘…π‘„π΅ [Corresponding angles] ...(1)
βˆ π‘ƒπ΄π‘† = βˆ π‘„π΅π‘… [Corresponding angles] ...(2)
∴ βˆ π‘ƒπ‘†π΄ = βˆ π‘„π‘…π΅ [Angle sum property of a triangle] ...(3)
Also, 𝑃𝑆 = 𝑄𝑅 [Opposite sides of the parallelogram PQRS] ...(4)
So, Δ𝐴𝑆𝑃 β‰… Δ𝐡𝑅𝑄 [ASA axiom, using (1), (3) and (4)]
Therefore, area of Δ𝑃𝑆𝐴 = area of Δ𝑄𝑅𝐡 [Congruent figures have equal areas] ...(5)
Now, π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ (𝑃𝑆𝐴) + π‘Žπ‘Ÿ (𝐴𝑆𝑅𝑄]
= π‘Žπ‘Ÿ (𝑄𝑅𝐡) + π‘Žπ‘Ÿ (𝐴𝑆𝑅𝑄]
= π‘Žπ‘Ÿ (𝐴𝐡𝑅𝑆)
So, π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ (𝐴𝐡𝑅𝑆)                                    Proved.

(ii) Now, Δ𝐴𝑋𝑆 and ||π‘”π‘š 𝐴𝐡𝑅𝑆 are on the same base AS and between same parallels AS and BR
∴ area of Δ𝐴𝑋𝑆 = 1/2
area of ABRS
β‡’ area of Δ𝐴𝑋𝑆 = 1/2
area of 𝑃𝑄𝑅𝑆 [ π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ (𝐴𝐡𝑅𝑆]
β‡’ ar of (𝐴𝑋𝑆) = 1/2
ar of (𝑃𝑄𝑅𝑆)                                                           Proved.

Q.6) A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
Sol.6) The field is divided in three triangles.
Since triangle APQ and parallelogram PQRS are on the same base PQ and between the same parallels PQ and RS.
∴ π‘Žπ‘Ÿ (𝐴𝑃𝑄) = 1/2 π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆)
β‡’ 2π‘Žπ‘Ÿ (𝐴𝑃𝑄) = π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆)
But π‘Žπ‘Ÿ (𝑃𝑄𝑅𝑆) = π‘Žπ‘Ÿ(𝐴𝑃𝑄) + π‘Žπ‘Ÿ (𝑃𝑆𝐴) + π‘Žπ‘Ÿ (𝐴𝑅𝑄)
β‡’ 2 π‘Žπ‘Ÿ (𝐴𝑃𝑄) = π‘Žπ‘Ÿ(𝐴𝑃𝑄) + π‘Žπ‘Ÿ(𝑃𝑆𝐴) + π‘Žπ‘Ÿ (𝐴𝑅𝑄)
β‡’ π‘Žπ‘Ÿ (𝐴𝑃𝑄) = π‘Žπ‘Ÿ(𝑃𝑆𝐴) + π‘Žπ‘Ÿ(𝐴𝑅𝑄)
Hence, area of Δ𝐴𝑃𝑄 = area of Δ𝑃𝑆𝐴 + area of Δ𝐴𝑅𝑄.
To sow wheat and pulses in equal portions of the field separately, farmer sow wheat in Ξ”𝐴𝑃𝑄 and pulses in other two triangles or pulses in Δ𝐴𝑃𝑄 and wheat in other two triangles.

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NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles

The above provided NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 9 Mathematics textbook online or you can easily download them in pdf.Β The answers to each question in Chapter 9 Areas of Parallelograms and Triangles of Mathematics Class 9 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 9 Areas of Parallelograms and Triangles Class 9 chapter of Mathematics so that it can be easier for students to understand all answers.Β These solutions of Chapter 9 Areas of Parallelograms and Triangles NCERT Questions given in your textbook for Class 9 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 9.

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