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Chapter 9 Areas of Parallelograms and Triangles Class 9 Mathematics NCERT Solutions
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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.
Sol.1) (i) Trapezium ABCD and π₯ππ·πΆ lie on the same DC and between the same parallel lines AB and DC.
(ii) Parallelogram PQRS and trapezium ππππ
lie on the same base SR but not between the same parallel lines.
(iii) Parallelogram πππ
π and π₯π
ππ lie on the same base QR and between the same parallel lines QR and PS.
(iv) Parallelogram ABCD and π₯πππ
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 πππ
π 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, π΄πΈ β₯ π·πΆ and πΆπΉ β₯ π΄π·. If π΄π΅ = 16 ππ, π΄πΈ = 8 ππ and πΆπΉ = 10 ππ, find π΄π·.
Sol.1) Area of parallelogram ABCD
= π΄π΅ Γ π΄πΈ
= 16 Γ 8 ππ2 = 128 ππ2
Also, area of parallelogram ABCD
= π΄π· Γ πΉπΆ = (π΄π· Γ 10)ππ2
β΄ π΄π· Γ 10 = 128
β π΄π· = 128/10 = 12.8 ππ
Q.2) If E, F, G, and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ππ (πΈπΉπΊπ») = 1/2 ππ (π΄π΅πΆπ·).
Sol.2) Given : A parallelogram ABCD Β· E, F, G, H are mid-points of sides AB, BC, CD, DA respectively.
To Prove : ππ (πΈπΉπΊπ») = 1/2 ππ (π΄π΅πΆπ·)
Construction : Join AC and HF.
Proof : In ΞABC, E is the mid-point of AB.
F is the mid-point of BC.
β πΈπΉ is parallel to AC and πΈπΉ = 1/2
π΄πΆ ... (i)
Similarly, in Ξπ΄π·πΆ, we can show that
π»πΊ || π΄πΆ and π»πΊ = 1/2
π΄πΆ ... (ii)
From (i) and (ii)
πΈπΉ || π»πΊ and πΈπΉ = π»πΊ
β΄ πΈπΉπΊπ» is a parallelogram. [One pour of opposite sides is equal and parallel]
In quadrilateral ABFH, we have
π»π΄ = πΉπ΅ πππ π»π΄ || πΉπ΅ [π΄π· = π΅πΆ β 1/2
π΄π· = 1/2
π΅πΆ β π»π΄ = πΉπ΅]
β΄ π΄π΅πΉπ» is a parallelogram. [One pair of opposite sides is equal and parallel]
Now, triangle HEF and parallelogram π»π΄π΅πΉ are on the same base HF and between the same parallels HF and AB.
β΄ Area of Ξπ»πΈπΉ = Β½ area of HABF ... (iii)
Similarly, area of Ξπ»πΊπΉ = Β½ area of HFCD ... (iv)
Adding (iii) and (iv),
Area of Ξπ»πΈπΉ + area of Ξπ»πΊπΉ = 1/2
(area of HABF + area of HFCD)
β ππ (πΈπΉπΊπ») = 1/2
ππ (π΄π΅πΆπ·) Proved.
Q.3) P and Q are any two points lying on the sides DC and AD respectively of a parallelogram π΄π΅πΆπ·. Show that ππ (π΄ππ΅) = ππ (π΅ππΆ).
Sol.3) Given : A parallelogram ABCD. P and Q are any points on DC and AD respectively.
To prove : ππ (π΄ππ΅) = ππ (π΅ππΆ)
Construction : Draw ππ || π΄π· and ππ
|| π΄π΅.
Proof : In parallelogram π΄π΅π
π, π΅π is the diagonal.
β΄ area of Ξπ΅ππ
= 1/2 area of π΄π΅π
π ... (i)
In parallelogram CDQR, CQ is a diagonal.
β΄ area of Ξπ
ππΆ = 1/2 area of πΆπ·ππ
... (ii)
Adding (i) and (ii), we have
area of Ξπ΅ππ
+ area of Ξπ
ππΆ
= 1/2 [area of π΄π΅π
π + area of CDQR]
β area of Ξπ΅ππΆ = 1/2 area of π΄π΅πΆπ· ... (iii)
Again, in parallelogram π·πππ΄, π΄π is a diagonal.
β΄ area of Ξπ΄ππ = 1/2 area of π·πππ΄ ... (iv)
In parallelogram BCPS, PB is a diagonal.
β΄ area of Ξπ΅ππ = 1/2 area of π΅πΆππ ... (v)
Adding (iv) and (v)
area of Ξπ΄ππ + area of Ξπ΅ππ = 1/2 (area of DPSA + area of BCPS)
β area of Ξπ΄ππ΅ = 1/2 (area of ABCD) ... (vi)
From (iii) and (vi), we have
area of Ξπ΄ππ΅ = area of Ξπ΅ππΆ. Proved.
Q.4) In the figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ππ (π΄ππ΅) + ππ (ππΆπ·) = 1/2 ππ (π΄π΅πΆπ·)
(ii) ππ (π΄ππ·) + ππ (ππ΅πΆ) = ππ(π΄ππ΅) + ππ (ππΆπ·)
Sol.4) Given : A parallelogram ABCD. P is a point inside it.
To prove : (i) ππ (π΄ππ΅) + ππ(ππΆπ·) = 1/2 ππ (π΄π΅πΆπ·)
(ii) ππ (π΄ππ·) + ππ (ππ΅πΆ) = ππ (π΄ππ΅) + ππ (ππΆπ·)
Construction : Draw EF through P parallel to AB, and GH through P parallel to AD.
Proof : In parallelogram FPGA, AP is a diagonal,
β΄ area of Ξπ΄ππΊ = area of Ξπ΄ππΉ ... (i)
In parallelogram BGPE, PB is a diagonal,
β΄ area of Ξπ΅ππΊ = area of ΞπΈππ΅ ... (ii)
In parallelogram DHPF, DP is a diagonal,
β΄ area of Ξπ·ππ» = area of ΞDPF ... (iii)
In parallelogram HCEP, CP is a diagonal,
β΄ area of ΞπΆππ» = area of ΞπΆππΈ ... (iv)
Adding (i), (ii), (iii) and (iv)
area of Ξπ΄ππΊ + area of Ξπ΅ππΊ + area of Ξπ·ππ» + area of ΞπΆππ»
= area of Ξπ΄ππΉ + area of ΞπΈππ΅ + area of Ξπ·ππΉ + area ΞπΆππΈ
β [area of Ξπ΄ππΊ + area of Ξπ΅ππΊ] + [area of Ξπ·ππ» + area of ΞπΆππ»]
= [area of Ξπ΄ππΉ + area of Ξπ·ππΉ] + [area of ΞEPB + area of ΞπΆππΈ]
β area of Ξπ΄ππ΅ + area of ΞπΆππ· = area of Ξπ΄ππ· + area of Ξπ΅ππΆ ... (v)
But area of parallelogram ABCD
= area of Ξπ΄ππ΅ + area of ΞπΆππ· + area of Ξπ΄ππ· + area of Ξπ΅ππΆ ... (vi)
From (v) and (vi)
area of Ξπ΄ππ΅ + area of ΞππΆπ· = 1/2 area of ABCD
or, ππ (π΄ππ΅) + ππ (ππΆπ·) = 1/2 ππ (π΄π΅πΆπ·) Proved.
(ii) From (v),
β ππ (π΄ππ·) + ππ (ππ΅πΆ) = ππ (π΄ππ΅) + ππ (πΆππ·) Proved.
Q.5) In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ππ (πππ
π) = ππ (π΄π΅π
π)
(ii) ππ (π΄ππ) = 1/2 ππ (πππ
π)
Sol.5) Given : PQRS and ABRS are parallelograms and X is any point on side BR.
To prove : (i) ππ (πππ π) = ππ (π΄π΅π π)
(ii) ππ (π΄ππ) = 1/2 ππ (πππ
π)
Proof : (i) In Ξπ΄ππ and BRQ, we have
β πππ΄ = β π
ππ΅ [Corresponding angles] ...(1)
β ππ΄π = β ππ΅π
[Corresponding angles] ...(2)
β΄ β πππ΄ = β ππ
π΅ [Angle sum property of a triangle] ...(3)
Also, ππ = ππ
[Opposite sides of the parallelogram PQRS] ...(4)
So, Ξπ΄ππ β
Ξπ΅π
π [ASA axiom, using (1), (3) and (4)]
Therefore, area of Ξπππ΄ = area of Ξππ
π΅ [Congruent figures have equal areas] ...(5)
Now, ππ (πππ
π) = ππ (πππ΄) + ππ (π΄ππ
π]
= ππ (ππ
π΅) + ππ (π΄ππ
π]
= ππ (π΄π΅π
π)
So, ππ (πππ
π) = ππ (π΄π΅π
π) Proved.
(ii) Now, Ξπ΄ππ and ||ππ π΄π΅π
π are on the same base AS and between same parallels AS and BR
β΄ area of Ξπ΄ππ = 1/2
area of ABRS
β area of Ξπ΄ππ = 1/2
area of πππ
π [ ππ (πππ
π) = ππ (π΄π΅π
π]
β ar of (π΄ππ) = 1/2
ar of (πππ
π) 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.
β΄ ππ (π΄ππ) = 1/2 ππ (πππ
π)
β 2ππ (π΄ππ) = ππ (πππ
π)
But ππ (πππ
π) = ππ(π΄ππ) + ππ (πππ΄) + ππ (π΄π
π)
β 2 ππ (π΄ππ) = ππ(π΄ππ) + ππ(πππ΄) + ππ (π΄π
π)
β ππ (π΄ππ) = ππ(πππ΄) + ππ(π΄π
π)
Hence, area of Ξπ΄ππ = area of Ξπππ΄ + area of Ξπ΄π
π.
To sow wheat and pulses in equal portions of the field separately, farmer sow wheat in Ξπ΄ππ and pulses in other two triangles or pulses in Ξπ΄ππ and wheat in other two triangles.
NCERT Solutions Class 9 Mathematics Chapter 1 Number Systems |
NCERT Solutions Class 9 Mathematics Chapter 2 Polynomials |
NCERT Solutions Class 9 Mathematics Chapter 3 Coordinate Geometry |
NCERT Solutions Class 9 Mathematics Chapter 4 Linear Equations In Two Variables |
NCERT Solutions Class 9 Mathematics Chapter 5 Introduction to Euclid's Geometry |
NCERT Solutions Class 9 Mathematics Chapter 6 Lines and Angles Variables |
NCERT Solutions Class 9 Mathematics Chapter 7 Triangles |
NCERT Solutions Class 9 Mathematics Chapter 8 Quadrilaterals |
NCERT Solutions Class 9 Mathematics Chapter 9 Areas of Parallelograms and Triangles |
NCERT Solutions Class 9 Mathematics Chapter 10 Circles |
NCERT Solutions Class 9 Mathematics Chapter 11 Construction |
NCERT Solutions Class 9 Mathematics Chapter 12 Herons Formula |
NCERT Solutions Class 9 Mathematics Chapter 13 Surface Area and Volume |
NCERT Solutions Class 9 Mathematics Chapter 14 Statistics |
NCERT Solutions Class 9 Mathematics Chapter 15 Probability |
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