CBSE Class 9

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4.

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4

Question 1.
Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Solution:
Given: parallelogram ABCD and rectangle ABEF are on same base AB, and area of both are equal.
In rectangle ABEF, AB = EF and in parallelogram ABCD,
CD = AB ⇒ AB + CD = AB + EF ….(i)
We know that, the perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.

∴ BE < SC and AF < AD On adding both, we get, BC + AD > BE + AF …(ii)
⇒ BC + AD + AB + CD > BE + AF + AB + CD (Adding AB + CD on both sides)

⇒ AB + BC+ CD + AD > AB + BE + EF + AF [Put the values from Eq. (i)]
Hence, the perimeter of the parallelogram is greater than the perimeter of the rectangle.

Question 2.
In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC). Can you now answer the question that you have left in the Introduction’ of this chapter, whether the field of Budhia has been actually divided into three pares of equal area?
[Remark Note that by taking BD = DE = EC, the ∆ ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the sameway, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∆ ABC into n triangles of equal areas.]
Solution:
Given: ABC is a triangle and D and E are two points on BC, such that

BD = DE = EC
Let AO be the perpendicular to BC.
∴ ar ( ∆ABD) = \(\frac { 1 }{ 2 }\) x BD x AO
ar (∆ADE) = \(\frac { 1 }{ 2 }\) x DE x AO
and ar(∆AEC) = \(\frac { 1 }{ 2 }\) x EC x AO
Since, BD = DE = EC (Given)
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Yes, altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.

Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).

Solution:
Given: ABCD, DCFE, and ABFE and parallelograms
In ∆ADE and ∆BCF,
AD = BC (∵ ABCD is a parallelogram)
DE – CF (∵ DCFE is a parallelogram)
and AE = BF (∵ ABFE is a parallelogram)
Hence ∆ADE = ∆BCF
∴ ar (∆ADE) = ar (∆BCF)

Question 4.
In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar(BPC) = ax(DPQ).[Hint Join AC.]

Solution:
Given: ABCD is a parallelogram and AD || CQ, and AQ = CQ. Join the line segment AC.
Now, ∆ APC and ∆ BPC lie on the same base PC and between the same parallels PC and AB, therefore
ar(∆ APC) = ar(∆ BPC) …(i)
AD = CQ and AD || CQ (Given)

Thus, in quadrilateral ACQD, one pair of opposite sides is equal and parallel.
∴ ADQC is a parallelogram.
We know that, diagonals of a parallelogram bisect each other.
∴ CP = DP and AP = PQ ….(ii)
In ∆ APC and ∆ DPQ, we have

Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that

Solution:
Join AD and EC. Let x be the side of ∆ ABC. Then
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Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
[Hint From A and C, draw perpendiculars to BD.]
Solution:
Given: ABCD is a quadrilateral whose diagonals intersect at P.
Draw two perpendiculars AE and CF from A and Con BD, respectively. Now,
LHS = ar (∆ APB) x ar (∆ CPD)


From Eqs. (i) and (ii), we get, LHS = RHS
i.e., ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)

Question 7.
P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that

Solution:
(i) Given: P and Q are mid-points of AB and BC. Also, R is mid-point of AP.
Since, P and 0 are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC (By mid-point theorem)
Draw RM || AC || PQ
Also, draw QG ⊥ RM and MH ⊥ AC
∵ PQ || RM || AC and PR = RA
∴ QM = MC
In ∆ QGM and ∆ MHC,

Question 8.
In figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that

(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(ACFG)
(vii) ar(BCED) = ar(ABMN) + ar(ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler, proof of this theorem in Class X.
Solution:
(i) In ∆ABD and ∆MBC,
BC = BD (These are the sides of square)
MB = AB
and ∠ MBC = 90° + ∠ ABC
= ∠DBC + ∠ABC
= ∠ABD
∴ ∆MBC = ∆ABD (By SAS rule)

(ii) From part (i), ar(∆ MBC) = ar (∆ ABD) …(i)
But ar(∆ ABD) = \(\frac { 1 }{ 2 }\) ar (BYXD) …(ii)
(∵ ∆ ABD and rectangle BYXD lie on the same base and between same parallel between lines.)
From Eqs. (i) and (ii), we get
ar (∆MBC) = \(\frac { 1 }{ 2 }\) ar (BYXD) .. .(iii)
⇒ ar (BYXD) = 2 ar (∆MBC)

(iii) Now ar (∆MBC) = \(\frac { 1 }{ 2 }\) ar (ABMN) …..(iv)
(∵ ∆MBC and square ABMN lie on the same base MB and between same parallels MB and NC)
From Eqs. (iii) and (iv), we get
ar (BYXD) = ar (ABMN)

(iv) In ∆ ACE and ∆FCS,
AC = FC
and CE = BC (These are the sides of square)
∠ FCB = 90° + ∠ ACB = ∠ BCE + ∠ACB = ∠ACE
So, ∆ FCB = ∆ ACE (By SAS rule)

(v) From Eqs. (iv), ar(AACE) = ar(AFCB) …(vi)
But ar(∆ACE) = \(\frac { 1 }{ 2 }\) ar(CVXE)
(∵ Both lie on the same base CE and between same parallel lines CE and AX.)
From Eqs. (vi) and (vii), we get
ar (∆ACE) = \(\frac { 1 }{ 2 }\) ar (CYXE)
= ar (∆FCB) …(vii)
⇒ ar (CYXE) = \(\frac { 1 }{ 2 }\) ar (∆ FCB)

(vi) Now, ar(AFCB) = \(\frac { 1 }{ 2 }\) ar (ACFG) …(ix)
(∵Both lie on same base CF and between same parallel lines CF and BG)
From Eqs. (viii) and (ix), we get
\(\frac { 1 }{ 2 }\) ar (ALFG) = \(\frac { 1 }{ 2 }\) ar (CYXE)
⇒ ar (ACFG) = ar (CYXE)

(vii) Now, ar (BCED) = ar (BYXD) + ar (CYXE)
= ar (ABMN) + ar (ACFG) [From part (iii) and (vi)]
We hope the NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2.

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2

Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

Solution:
We know that,
Area of parallelogram = Base x Altitude
Given, AE = 8 cm CF = 10 cm and AB = 16cm
∴ ar (parallelogramABCD) = DC x AE
= 16 x 8 cm2 (∵ AE = 8 cm)…(i)
and ar (parallelogram ABCD) = AD x CF – AD x 10 ( ∵ CF = 10 cm)
From Eq. (i), we have,
16 x 8 = AD x 10

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = \(\frac { 1 }{ 2 }\) ar (ABCD).
Solution:
Given: E,F, G and H are respectively the mid-points of the sides AB, BC, CD and AD. Joint if, it will parallel to CD and AB.
Now, parallelogram HDCF and triangle HGF stand on the same base HF and lie between the same parallel lines DC and HF.

Similarly,parallelogram ABFH and triangle HEF stand on the same base HF and lie between the same parallel lines HF and AB.
ar (∆HEF) = \(\frac { 1 }{ 2 }\) ar (∆BFH) …(ii)
On adding Eqs. (i) and (ii), we get
ar (∆HGF) + ar(∆HEF) = \(\frac { 1 }{ 2 }\) [ar (HDCF) + ar (ABFH)]
⇒ ar (EFGH) = \(\frac { 1 }{ 2 }\) ar (ABCD)

Question 3.
P and Q are any two points lying on the sides DC and AD, respectively of a parallelogram ABCD. Show that ar (APB) = ar(BQC).
Solution:
Given: a parallelogram ABCD. P and Q are any two points lying on the sides DC and AD, respectively.

Now, parallelogram ABCD and ABQC stand on the same base BC and lie between the same parallel lines BC and AD.
∴ ar (∆ BQC) = \(\frac { 1 }{ 2 }\) ar(ABCD) …(i)
Similarly, ∆ APB and parallelogram ABCD stand on the same base AB and lie between the same parallels AB and CD.
∴ ar (∆ APB) = \(\frac { 1 }{ 2 }\) ar (ABCD) ….(ii)
From Eqs. (i) and (ii), we get
ar (∆ APB) = ar (∆ BQC)

Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = \(\frac { 1 }{ 2 }\) ar (ABCD)
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)

Solution:
Given: ABCD is a parallelogram. So, AB || CD, AD |[ BC.
(i) Now, draw MPR parallel to AB and CD both and also draw a perpendicular PS on AB.
∵ MR || AB and AM || BR

∴ ABRM is a parallelogram so
∴ ar (|| gm ABRM) = AB x PS …(i)
and ar (∆ APB) = \(\frac { 1 }{ 2 }\) x AB x PS
ar (∆ APB) = \(\frac { 1 }{ 2 }\) ar (|| gm ABRM)
ar (∆PCD) = \(\frac { 1 }{ 2 }\) ar (|| gm MRCD)
Now, ar (∆APB) + ar (∆ PCD) = \(\frac { 1 }{ 2 }\) ar (|| gm ABRM) + \(\frac { 1 }{ 2 }\) ar (||gm MRCD)
= \(\frac { 1 }{ 2 }\) ar ( || gm ABCD) …(ii)
(ii) Similarly, we can draw a line through P parallel to AD and through the point P draw perpendicular on AD, we cah prove that
ar (∆APD) + ar (∆PBC) = \(\frac { 1 }{ 2 }\) ar (|| gm ABCD) …(iii)
From Eqs. (ii) and (iii), we get
ar (∆APD) + ar (∆PBQ = ar (∆APB) + ar (∆PCD)
Hence proved.

Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \(\frac { 1 }{ 2 }\) ar (PQRS)

Solution:
Given: PQRS and ABRS both are parallelogram and x is any point on BR.
(i) Here, parallelogram PQRS and ABRS lies on the same base SR and between the same parallel lines SR and PB.
∴ ar (PQRS) = ar (ABRS) …(i)
(ii) Again, in parallelogram ABRS, ∆ AXS and parallelogram lies on the same base AS and between the same parallel lines AS and BR.
∴ ar (∆AXS) = \(\frac { 1 }{ 2 }\)ar (∆BRS) …(ii)
Now, from Eqs. (i) and (ii), we get
ar (∆AXS) = \(\frac { 1 }{ 2 }\) ar (∆ PQRS)

Question 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.
Solution:
Given: PQRS is a parallelogram and A is any point as RS. Now, join PA and PQ. Thus, the field will be divided into three parts and each part is in the shape of a triangle.
Since, the AAPQ and parallelogram PQRS lie on the same base PQ and between same parallel lines PQ and SR.

∴ ar (∆APQ) = \(\frac { 1 }{ 2 }\) ar (PQRS) ….(i)
Then, remaining
∴ ar (∆ASP) + ar (∆ARQ) = \(\frac { 1 }{ 2 }\) ar (PQRS) ….(ii)
Now, from Eqs. (i) and (ii), we get
ar (∆APQ) = ar (∆ASP) + ar (∆ARQ)
So, farmer has two options.
Either the farmer should sow wheat and pulses in ∆APS and ∆AQR or in ar [∆APQ and (∆APS and ∆AQR)] separately.
We hope the NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2, drop a comment below and we will get back to you at the earliest.