NCERT Solutions for Class 9 Maths Exercise 9.4

NCERT Solutions for Class 9 Maths Exercise 9.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Areas of Parallelograms and Triangles Download as PDF

NCERT Solutions for Class 9 Mathematics Areas of Parallelograms and Triangles

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.

Ans. Given: Parallelogram ABCD and rectangle ABEF are on same base AB and between the same parallels AB and CF.

ar (gm ABCD) = ar (rect. ABEF)

To prove: AB + BC + CD + AD > AB + BE + EF + AF

Proof: AB = CD [ opposites sides of a parallelogram are always equal]

AB = EF [ opposites sides of a rectangle are always equal]

CD = EF

Adding AB both sides,

AB + CD = AB + EF ……….(i)

Off all the segments that can be drawn to a given line from a point not lying on it, the perpendicular segment is the shortest.

BE < BC and AF < AD

BC > BE and AD > AF

BC + AD > BE + AF ……….(ii)

From eq. (i) and (ii),

AB + CD + BC + AD = AB + EF + BE + AF

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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 know answer the question that you have left in the ‘introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

Ans. In ABC, points D and E divides BC in three equal parts such that BD = DE = EC.

BD = DE = EC = BC

Draw AFBC

ar (ABC) = ……….(i)

and ar (ABD) = ……….(ii)

= =

= ar (ABC) ……….(iii)

And ar (AEC) = ar (ABC) ……….(iv)

From (ii), (iii) and (iv),

ar (ABD) = ar (ADE) = ar (AEC)

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NCERT Solutions for Class 9 Maths Exercise 9.4

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

Ans. As we know that opposite sides of a parallelogram are always equal.

In parallelogram ABFE, AE = BF and AB = EF

In parallelogram DCFE, DE = CF and DC = EF

In parallelogram ABCD, AD = BC and AB = DC

Now in ADE and BCF,

AE = BF [Opposite sides of parallelogram ABFE]

DE = CF [Opposite sides of parallelogram DCFE]

And AD = BC [Opposite sides of parallelogram ABCD]

ADE BCF [By SSS congruency]

ar (ADE) = ar (BCF)

[ Area of two congruent figures is always equal]

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NCERT Solutions for Class 9 Maths Exercise 9.4

4. In figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersects DC at P, show that ar (BPC) = ar (DPQ).

Ans. Join A and C.

APC and BPC are on the same base PC and between the same parallels PC and AB.

ar (APC) = ar (BPC) ……….(i)

Now ACBD is a parallelogram.

AD = BC [opposite sides of a parallelogram are always equal]

Also BC = CQ [given]

AD = CQ

Now AD CQ [Since CQ is the extension of BC]

And AD = CQ

ADQC is a parallelogram.

[ If one pair of opposite sides of a quadrilateral is equal and parallel then it is a parallelogram]

Since diagonals of a parallelogram bisect each other.

AP = PQ and CP = DP

Now in APC and DPQ,

AP = PQ [Proved above]

APC = DPQ [Vertically opposite angles]

PC = PD [Prove above]

APC DPQ ……….(ii)

ar (APC) = ar (DPQ) [area of congruent figures is always equal]

From eq. (i) and (ii),

ar (BPC) = ar (DPQ)

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NCERT Solutions for Class 9 Maths Exercise 9.4

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

(i) ar (BDE) = ar (ABC)

(ii) ar (BDE) = ar (BAE)

(iii) ar (ABC) = 2 ar (BEC)

(iv) ar (BFE) = ar (AFD)

(v) ar (BFE) = 2 ar (FED)

(vi) ar (FED) = ar (AFC)

Ans. Join EC and AD.

Since ABC is an equilateral triangle.

A = B = C =

Also BDE is an equilateral triangle.

B = D = E =

If we take two lines, AC and BE and BC as a transversal.

Then B = C = [Alternate angles]

BE AC

Similarly, for lines AB and DE and BF as transversal.

Then B = C = [Alternate angles]

BE AC

(i) Area of equilateral triangle BDE = (BD)² ……….(i)

Area of equilateral triangle ABC = (BC)² ……….(ii)

Dividing eq. (i) by (ii),

[ BD = DC]

ar (BDE) = ar (ABC)

(ii) In BEC, ED is the median.

ar (BEC) = ar (BAE) ……….(i)

[Median divides the triangle in two triangles having equal area]

Now BE AC

And BEC and BAE are on the same base BE and between the same parallels BE and AC.

ar (BEC) = ar (BAE) ……….(ii)

Using eq. (i) and (ii), we get

Ar (BDE) = ar (BAE)

(iii) We have ar (BDE) = ar (ABC) [Proved in part (i)] ……….(iii)

ar (BDE) = ar (BAE) [Proved in part (ii)]

ar (BDE) = ar (BEC) [Using eq. (iii)] ……….(iv)

From eq. (iii) and (iv), we het

ar (ABC) = ar (BEC)

ar (ABC) = 2 ar (BEC)

(iv) BDE and AED are on the same base DE and between same parallels AB and DE.

ar (BDE) = ar (AED)

Subtracting FED from both the sides,

ar (BDE) – ar (FED) = ar (AED) – ar (FED)

ar (BFE) = ar (AFD) ……….(v)

(v) An in equilateral triangle, median drawn is also perpendicular to the side,

AD BC

Now ar (AFD) = ……….(vi)

Draw EG BC

ar (FED) = ……….(vii)

Dividing eq. (vi) by (vii), we get

[Altitude of equilateral triangle = side]

[D is the mid-point of BC]

ar (AFD) = 2 ar (FED) ……(viii)

Using the value of eq. (viii) in eq. (v),

Ar (BFE) = 2 ar (FED)

(vi) ar (AFC) = ar (AFD) + ar (ADC) = 2 ar (FED) + ar (ABC) [using (v)

= 2 ar (FED) + [4 ar (BDE)] [Using result of part (i)]

= 2 ar (FED) + 2 ar (BDE) = 2 ar (FED) + 2 ar (AED)

[BDE and AED are on the same base and between same parallels]

= 2 ar (FED) + 2 [ar (AFD) + ar (FED)]

= 2 ar (FED) + 2 ar (AFD) + 2 ar (FED) [Using (viii)]

= 4 ar (FED) + 4 ar (FED)

ar (AFC) = 8 ar (FED)

ar (FED) = ar (AFC)

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NCERT Solutions for Class 9 Maths Exercise 9.4

6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar (APB) ar (CPD) = ar (APD) ar (BPC)

Ans. Given: A quadrilateral ABCD, in which diagonals AC and BD intersect each other at point E.

To Prove: ar (AED) ar (BEC)

= ar (ABE) ar (CDE)

Construction: From A, draw AM BD and from C, draw CN BD.

Proof: ar (ABE) = ……….(i)

And ar (AED) = ……….(ii)

Dividing eq. (ii) by (i), we get,

……….(iii)

Similarly ……….(iv)

From eq. (iii) and (iv), we get

=

ar (AED) ar (BEC) = ar (ABE) ar (CDE)

Hence proved.

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NCERT Solutions for Class 9 Maths Exercise 9.4

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

(i) ar (PRQ) = ar (ARC)

(ii) ar (RQC) = ar (ABC)

(iii) ar (PBQ) = ar (ARC)

Ans. (i) PC is the median of ABC.

ar (BPC) = ar (APC) ……….(i)

RC is the median of APC.

ar (ARC) = ar (APC) ……….(ii)

[Median divides the triangle into two triangles of equal area]

PQ is the median of BPC.

ar (PQC) = ar (BPC) ……….(iii)

From eq. (i) and (iii), we get,

ar (PQC) = ar (APC) ……….(iv)

From eq. (ii) and (iv), we get,

ar (PQC) = ar (ARC) ……….(v)

We are given that P and Q are the mid-points of AB and BC respectively.

PQ AC and PA = AC

ar (APQ) = ar (PQC) ……….(vi) [triangles between same parallel are equal in area]

From eq. (v) and (vi), we get

ar (APQ) = ar (ARC) ……….(vii)

R is the mid-point of AP. Therefore RQ is the median of APQ.

ar (PRQ) = ar (APQ) ……….(viii)

From (vii) and (viii), we get,

ar (PRQ) = ar (ARC)

(ii) PQ is the median of BPC

ar (PQC) = ar (BPC) = ar (ABC) = ar (ABC) ……….(ix)

Also ar (PRC) = ar (APC) [Using (iv)]

ar (PRC) = ar (ABC) = ar (ABC) ……….(x)

Adding eq. (ix) and (x), we get,

ar (PQC) + ar (PRC) = ar (ABC)

ar (quad. PQCR) = ar (ABC) ……….(xi)

Subtracting ar (PRQ) from the both sides,

ar (quad. PQCR) – ar (PRQ) = ar (ABC) – ar (PRQ)

ar (RQC) = ar (ABC) – ar (ARC) [Using result (i)]

ar (ARC) = ar (ABC) – ar (APC)

ar (RQC) = ar (ABC) – ar (APC)

ar (RQC) = ar (ABC) – ar (ABC) [PC is median of ABC]

ar (RQC) = ar (ABC) – ar (ABC)

ar (RQC) = ar (ABC)

ar (RQC) = ar (ABC)

(iii) ar (PRQ) = ar (ARC) [Using result (i)] 2 ar (PRQ) = ar (ARC) ..(xii)

ar (PRQ) = ar (APQ) [RQ is the median of APQ] ……….(xiii)

But ar (APQ) = ar (PQC) [Using reason of eq. (vi)] ……….(xiv)

From eq. (xiii) and (xiv), we get,

ar (PRQ) = ar (PQC) ……….(xv)

But ar (BPQ) = ar (PQC) [PQ is the median of BPC] ……….(xvi)

From eq. (xv) and (xvi), we get,

ar (PRQ) = ar (BPQ) ……….(xvii)

Now from (xii) and (xvii), we get,

2 = ar (ARC) ar (BPQ) = ar (ARC)

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NCERT Solutions for Class 9 Maths Exercise 9.4

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 AXDE meets BC at Y. Show that:

(i) MBC ABD

(ii) ar (BYXD) = 2 ar (MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) FCB ACE

(v) ar (CYXE) = 2 ar (FCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

Ans. (i) ABM = CBD =

Adding ABC both sides, we get,

ABM + ABC = CBD + ABC MBC = ABD ……….(i)

Now in MBC and ABD,

MB = AB [equal sides of square ABMN]

BC = BD [sides of square BCED]

MBC = ABD [proved above]

MBC ABD [By SAS congruency]

(ii) From above, MBC ABD

ar (MBC) = ar (ABD) ar (MBC) = ar (trap. ABDX) – ar (ADX)

ar (MBC) = (BD + AX) BY – DX.AX

ar (MBC) = BD.BY + AX.BY – DX.AX

ar (MBC) = BD.BY + AX (BY – DX)

ar (MBC) = BD.BY + AX. 0 [BY = DX]

ar (MBC) = BD.BY

2 ar (MBC) = BD.BY 2 ar (MBC) = ar (rect. BYXD)

Hence ar (BYXD) = 2 ar (MBC)

(iii) Join AM. ABMN is a square.

Therefore, NA MB AC MB

Now AMB and MBC are on the same base and between the same parallels MB and AC.

ar (AMB) = ar (MBC) ……….(ii)

From result (ii), we have ar (BYXD) = 2 ar (MBC) ……….(iii)

Using eq. (ii) and (iii), we get, ar (BYXD) = 2 ar (AMB)

ar (BYXD) = ar (square ABMN)

[Diagonal AM of square ABMN divides it in two triangles of equal area]

(iv) In FCB and ACE,

FC = AC [sides of square ACFG]

BC = CE [sides of square BCED]

BCF = ACE [ ACF = BCE = ]

Adding ACB both sides,

BCF + ACB = ACE + ACB BCF = ACE

FCB ACE [By SAS congruency]

(v) From (iv), we have, FCB ACE

ar (FCB) = ar (ACE) ar (FCB) = ar (trap. ACEX) – ar (AEX)

ar (FCB) = (CE + AX) CY – XE.AX

ar (FCB) = CE.CY + AX.CY – XE.AX

ar (FCB) = CE.CY + AX (CY – XE)

ar (FCB) = CE.CY + AX. 0 [CY = XE]

ar (FCB) = CE.CY

2 ar (FCB) = CE.CY 2 ar (FCB) = ar (rect. CYXE)

Hence ar (BYXD) = 2 ar (FCB)

NCERT Solutions for Class 9 Maths Exercise 9.4

(vi) Join AF. ACFG is a square.

FC AG FC AB

Now ACF and FCB are on the same base FC and between the same parallels FC and AB.

ar (ACF) = ar (FCB) ……….(v)

From result (v), we get, ar (CYXE) = 2 ar (FCB) ……….(vi)

Using eq. (v) in (vi), we get, ar (CYXE) = 2 ar (ACF)

Diagonal AF of square ACFG divides it in two triangles of equal area.

ar (CYXE) = ar (sq. ACFG) ……….(vii)

(vii) Adding eq. (iv) and (vii), we get,

ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG)

ar (BCED) = ar (ABMN) + ar (ACFG)

NCERT Solutions for Class 9 Maths Exercise 9.4

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