Gujarat Board GSEB Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

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 the same base AB and have equal areas.

To prove: The perimeter of the parallelogram ABCD is greater than that of rectangle ABEF. Proof: Here the parallelogram ABCD and the rectangle ABEF are on the same base AB and between the same parallels AB and FC. Then, perimeter of the parallelogram ABCD = 2(AB + AD)

and, the perimeter of the rectangle ABEF = 2(AB + AF).

In Î”ADF, âˆ AFD = 90Â°

âˆ´ âˆ ADF is an acute angle. (< 90Â°) [Angle sum property of a triangle ] â‡’ âˆ AFD > âˆ ADF

â‡’ AD > AF

The side opposite to the greater angle of a triangle is longer.

â‡’ AB + AD > AB + AF

â‡’ 2(AB + AD) > 2(AB + AF)

âˆ´ Perimeter of the parallelogram ABCD > Perimeter of the rectangle ABEF.

Question 2.

In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(Î”ABD) = ar(Î”ADE) = ar(Î”AEC). Can you now answer the question that you have left in the â€˜Introduction, of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide Î”ABC into n triangles of equal areas.]

Solution:

Let AM âŠ¥ BC. Then, the height of each of Î”ABD, Î”ADE and Î”AEC is the same.

âˆ´ ar(Î”ABD) = \(\frac {1}{2}\) x BD x AM ……(1)

ar(Î”ADE) = \(\frac {1}{2}\) x DE x AM ……(2)

and, ar(Î”AEC) = \(\frac {1}{2}\) x EC x AM ……(3)

âˆ´ BD = DE = EC

âˆ´ From (1), (2) and (3),

ar(Î”ABD) = ar(Î”ADE) = ar(Î”AEC)

Yes, the field of Budhia has been actually divided into three parts of equal area.

Question 3.

In figure, ABCD, DCFE and ABFE are parallelograms. Show that

ar(Î”ADE) = ar(Î”BCF)

Solution:

âˆ´ ABCD is a || gm.

âˆ´ AD || BC …….(1)

[Opposite sides of a || gm are equal Again, DCFE is a || gm.]

âˆ´ DE || CF ……(2)

[Opposite sides of a || gm are equal Also, ABFE is a || gm]

âˆ´ AE = BF ……(3)

[Opposite sides of a || gm are equal From (1), (2) and (3)]

Î”ADE â‰… Î”BCFÂ [By SSS congruency]

âˆ´ ar(Î”ADE) = ar(Î”BCF)

âˆ´ Two congruent triangles are equal in areas.

Question 4.

In the 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(Î”BPC) = ar(Î”DPQ).

Solution:

Join AC.

âˆ´ AQAC and AQDC are on the same base QC and between the same parallels AD and QC.

âˆ´ ar(Î”QAC) = ar(Î”QDC) ……(1)

Two triangles on the same base (or equal bases) and between the same parallels are equal in areas.

â‡’ ar(Î”QAC) â‰… ar(Î”QPC) = ar(Î”QDC) â‰… ar(Î”QPC)

Subtracting the same areas from both sides.

â‡’ ar(Î”PAC) = ar(Î”QDP) ……(2)

âˆ´ Î”PAC and Î”PBC are on the same base PC and between the same parallels AB and DC.

âˆ´ ar(Î”PAC) = ar(Î”PBC) …….(3)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

From (2) and (3), we get

ar(Î”PBC) = ar(Î”QDP)

âˆ´ ar(Î”BPC) = ar(Î”DPQ)

Question 5.

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

(i) ar(Î”BDE) = \(\frac {1}{4}\) ar(Î”ABC)

(ii) ar(Î”BDE) = \(\frac {1}{2}\) ar(Î”BAE)

(iii) ar(Î”ABC) = 2ar(Î”BEC)

(iv) ar(Î”BFE) = ar(Î”AFD)

(v) ar(Î”BFE) = 2ar(Î”FED)

(vi) ar(Î”FED) = \(\frac {1}{8}\) ar(AAFC)

[Hint: Join EC and AD. Show that BE || AC and DE || AB, etc.]

Solution:

âˆ´ Î”ABC is an equilateral triangle.

âˆ´ âˆ ABC = âˆ BCA = âˆ CAB = 60Â° ……(1)

âˆ´ Î”BDE is an equilateral triangle.

âˆ´ âˆ BDE = âˆ DEB = âˆ EBD = 60Â° ……(2)

â‡’ âˆ ABC = âˆ BDE

AB || DE …….(3)

âˆ´ Alternate interior angles. Similarly, âˆ ACB = âˆ EBC

âˆ´ AC || BE …….(4) (each 60Â°)

Now, Î”CBA and Î”CEA are on the same base AC and between the same parallels.

âˆ´ ar(Î”CBA) = ar(Î”CEA)

Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

â‡’ ar(Î”ABC) = ar(Î”CDA) + ar(Î”CED) + ar(Î”ADE) …….(5)

In Î”ABC, AD is a medan.

âˆ´ ar(Î”ABD) = ar(Î”ACD) = \(\frac {1}{2}\) ar(Î”ABC) …….(6)

A median of a triangle divides it into two triangles of equal area.

In Î”EBC, ED is a median.

âˆ´ ar(Î”ECD) = ar(Î”EBD) = \(\frac {1}{2}\) ar(Î”EBC) ……(7)

âˆ´ A median of a triangle divides it into two triangles of equal area.

âˆ´ Î”DEA and ADBE are on the same base DE and between the same parallels AB and DE.

âˆ´ ar(Î”DEA) = ar(Î”DBE) ………(8)

âˆ´ Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Using (6), (7) and (8), in (5) gives

ar(Î”ABC) = \(\frac {1}{2}\) ar(Î”ABC) + ar(Î”BDE) + ar(Î”BDE)

â‡’ \(\frac {1}{2}\) ar(Î”ABC) = 2ar(Î”BDE)

â‡’ ar(Î”BDE) = \(\frac {1}{4}\) ar(Î”ABC)

(ii) âˆ´ Î”BAE and Î”BCE are on the same base BE and between the same parallels BE and AC.

âˆ´ ar(Î”BAE) = ar(Î”BCE)

[âˆ´ Two triangles on the same base (or equal bases) and between the same parallels are equal in area]

â‡’ \(\frac {1}{2}\) ar(Î”BAE) = 2ar(Î”BDE) | From (7)

â‡’ ar(Î”BDE) = \(\frac {1}{4}\) ar(Î”BAE)

(iii) 2ar(Î”BEC) = 2ar(Î”BDE) [From (7)]

= 4ar(Î”BDE)

= ar(Î”ABC) [From (i)]

(iv) âˆ´ Î”EBD and Î”EAD are on the same base ED and between the same parallels AB and DE.

âˆ´ ar(Î”EBD) = ar(Î”EAD)

[âˆ´ Two triangles on the same base (or equal bases) and between the same parallels are equal in area]

â‡’ ar(Î”EBD) – ar(Î”EFD) = ar(Î”EAD) – ar(Î”EFD)

[Subtracting the same areas from both sides]

â‡’ ar(Î”BFE) = ar(Î”AFD)

(v) ar(ABDE) = \(\frac {1}{4}\) ar(Î”ABC)Â [From (i)]

= \(\frac {1}{4}\) .2 ar(Î”ABD)

= \(\frac {1}{4}\) ar(Î”ABD)

âˆ´ Bases of Î”BDE and Î”ABD are the same.

âˆ´ Altitude of ABDE = \(\frac {1}{2}\) Altitude of Î”ABD ……..(9)

ar(Î”BEF) = ar(Î”AFD) ……..(10) [From (iv)]

But Altitude of Î”BDE = Altitude of Î”BEF (Considering common vertex)

And, altitude of Î”ABD = Altitude of Î”AFD (Considering common vertex A)

Altitude of Î”BEF = \(\frac {1}{2}\) Altitude of Î”AFD ………(11)

From (10) and (11),

BF = 2FD ……(12)

In Î”BFE and Î”FED,

âˆ´ BF = 2FD

and alt. (Î”BFE) = alt. (Î”FED)

âˆ´ ar(Î”BFE) = 2ar(Î”FED).

(vi) Let the altitude of Î”ABD be h.

Then, altitude of Î”BED = \(\frac {h}{2}\) [From (9)]

Now, ar(Î”FED) = \(\frac {1}{2}\). FD. \(\frac {h}{2}\) = \(\frac {FD.h}{4}\) …….(13)

ar(Î”AFC) = \(\frac {1}{2}\) . FC .h

= \(\frac {1}{2}\) (FD + DC) h 2

= \(\frac {1}{2}\) (FD + BD) h 2

= \(\frac {1}{2}\) (FD + BF + FD) h 2

= \(\frac {1}{2}\) (2FD + BF) h 2

= \(\frac {1}{2}\) (2FD + 2FD) h [From (12) ]

= 2 . FD . h ……(14)

From (13) and (14), we obtain

ar(Î”FED) = \(\frac {1}{8}\) ar(Î”AFC)

Question 6.

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

ar(Î”APB) x ar(Î”CPD) = ar(Î”APD) x ar(Î”BPC)

[Hint: From A and C, draw perpendiculars to BD]

Solution:

Construction: From A and C, draw perpendiculars AE and CF respectively to BD.

Proof: ar(Î”APB) x ar(Î”CPD)

= \(\frac {(PB) (AE)}{2}\) x \(\frac {(DP x CE)}{2}\)

= \(\frac {1}{4}\) (PB) (AE) (DP) (CF) ……..(1)

ar(Î”APD) x ar(Î”BPC)

= \(\frac {(DP) (AE)}{2}\)Â = \(\frac {(PB) (CF)}{2}\)

= \(\frac {1}{4}\) (PB) (AE) (DP) (CF) ……(2)

From (1) and (2), we get

ar(Î”APB) x ar(Î”CPD)

= ar(Î”APD) x ar(Î”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:

1. ar(APRQ) = \(\frac {1}{2}\) ar(Î”ARC)

2. ar(ARQC) = \(\frac {3}{8}\) ar(Î”ABC)

3. ar(Î”PBQ) = ar(Î”ARC)

Solution:

Construction: Join AQ and CP.

Proof: 1.Â ar(Î”PRQ) = ar(Î”ARQ)

(âˆ´ a median of a A divides it into two triangles of equal area)

= \(\frac {1}{2}\) ar(Î”APQ) = \(\frac {1}{2}\) ar(Î”BPQ)

= \(\frac {1}{2}\) ar(Î”CPQ) = \(\frac {1}{2}\) . \(\frac {1}{2}\) ar(Î”BPC)

= \(\frac {1}{4}\) ar(Î”BPC) = \(\frac {1}{4}\) . \(\frac {1}{2}\) ar(Î”ABC)

= \(\frac {1}{8}\) ar(Î”ABC) ………(1)

\(\frac {1}{2}\) ar(Î”ARC) = \(\frac {1}{2}\) . \(\frac {1}{2}\) ar(Î”APC)

= \(\frac {1}{4}\) ar(Î”APC) = \(\frac {1}{4}\).\(\frac {1}{2}\) ar(Î”ABC)

= \(\frac {1}{8}\)ar(Î”ABC) ………..(2)

From (1) and (2), we have

ar(Î”PRQ) = \(\frac {1}{2}\) ar(Î”ARC)

2. ar(Î”RQC) = ar(Î”RBQ)

(âˆ´ a median of a A divides it into triangles of equal areas)

= ar(Î”PRQ) + ar(Î”BPQ)

= \(\frac {1}{8}\) ar(Î”ABC) + \(\frac {1}{2}\) ar(Î”PBC) [Using (1)]

= \(\frac {1}{8}\) ar(Î”ABC) + \(\frac {1}{2}\).\(\frac {1}{2}\) ar(AABC)

= \(\frac {1}{8}\) ar(Î”ABC) + \(\frac {1}{4}\) ar(Î”ABC)

= \(\frac {3}{8}\) ar(Î”ABC)

3. ar(Î”PBQ) = \(\frac {1}{4}\) ar(Î”ABC) …….(3) [From (ii)]

ar(Î”ARC) = \(\frac {1}{4}\) ar(Î”ABC) …….(4) [From (i) ]

From (3) and (4), we get ar(Î”PBQ) = ar(Î”ARC)

Question 8.

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

(i) Î”MBC – Î”ABD

(ii) ar(BYXD) = 2ar(Î”MBC)

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

(iv) Î”FCB = Î”ACE

(v) ar(CYXE) = 2ar(Î”FCB)

(vi) ar(CYXE) = ar(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 Î”MBC and Î”ABD,

MB = AB …(1) [Sides of a square]

BC = BD …(2) [Sides of a square]

âˆ MBA = âˆ CBD [Each = 90Â°]

â‡’ âˆ MBA + âˆ ABC = âˆ CBD + âˆ ABC

[Adding âˆ ABC to both sides]

â‡’ âˆ MBC = âˆ ABD …….(3)

In view of (1), (2) and (3),

â‡’ Î”MBC = Î”ABD [By congruence SAS rule]

(ii) ar(BYXD) = 2 ar(Î”ABD) (âˆ´ BD || AX and common base is BD)

= ar(BYXD) = 2ar(Î”MBC)

[From part (i), ar(Î”MBC) = ar(Î”ABD)]

(iii) âˆ´ ar(BYXD) = 2ar(Î”ABD)

and ar(Î”BMN) = 2ar(Î”MBC)

(âˆ´ MB || NC, and common base is MB)

= 2ar(Î”ABD) | From (i)

âˆ´ ar(BYXD) = ar(ABMN)

(iv) In Î”FCB and Î”ACE,

FC = AC [Sides of a square]

CB = CE [Sides of a square]

âˆ FCA = âˆ BCE [Each = 90Â°]

â‡’ âˆ FCA + âˆ ACB = âˆ BCE + âˆ ACB

(Adding the same on both sides)

â‡’ âˆ FCB = âˆ ACE

âˆ´ Î”FCB = Î”ACE [SAS congruence rule]

(v) ar(CYXE) = 2ar(Î”ACE) = 2ar(Î”FCB) [From (iv)]

âˆ´ AFCB = AACE

âˆ´ ar(AFCB) = ar(AACE)

Congruent As have equal areas

(vi) âˆ´ ar(CYXE) = 2ar(AACE) = 2ar(AFCB) And, ar(ACFG) = 2ar(AFCB)

(âˆ´ BG || CF and CF is the common base)

âˆ´ ar(CYXE) = ar(ACFG)

(vii) ar(BCED = ar(CYXE) + ar(BYXD)

= ar(ACFG) + ar(ABMN) [From (iii) and (vi)]

= ar(ABMN) + ar(ACFG)