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# Class 9 RD Sharma Solutions – Chapter 15 Areas of Parallelograms and Triangles- Exercise 15.3 | Set 2

• Last Updated : 09 Mar, 2022

### Question11. If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.

Solution:

According to the question

ABCD is a parallelogram and P is any point in the interior of parallelogram

Prove: ar(Î”APB) < 1/2 ar(|| gm ABCD)

Proof:

Construction: Draw DN âŠ¥ AB and PM âŠ¥ AB

Now, we find the area of âˆ¥gm ABCD

= AB Ã— DN,

Now, the area of Î”APB

= (1/2) (AB Ã— PM)

Now, PM < DN

AB Ã— PM < AB Ã— DN

(1/2)(AB Ã— PM) < (1/2)(AB Ã— DN)

ar(Î”APB) < 1/2 ar(âˆ¥ gm ABCD)

Hence proved

### Question 12. If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(Î”BGC) = 2ar(Î”AGC).

Solution:

According to the question

AD is a median of a triangle ABC

Now, construct AM âŠ¥ BC

It is given that AD is the median of Î”ABC

So, BD = DC

BD = AM = DC Ã— AM

(1/2)(BD Ã— AM) = (1/2)(DC Ã— AM)

ar(Î”ABD) = ar(Î”ACD)                 ….(i)

In Î”BGC,

GD is the median

So, ar(Î”BGD) = ar(Î”CGD)          …..(ii)

In Î”ACD,

CG is the median

So, ar(Î”AGC) = ar(Î”CGD)        ……..(iii)

From equation (ii) and (iii), we get

ar(Î”BGD) = ar(Î”AGC)

But, ar(Î”BGC) = 2ar(Î”BGD)

Hence, ar(Î”BGC) = 2ar(Î”AGC)

Hence proved

### Question 13. A point D is taken on the side BC of a Î”ABC, such that BD = 2DC. Prove that ar(Î”ABD) = 2ar(Î”ADC).

Solution:

According to the question

BD = 2DC(In Î”ABC)

Proof:

Now draw a point E on BD such that BE = ED

Since, BE = ED and BD = 2 DC

So, BE = ED = DC

As we know that, in triangles, median divides the triangle into two equal triangles.

In Î”ABD,

AE is the median.

So, ar(Î”ABD) = 2ar(Î”AED)            …..(i)

In Î”AEC,

From eq(i) and (ii), we get

Hence proved

### (ii) ar(Î”ABP) = 2ar(Î”CBP).

Solution:

According to the question

ABCD is the parallelogram, whose diagonals intersect at O

So, AO = OC and BO = OD

(i) Prove that ar(Î”ADO) = ar(Î”CDO)

Proof:

In Î”DAC,

DO is a median.

Hence proved

(ii) Prove that ar(Î”ABP) = 2ar(Î”CBP)

Proof:

In Î”BAC,

BO is a median.

So, ar(Î”BAO) = ar(Î”BCO)       …..(i)

In Î”PAC,

PO is a median.

So, ar(Î”PAO) = ar(Î”PCO)          …..(ii)

Now, subtract eq(ii) from (i), we get

ar(Î”BAO) âˆ’ ar(Î”PAO) = ar(Î”BCO) âˆ’ ar(Î”PCO)

ar(Î”ABP) = 2ar(Î”CBP)

Hence proved

### (ii) If the area of Î”DFB = 3 cm2, find the area of âˆ¥ gm ABCD.

Solution:

According to the question

ABCD is a parallelogram in which BC is produced to E

such that CE = BC and AE intersects CD at F

(i) Prove that ar(Î”ADF) = ar(Î”ECF)

Proof:

âˆ ADF = âˆ ECF                     (Alternate interior angles)

âˆ DFA = âˆ CFA                      (Vertically opposite angle)

So, by AAS congruence

By c.p.c.t

DF = CF

Hence proved

(ii) From the question area of Î”DFB = 3 cm2

Find the area of ||gm ABCD

Now, DF = CF (Proved above)

BF is a median in Î” BCD.

ar(Î”BCD) = 2ar(Î”DFB)

ar(Î”BCD) = 2 Ã— 3 cm2 = 6 cm2

Now we find the area of a parallelogram = 2ar(Î”BCD)

= 2 Ã— 6 cm2 = 12 cm2

Hence the area of parallelogram is 12 cm2

### Question 16. ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(Î”POA) = ar(Î”QOC).

Solution:

Prove: ar(Î”POA) = ar(Î”QOC)

In Î”POA and QOC,

âˆ AOP = âˆ COQ  (Vertically opposite angle)

AO = OC

âˆ PAC = âˆ QCA   (Alternate angle)

So, by ASA congruence criterion, we have

Î”POA â‰… Î”QOC

Hence ar(Î”POA) = ar(Î”QOC)

Hence proved

### Question 17. ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.

Solution:

Prove: ar(âˆ¥gm AECF) = 1/3 ar(||gm ABCD)

Proof:

First construct FG âŠ¥ AB

Now, according to the question

BE = 2 EA and DF = 2FC

AB – AE = 2 AE and DC – FC = 2 FC

AB = 3 AE and DC = 3 FC

AE = (1/3) AB and FC = (1/3)DC            …….(i)

But AB = DC

So, AE = FC                       (Opposite sides of the parallelogram)

AE = FC and AE âˆ¥ FC

So, AECF is a parallelogram

Now, we find the area of parallelogram AECF

= AE Ã— FG

Put the value of AE from eq(i), we get

ar(||gm AECF) = 1/3 AB Ã— FG

3ar (||gm AECF) = AB Ã— FG          ….(ii)

and ar(||gm ABCD) = AB Ã— FG            ….(iii)

From eq(ii) and (iii), we get

3ar(||gm AECF) = ar(||gm ABCD)

Hence ar(||gm AECF) = 1/3 ar(||gm ABCD)

Hence proved

### (iii) ar(Î”RQC) = 3/8 ar(Î”ABC).

Solution:

According to the question

ABC is a triangle, in which P, Q, and R are the mid-points of AB, BC and AP

(i) Prove that ar(Î”PBQ) = ar(Î”ARC)

Proof:

CR is the median of Î”CAP

So, ar(Î”CRA) = (1/2) ar(Î”CAP)                ….(i)

CP is the median of a Î”CAB

So, ar(Î”CAP) = ar(Î”CPB)                 ….(ii)

So, from eq(i) and (ii), we conclude that

ar(Î”ARC) = (1/2) ar(Î”CPB)                     ….(iii)

Now, PQ is the median of a Î”PBC

So, ar(Î”CPB) = 2ar(Î”PBQ)         ….(iv)

From eq(iii) and (iv),  we conclude that

ar(Î”ARC) = ar(Î”PBQ)                ….(v)

Hence proved

(ii) Now QP and QR are the medians of triangles QAB and QAP

So, ar(Î”QAP) = ar(Î”QBP)                    ….(vi)

And ar(Î”QAP) = 2ar(Î”QRP)                ….(vii)

From eq(vi) and (vii),  we conclude that

ar(Î”PRQ) = (1/2) ar(Î”PBQ)                  ….(viii)

And from eq (v) and (viii), we conclude that

ar(Î”PRQ) = (1/2) ar(Î”ARC)

Hence proved

(iii) Now, LR is a median of Î”CAP

= 1/2 Ã— (1/2) ar(Î”ABC)

= (1/4) ar(Î”ABC)

and RQ is the median of Î”RBC.

So, ar(Î”RQC) = (1/2) ar(Î”RBC)

= (1/2) {ar(Î”ABC) âˆ’ ar(Î”ARC)}

= (1/2) {ar(Î”ABC) â€“ (1/4) ar(Î”ABC)}

= (3/8) ar(Î”ABC)

Hence proved

### Question 19. ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(ii) ar(Î”EGB) = (1/6) ar(ABCD)

(iii) ar(Î”EFC) = (1/2) ar(Î”EBF)

(iv) ar(Î”EGB) = 3/2 Ã— ar(Î”EFC)

(v) Find what portion of the area of parallelogram is the area of Î”EFG.

Solution:

According to the question

ABCD is a parallelogram

So, AG = 2 GB, CE = 2 DE and BF = 2 FC

Now, draw a parallel line to AB that pass through point F and a perpendicular line to AB from C.

(i) Prove that ar(ADEG) = ar(GBCE)

Since it is given that ABCD is a parallelogram

So, AB = CD and AD = BC

Now, let us consider the two trapezium ADEG and GBCE

Since AB = DC, EC = 2DE, AG = 2GB

So, ED = (1/3) CD = (1/3) AB and EC = (2/3) CD = (2/3) AB

AG = (2/3) AB and BG = (1/3) AB

So, DE + AG = (1/3) AB + (2/3) AB = AB and EC + BG = (2/3) AB + (1/3) AB = AB

Since these two trapeziums have same heights, also their sum of two parallel sides are equal,

So, the area of trapezium = sum of parallel side/2 x height

Hence proved

(ii) Prove that ar(Î”EGB) = (1/6) ar(ABCD)

From the above we know that

BG = (1/2) AB

So, ar(Î”EGB) = (1/2) Ã— GB Ã— height

ar(Î”EGB) = (1/2) Ã— (1/3) Ã— AB Ã— height

ar(Î”EGB) = (1/6) Ã— AB Ã— height

Then, ar(Î”EGB) = (1/6) ar(ABCD)

Hence proved

(iii) Prove that ar(Î”EFC) = (1/2) ar(Î”EBF)

As we know that the height of triangle EFC and EBF are equal,

So, ar(Î”EFC) = (1/2) Ã— FC Ã— height

ar(Î”EFC) = (1/2) Ã— (1/2) Ã— FB Ã— height

ar(Î”EFC) = (1/2) ar(EBF)

Then, ar(Î”EFC) = (1/2) ar(Î”EBF)

Hence proved

(iv) Prove that ar(Î”EGB) = 3/2 Ã— ar(Î”EFC)

Let us consider the trapezium EGBC

So, ar(EGBC) = ar(Î”EGB) + ar(Î”EBF) + ar(Î”EFC)

Now put all these values, we get

(1/2) ar(ABCD) = (1/6) ar(ABCD) + 2ar(Î”EFC) + ar(Î”EFC)

(1/3) ar(ABCD) = 3 ar(Î”EFC)

ar(Î”EFC) = (1/9) ar(ABCD)

Now from option (ii), we get,

ar(Î”EGB) = (1/6) ar(Î”EFC)

ar(Î”EGB) = (3/2) Ã— (1/9) ar(ABCD)

ar(Î”EGB) = (3/2) ar(Î”EFC)

Hence, ar(Î”EGB) = (3/2) ar(Î”EFC)

Hence proved

(v) From the figure, we have FB = 2CF,

So, Let us considered CF = x and FB = 2x.

Now, we take triangle CFI and CBH which are similar triangle.

So, by the property of similar triangle, we get

CI = k and IH = 2k

Now, we take Î”EGF in which,

ar(Î”EFG) = ar(Î”ESF) + ar(Î”SGF)

ar(Î”EFG) = (1/2) SF Ã— k + (1/2) SF Ã— 2k

ar(Î”EFG) = (3/2) SF Ã— k             …….(i)

Now,

ar(Î”EGBC) = ar(SGBF) + ar(ESFC)

ar(Î”EGBC) = (1/2)(SF + GB) Ã— 2k + (1/2)(SF + EC) Ã— k

ar(Î”EGBC) = (3/2) k Ã— SF + (GB + (1/2)EC) Ã— k

ar(Î”EGBC) = (3/2) k Ã— SF + (1/3 AB + (1/2) Ã— (2/3) AB) Ã— k

(1/2) ar(Î”ABCD) = (3/2) k Ã— SF + (2/3) AB Ã— k

ar(Î”ABCD) = 3k Ã— SF + (4/3) AB Ã— k

ar(Î”ABCD) = 3k Ã— SF + 4/9 ar(ABCD)

k Ã— SF = (5/27)ar(ABCD)                …….(ii)

From eq(i) and (ii), we conclude that

ar(Î”EFG) = (3/2) Ã— (5/27) ar(ABCD)

ar(Î”EFG) = (5/18) ar(ABCD)

Hence proved

### (iii) Prove that ar(BCZY) = ar(Î”EDZ)

Solution :

According to the question

CD || AE and CY || BA

(i) Here, triangle BCA and triangle BYA are on the same base BA and

between same parallel lines BA and CY.

So, ar(Î”BCA) = ar(Î”BYA)       ….(i)

Now as we know that ar(Î”BCA) = ar(Î”CBX) + ar(Î”BXA)

ar(Î”BYA) = ar(Î”BXA) + ar(Î”AXY)

So, put all these values in eq(i), we get

ar(Î”CBX) + ar(Î”BXA) = ar(Î”BXA) + ar(Î”AXY)

So, ar(Î”CBX) = ar(Î”AXY)

(ii) Here, triangles ACE and ADE are on the same base AE and

between same parallels CD and AE

Now as we know that ar(Î”ACE) = ar(Î”CZA) + ar(Î”AZE)

and ar(Î”ADE) = ar(Î”AZE) + ar(Î”DZE)

So, put all these values in eq(ii), we get

ar(Î”CZA) + ar(Î”AZE) = ar(Î”AZE) + ar(Î”DZE)

ar(Î”CZA) = ar(Î”DZE)                  ……(iii)

Hence proved

(iii) As we know that ar(Î”CBX) = ar(Î”AXY)

Now, add ar(Î”CYG) on both sides, we get

ar(Î”CBX) + ar(Î”CYZ) = ar(Î”CAY) + ar(Î”CYZ)

ar(BCZY) = ar(Î”CZA)                ….(iv)

From eq(iii) and (iv), we conclude that

ar(BCZY) = ar(Î”DZE)

Hence proved

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