RD Sharma Solutions Class 9 Areas Of Parallelograms And ... · RD Sharma Solutions Class 9 Areas Of Parallelograms And Triangles Ex 15.1 RD Sharma Solutions Class 9 Chapter 15 Ex

RD Sharma Solutions Class 9 Areas Of Parallelograms And Triangles Ex 15.1

RD Sharma Solutions Class 9 Chapter 15 Ex 15.1

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 two

parallels:

SOLUTION :

(i) A PCD and trapezium ABCD are on the same base CD and between the same parallels AB and DC.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and APQR are between the same parallels AD and BC but they are not on the same base.

(iv) AQQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelograms PQRS and trapezium SMNR are on the same base but not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, APSD are between the same parallels,

(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, APSD are between the same parallels.



Q1. I f figure, ABCD is a parallelogram, AE _L DC and CF _L AD. IfA B = 16 cm, AE = 8 cm, and CF =10 cm, Find AD.

Given that.

In parallelogram ABCD, CD = AB = 16 cm [v Opposite side o f a parallelogram are equal]

We know that.

Area of parallelogram = Base x Corresponding altitude

Area of parallelogram ABCD = CD x AE = AD x CF

16 cm x cm = AD x 10 cm

AD = ^ j fT cm = 12.8 cm

Thus, The length of AD is 12.8 cm.

Q2. InQ 1, i f AD = 6 cm, CF = 10 cm, and AE = 8 cm, Find AB.

Solution:

We know that.

Area of a parallelogram ABCD = AD x C F...................(1)

Again area of parallelogram ABCD = CD x AE...................(2)

Compare equation(l) and equation(2)

AD X CF = CD X AE

= > 6 x l O = D x 8

= > £ > = “ = 7.5 cm

A B = D C = 7.5 cm [v Opposite side o f a parallelogram are equal]

Q3. Let ABCD be a parallelogram o f area 124 cm2 . I f E and F are the mid-points o f sides AB and CD respectively, then find the area o f parallelogram AEFD.

Solution:

Given,

Area of a parallelogram ABCD = 124 cm 2

Construction: Draw AP_LDC

Proof-

Area of a parallelogram AFED = DF x A P ...................(1)

And area of parallelogram EBCF = FC x AP...................(2)

And DF = F C ...................(3) [F is the midpoint o f DC]

Compare equation (1), (2) and (3)

Area of parallelogram AEFD = Area of parallelogram EBCF

Area o f parallelogram A E F D = ------------------ -̂----------------- = — = 62 cmr

Q 4. I f ABCD is a parallelogram, then prove that

Ar(AABD) =Ar(ABCD) =Ar{AABC) =A r(A A C D )=\A r (I/3"1 ABCD).

Solution:

Given:-

ABCD is a parallelogram.

To prove: - Ar ( A ABD) = Ar (ABCD) = Ar ( A ABC) = Ar ( A ACD) = \Ar ( / / 9mA B C D ).

Proof:- We know that diagonal of a parallelogram divides it into two equilaterals.

Since, AC is the diagonal.

Then, Ar ( A ABC) = Ar ( A ACD) = \Ar { / / 3m ABCD)...................(1)

Since, BD is the diagonal.

Then, Ar ( A ABD) = Ar ( A BCD) = \Ar { / / gmA B C D ) ...................(2)

Compare equation (1) and (2)

Ar ( A ABC) = Ar (AACD) = Ar ( A ABD) = Ar {ABCD) = \Ar (//«"*ABCD)..



Q 1. In figure, compute Ifte area o f quadrilateral ABCD.

§O)

A B

Solution:

Given:

DC = 17 cm, AD = 9 cm and BC = 8 cm

In ABCD we have

CD2 = BD2 + BC2

=> 172 = BD2 + 82

=► R D 2 = 289 - 64

=>BD = 15

In AABD we have

4 .B 2 + A D 2 = B D 2

=> 152 = AB2 + 92

=>• A S 2 = 225 - 81 = 144

=> AB = 12

ar (quad ABCD ) = a r (AABD) + ar (ABCD)

ar (quad ABCD) = | (1 2 x 9 ) + | ( 8 x 17) = 54 + 68 = 122 cm2

ar (quad ABCD) = -| (1 2 x 9 ) + | ( 8 x 15) = 54 + 60 = 114 cm2

Q2. In figure, PQRS is a square and T and U are, respectively, tite midpoints o f PS and Q R. Find the area o f A O T S if PQ = 8 cm.

From the figure,

T and U are mid points of PS and QR respectively

:.TU\\PQ

=> TO\\PQ

T h u s , in A PQ S, T is the mid point of PS and T0||PQ

TO = \PQ = 4 cm

Also, T S = | PS = 4 cm

ar (AOTS) = { ( T O x TS) = \(A x 4 ) cm2 = 8 cm2

Q3. Compute the area o f trapezium PQRS in figure

We have,

ar (trap. PQRS) = ar (red. PSRT) + ar ( A QRT)

=> ar (trap. PQRS) = PT x RT + (QT x RT)

= = 8 x R T + | ( 8 x RT) = 12 x RT

In A QRT , we have

Q R 2 = Q T 2 + R T 2

=>■ R T 2 = Q R 2 - Q T 2

=>• R T 2 = 172 - 82 = 225

=> R T = 15

H e n ce , area o f trapez ium = 12 x 15 c m 2 = 180 c m 2

Q4. In figure, Z .A O B = 9 0 ° , A C = B C , OA = 12 cm and 0C = 6.5cm .Find the area o fA A O B .

Since, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices

C A = C B = O C

=>• C A = C B — 6 .5 cm

=>• A B = 13 cm

In right angled triangle O A B , we have

A B 2 = O B 2 + O A 2

=> 132 = O B 2 + 122

=> O B 2 = 132 - 122 = 1 6 9 - 1 4 4 = 25

=>• O B = 5

a r ( A A O B ) = | ( 1 2 x 5 ) = 3 0 cm 2

Q5. In figure, ABCD is a trapezium In which AB = 7 cm, A D = B C = 5 cm, DC = x cm , and distance between AB and DC Is 4 cm . Find die value o fx and area o f trapezium ABCD.

Draw AL _L DC, BM _L DC then,

A L = BM = 4 cm and LM = 7 cm.

In A A D L , we have

A D 2 = A L 2 + D L 2

=> 25 = 16 + D L 2

=> D L = 3 cm

Similarly, MC = \ / B C 2 — B M 2 = \ / 2 5 — 16 = 3 cm

; . x = C D = C M + M L + L D = (3 + 7 + 3 ) cm = 13 cm

a r (tra p . A B C D ) = ^ (A B + C D ) x A L = ^ (7 + 1 3 ) x 4 cm 2 = 4 0 cmi2

Q 6. In figure, OCDE is a rectangle inscribed in a quadrant o f a circle o f radius 10 cm. IfO E = 2 i / 5 cm , find die area o f the rectangle.

Given OD = 10 cm and OE = 2-^/5 cm

By using Pythagoras theorem

OD2 = OE2 + DE2

=>DE = VOD2 - OE2 = ^ 102 - (2^/5)2 = 4>/5 cm

Area of rectangle OCDE = OE x DE= 2-\/5 x 4\/5 c m 2 = 40 c m 2

Q 7. In figure, ABCD is a trapezium in which A B II DC. Prove thatar (AAOD) — ar(ABOC)

Given: ABCD is a trapezium in which AB || DC

To prove: or (AAOD) = ar (ABOC)

Proof:- Since, A ADC and A BDC are on the same base DC and between same parallels AB and DC

Then, ar(AADC) = a r (A BDC)

=> ar (AAOD) + ar (ADOC) = ar (ABOC) + ar (ADOC)

=> ar (AAOD) = ar (ABOC)

Q 8. In figure, ABCD, A B FE and CDEFare parallelograms. Prove that ar (A A D E ) = ar (A B C F ) .

Given that

ABCD is parallelogram => AD = BC

CDEF is parallelogram => DE = CF ABFE is parallelogram => AE = BF

Thus, in A s ADF and B C F , we have

AD = BC, DE = CF and AE = BF

So, by SSS criterion of congruence, we have

AADE ^ A BCF

ar (AADE) = ar (ABCF)

Q 9. Diagonals A C and BD o f a quadrilateral ABCD Intersect each other at P. Show that:nr (AAPB) x ar (ACPD) = ar (AAPD) x ar (ABPC).

Solution:

Construction: - Draw BQ _LAC and DR _L AC

Proof:-

L.H.S

= ar (A APB) x ar (ACDP)

= \[(AP x BQ)] x ( I x PC x DR)

= (| x PC x BQ) x ( } x AP x DR)

= ar (AAPD) x ar (ABPC).

= R.H.S

Hence proved.

Q 10. In figure, AB C andABD are two Mangles on Me baseAB. If line segment CD is bisected b y AB at 0, show Matar (AABC) = ar (A ABD).

Given that CD is bisected by AB at 0

To prove: ar (AABC) = ar (AABD).

Construction: Draw CP _L AB and DQ _L AB .

Proof:

ar (AABC) = \ x AB x C P ............(1)

ar (AABD) = \ x AB x D Q ........... (2)

In A CPO and ADQO

Z.CPO = /DQO [each 90°]Given that, CO = OD

/.COP = ZDOQ [Vertically opposite angles are equal]

Th e n , ACPO = A DQO [By AAS condition]

.-. CP = DQ (3) [c. p. c. t]Compare equation (1), (2) and (3)

ar (AABC) = ar (AABD).

Q 11. IfP is any point in the interior o f a parallelogram ABCD, Men prove Mat area o f Me triangle APB is less Man half Me area o f parallelogram.

Solution:

Draw DN _L AB and PM ± AB

N ow ,

ar (If"1 ABCD) = AB x DN, ar (AAPB) = \(AB x PM)

N o w ,P M < DN

=> AB x PM < AB x DN

=> \(AB x PM) < \(AB x DN)

=4- ar (AAPB) < ±ar (W^ABCD)

Q 12. If AD is a median o f a Mangle ABC, then prove Mat triangles ADB and ADC are equal In area. If G Is the m idpoint o f the median AD, prove thatar (A BGC) = 2 a r (A AGC).

Solution:

Draw AM _L BC

Since, AD is the median of A ABC .'. BD = DC

=>• BD = AM = DC x AM

=> \{BD x AM) = ±(DC x AM)

=> o r {AABD) = a r ( A 4 C D ) .................... (1 )

In A BG C , GD is the median

ar {ABGD) = ar (A C G D )............... (2)

In A A C D , CG is the median

ar (AAGC) = ar (AC G D )................(3)

From (2) and (3) we have,

ar {ABGD) = ar (A AGC)

But, ar ( A BGC) = 2ar {ABGD)

:. ar (ABGC) = 2ar (AAGC)

Q 13. A point D Is taken on die side BC o f a A ABC, such that B D = 2 D C . Prove that ar{AABD) = 2ar {A ADC).

Solution:

Given that.

In A ABC, BD = 2DC

To prove: ar {AABD) = 2 a r {AADC).

Construction:

Take a point E on BD such that BE = ED

Proof: Since, BE = ED and BD = 2 DC

Then, BE = ED = DC

We know that median of triangle divides it into two equal triangles.

In A A B D , AE is the median .

Th e n , ar {AABD) = 2ar (AA E D ).......... (1)

In A A E C , AD is the median .

Then, ar (AAED) = 2ar (AA D C ) ............. (2)

Compare equation 1 and 2

ar {AABD) = 2ar (AADC).

Q 14. ABCD Is a parallelogram whose diagonals intersect at 0 .IfP is any point on BO, prove that:

(0. ar (AADO) = ar (ACDO).00. ar {AABP) = 2ar {ACBP).

Solution:

Given that ABCD is the parallelogram

To Prove: (i) ar ( A ADO) = ar {ACDO).

(ii) ar {AABP) = 2ar {ACBP).

Proof: we know that diagonals of parallelogram bisect each other

AO = OC and BO = OD

(i) . In A D A C , since DO is a median .

Then ar {AADO) = ar {ACDO).

( i i ) . In A B A C , since BO is a median .

Then ar ( A BAO) = ar ( A B C O ) ............... (1 )

In A P A C , since PO is a m edian.

Then o r ( APAO ) = ar (AP C O ) ............... (2 )

Subtract equation 2 from 1.

=> ar (ABAO) - ar (APAO) = ar (ABCO) - ar (APCO)

=> ar (AABP) = 2ar (ACBP).

Q 15. ABCD Is a parallelogram In which BC Is produced to E such that C E = BC. A E Intersects CD at F.

(I) . Prove that ar (AADF) = ar (A ECF).

(ii) . If the area o f A DFB = 3 cm2, find the area of\ I9”1 AB CD .

Solution:

In triangles ADF and ECF, we have

AADF = AECF [.Alternate interior angles, Since AD\\BE\

AD = EC [since AD = BC — CE\

And ADF A = AC FA [Vertically opposite angles]

So, by AAS congruence criterion, we have

A ADF ^ A ECF

=> ar (AADF) = ar (AECF) and DF = C F .

Now, DF = CF

=> BF is a median in A BCD.

=> ar (ABCD) = 2or (ABDF)

=>• ar (ABCD) = 2 x 3 cm2 = 6 cm2

Flence, area of a parallelogram = 2 o r (ABCD) = 2 x 6 cm2 = 12 c m 2

Q 16. ABCD is a parallelogram whose diagonals A C andBD Intersect a tO .A line through 0 intersects AB at P and DC at Q. Prove thatar (APOA) = ar (AQOC).

Solution:

In triangles POA and QOC, we have

AAOP = ACOQ

AO = OC

APAC = AQCA

So, by ASA congruence criterion , we have

A POA 9* A QOC

=> ar (APOA) = ar (AQOC).

Q 17. ABCD Is a parallelogram. E is a point on BA such that B E=2EA and F is point on DC such that D F=2FC . Prove that A EC FIs a parallelogram whose area is one third o f the area o f parallelogram ABCD.

Solution:

Draw FG _L AB

We have,

BE = 2 EA and DF = 2FC

=>■ AB - AE = 2 AE and DC - FC = 2 FC

=>■ AB = 3 AE and DC = 3 FC

=> AE = \AB and FC = \ D C ............. (1)

But AB = DC

Then, AE = FC [opposite sides o f \ \sm]

Thus, AE = FC and AE || FC

Then, AECF is a parallelogram

Now, area of parallelogram AECF = AE x FG

=> ar (| Is"* AECF) = \AB x FG from(1)

=> 3 ar (| I9”1 AECF) = AB x F G ........(2 )

And ar (||®m ABCD) = AB x F G ........(3 )


=> 3ar (||9m AECF) = ar (||*™ ABCD)

=► ar (| Is™ AECF) = |ar (H*"* ABCD)

Q 18. In a triangle ABC, P and Q are respectively the m id points o f AB and BC and R Is the m id point ofAP. Prove that:

0) . ar (APBQ) = ar (AARC).

00 . ar (APRQ)= ja r (AARC).

Oil). ar (A RQC)= fo r (A ABC).

Solution:

We know that each median of a triangle divides it into two triangles of equal area.

(i) . Since CR is the median of A CAP

ar ( A CRA) = ja r (ACAP) ........... (1)

A ls o , CP is the median of a A CAB

ar ( A CAP) = ar ( A C P B ) ............ (2 )

From 1 and 2 , we get

ar ( A ARC) = ja r ( A C P B )............(3 )

PQ is the median of a A PBC

ar ( A CPB) = 2 a r ( A P B Q )............ (4 )

From 3 and 4, we get

ar (AARC) = ar (AP B Q )............(5 )

( i i ) . Since QP and QR medians of triangles QAB and QAP respectively

ar (AQAP) = ar ( A Q B P ) ............... (6 )

And ar ( A QAP) = 2ar ( A Q R P ) ................(7 )


ar (APRQ) = \ar{A P B Q )............(8)


ar {APRQ) = \ar{AARC)

(i ii ) . Since, LR is a median of A CAP

ar (A ARC) = \ar (ACAD)

= | x \ar{AABC)

= \ar (A ABC)

Since RQ is the median of A RBC.

ar (ARQC) = ja r (ARBC)

— j {a r [AABC) — ar ( A ARC)}

= j{a r (A A B C ) - ja r {A AB C )}

= ja r (AABC)

Q 19. ABCD is a parallelogram. G is a point on AB such that A G = 2GB and E is point on DC such that C E = 2DEand F is die point ofB C such that B F=2FC . Prove that:

0) . ar (ADEG) = ar {GBCE).

00 . ar ( A EGB)= ja r (ABCD).

010 ■ ar (AE F C )= jar (AEBF).

Ov) . ar (AEGB) = f x ar (AEFC)

(v ) . Find what portion o f the area o f parallelogram is the area o f A EFG.

Solution:Given: ABCD is a parallelogram

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

To prove:

(i) . ar (ADEG) = ar (GBCE).

(ii) . ar (AEGB)= ja r (ABCD).

(i ii ) . ar (AEFC)= ja r (AEBF).

(iv) . ar ( A EGB) = § x ar ( A EFC)

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

Construction: Draw a parallel line to AB through point F and a perpendicular line to AB from C

Proof:

(i) .Since ABCD is a parallelogram

So, AB = CD and AD = BC

Consider the two trapezium s ADEG and GBCE

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

=> ED = jC D = jA B and EC = j CD = jA B

=>■ AG = jA B and BG = jA B

So, DE + A G = jA B + jA B = AB and EC + BG = jA B + jA B = AB

Since the two trapezium ADEG and GBCE have same height and their sum of two parallel sides are equal

. sum o f parallel sides . . .Since Area of trapezium = ---------------- -̂--------------- x height

So, ar (ADEG) = ar (GBCE).

(ii) . Since we know from above that

BG = jA B . So

ar (A EGB) = | x GB x Height

ar (AEGB) = j x | x AB x Height

ar ( A EGB) — j x AB x Height

ar (AEGB)= ja r (ABCD).

(i ii ) . Since height if triangle EFC and EBF are equal.So

ar (AEFC) — j x FC x Height

ar (AEFC) = j x j x FB x Height

ar (AEFC) = ja r (EBF)

Flence, ar (AEFC)= ja r (AEBF).

( iv ) . Consider the trapezium in which

ar (EGBC) = ar (AEGB) + ar (AEBF) + ar (AEFC)

=> ja r (ABCD) = ja r (ABCD) + 2ar (AEFC) + ar (AEFC)

=> ja r (ABCD) = 3ar (AEFC)

=> ar (AEFC) = ja r (ABCD)

Now from (ii)part we have

or (AEGB) = \ar (AEFC)

ar (AEGB ) = f x |or (ABCD)

ar (AEGB) = f ar (AEFC)

or (AEGB) = fa r (A EFC)

( v ) . In the figure it is given that FB = 2CF .Let CF = x and FB = 2x.

Now consider the two triangles CFI and CBFI which are similar triangle.

So by the property of similar triangle Cl = k and IH = 2k

Now consider the triangle EGF in which

or (AEFG) = ar (AESF) + ar (ASGF)

ar (AEFG) = \ SF x k + jS F x 2k

ar(AEFG) = f SF x k ........(i)

N ow ,

ar (AEGBC) = ar (SGBF) + or (ESFC)

ar (AEGBC) = \ (SF + GB) x 2 k + j ( S F + EC) x k

ar (AEGBC) = § k x S F + (GB + \EC) x k

ar (AEGBC) = f k x S F + AB + \ x f A b) x k

ja r (AABCD) = f k x S F + f AB x k

=> ar (AABCD) = 3k x SF + jA B x k [Multiply both sides by 2]

=> or (AABCD) = 3k x SF + ja r (ABCD)

=> fc x SF = ^ a r (A B C D )............(2)

From 1 and 2 we have,

ar(AEFG) = f x ^ o r (ABCD)

ar(AEFG) = ^ a r (ABCD)

Q 20. In figure, CD U AEand CYH B A

(I) . Name a triangle equal In area of ACBX (if). Prove tftatar (AZDE) = ar (ACZA).

(Hi). Prove thatar (BCZY) = ar (AEDZ).

Since, triangle BCA and triangle BYA are on the same base BA and between same parallel s BA and CY.

Then ar (ABCA) — ar (ABYA)

=> or (ACBX) + ar (ABXA) = or (ABXA) + ar (AAXY)

=> or (ACBX) = ar (A A X Y ) ..........(1)

Since, triangles ACE and ADE are on the same base AE and between same parallels CD and AE

Then, ar (AACE) = ar (AADE)

ar (ACZA) + ar (AAZE) = ar (AAZE) + ar (ADZE)

ar (ACZA) = or (A D Z E )..........(2)

Adding or (ACYG) on both sides, we get

=> or (ACBX) + ar (ACYZ) = ar (ACAY) + ar (ACYZ)

=> or (BCZY) = ar (A C Z A )........(3)


=> ar {BCZY) = ar {ADZE)

Q 21. In figure, PSDA is a parallelogram in which PQ = QR =RS and A P I! BQIICR. Prove that ar (A PQE) = ar (A CFD).

Given that PSDA is a parallelogram

Since, AP || BQ || OR ||DS and AD ||PS

Therefore, PQ = CD (equ. 1)

In triangle BED, C is the midpoint of BD and OF || BE

Therefore, F is the midpoint of ED

=> EF = PE

Smiliarly,

EF = PE

Therefore, PE = FD (equ . 2)

In triangles PQE and CFD, we have

PE = FD

Therefore, Z EPQ = Z F D C [Alternate angles]

So, by SAS criterion, we have

A PQE ^ A DCF

=> ar (APQE) = ar (ADCF)

Q 22. In figure, ABCD is a trapezium in which A B I I DC and D C =40 cm andAB = 60 cm .if X and Y are, respectively, the m idpoints o f AD and B C , prove that:

( i ) . X Y = 50 cm

0 0 . D CYXis a trapezium

OiO ■ ar {trap. D C Y X )=^ar {XYBA).

(i) Join DY and produce it to meet AB produced at P.

In triangles BYP and CYD we have,

Z B Y P = Z CYD [Vertically opposite angles]

Z D C Y = Z PBY [Since, DC ||AP]

And BY = CY

So, by ASA congruence criterion, we have

(A BYP) =* (A CYD)

=> DY = Yp and DC = BP

=> Y is the midpoint of DP

A ls o , x is the mid point of AD

Therefore, XY ||AP and XY ||| AP

=>XY = | {AB + BP)

= > X Y = | {AB + DC)

^ X Y = |(60 + 40)

= 50 cm

(ii) We have, XY ||AP

=+ XY || AB and AB ||DC

=+ XY || DC

=>■ DCYX is a trapezium

(iii) Since x and y are the mid points of Ad and BC respectively.

Therefore, trapezium DCYX and ABYX are of the same height say h cm

Now,

ar {trap. DCXY) = \ {DC + XY) x h

=>• ar {trap. DCXY ) = ^(50 + 40) x h cm2 = 45/i cm2

=+ ar {trap. ABYX) = \ {AB + XY) x h

=> ar {trap. ABYX) = |(60 + 50) x h cm2 = 55h cm2ar(trap.DCYX) _ 45ft _ ar(trap.AEYX ) 55 h 11

=> ar {trap. DCYX) = -jjar {trap. ABYX)

Q 23. In figure ABC and BDE are two equilateral triangles such that D is the midpoint o f BC. A E intersects BC in F. Prove that

(0. ar {ABDE)= \ar {AABC).

00. ar (A BDE)= \ar (ABAE).

OH). ar (ABFE)=ar {AAFD).

0v). ar (AABC)=2ar (A BEC).

(v ). ar (AFED)=\ar (AAFC).

(vt) . ar (ABFE)=2ar (AEFD).

Given that ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = -| = DE = BE

(i) We have,

ar {AABC) = ^ x2 andar{ABDE) = ^ ( f ) 2 = j x ^ -x 2

Therefore, ar (ABDE)= ja r {AABC).

( i i ) . It is given that triangles, ABC and BED are equilateral triangles

Z ACB = ZD B E = 60°

=> BE || AC (Since, alternative angles are equal)

Trinagles BAF and BEC are on the same base BE and between same parallels BF and AC .

Therefore, ar ( A BAE) = ar ( A BEC)

=> ar {ABAE) = 2ar ( A BDE) [Since, ED is a median of triangle E B C ; ar ( A BEC) = 2ar ( A BDE) ]

:. ar (A BDE) = \ar (ABAE)

(iii) Since, triangles ABC and BDE are equilateral triangles

AABC = 60° and ZBDE = 60°

AABC = ABDE

=>AB ||DE (since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between same parallels AB and DE.

Therefore, o r ( A BED) = ar {AAED)

=> ar {ABED) - ar (AEFD) = ar {A A E D )-ar (AEFD)

=> ar (ABEF) = ar (AAFD)

(iv) Since ED is the median of triangle BEC

Therefore, ar ( A BEC) = 2ar ( A BDE)

=> ar ( A BEC) = 2 x ja r ( A ABC) [From 1 , ar ( A BDE) = ja r ( A ABC)]

=> ar {ABEC) = \ar {AABC)

=> ar (AABC) = 2ar {ABEC)

(v) ar (AAFC) = ar {AAFD) + ar (AADC)

=>• a r {ABFE) + -| a r (A .A .B C ) [using part (iii), and AD is the median of triangle ABC ]

= ar {ABFE) + -| x 4 a r ( A BDE) (using part (i))

= ar {ABFE) = 2ar (AFED) • • • ■ (3)

ar (ABDE) — ar {ABFE) + ar (AFED)

=> 2o r ( A FED) + ar ( A FED)

=>3ar{AFED ).............(4 )

From 2,3 and 4 we g e t,

ar {AAFC) = 2ar (AFED) + 2 x 3ar (AFED) = 8ar {AFED)

ar {AFED) = \ar {AAFC)

(vi) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let FI be the height of vertex A, corresponding to the side BC in triangle ABC

From part (i)

ar{ABDE) = \ar{AABC)

=> \ x BD x h = j x BC x h)

=> BD x h = \{2BD x H)

^ h = \ H ............(1)

From part (iii)

ar {ABFE) = ar {AAFD)

ar {ABFE) = \ x FD x H

ar {ABFE) = \ x F D x 2 h

ar {ABFE) = 2 { \ x F D x h )

ar {ABFE) = 2ar (AEFD)

Q 24. D is the midpoint o f side BC o f A ABC end E is the midpoint ofBD. If 0 is the midpoint ofAE, Prove thet ar (A BOE)=

Solution:

Given that

D is the midpoint of sides BC of triangle ABC

E is the midpoint of BD and 0 is the midpoint of AE

Since AD and AE are the medians of triangles, ABC and ABD respectively

ar {AABD) = \ar (AA B C ) ............(1)

ar {AABE) = \ar (AA B D ) ............(2)

OB is the median of triangle ABE

T h e r e f o r e ,ar ( A BOE) = -| a r ( A ABE)

From 1,2 and 3 , we have

ar ( A BOE) = | a r ( A ABC)

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Q 25. In figure, X and Y are the m id points o f A C and AB respectively, QPIIBC and CYQ and BXP are straight lines. Provethatar {A A B P )= a r (A A C Q ).

Since X and Y are the mid points of A C and AB respectively.

Therefore, XY || BC

Clearly, triangles BYC and BXC are on the same base BC and between the same parallels Xy and BC

ar (A BYC) = ar (ABXC)

=> ar (ABYC) - ar (ABOC) = ar (ABXC) - ar (ABOC)

=> ar (ABOY) = ar {ACOX)

=> ar (ABOY) + ar (AXOY) = ar (A COX) + ar (AXOY)

=> ar (ABXY) = ar (A CXY) (2)

We observed that the quadrilaterals XYAP and XYAQ are on the same base XY and between same parallels XY and PQ.

ar {quad. XYAP) = ar {quadXYQA)..........(2)Adding 1 and 2 , we get

ar (A BXY) + ar {quad. XYAP) = ar (ACXY) + ar {quadXYQA)

=> ar {AABP) = ar {AACQ)

Q 26. In figure, ABCD and AEFD are two parallelograms. Prove that

0 ).P E = F Q

00. ar {AAPE) : ar (APFA) = ar (A QFD) : ar (APFD)

010. ar (APEA)= ar (AQFD).

Given that, ABCD and AEFD are two parallelograms

(i) . In triangles, EPA and FQD

/.PEA = /QFD [corresponding angles]

/EPA = /FQD [corresponding angles]

PA = QD [opposite sides of parallelogram]

Then, A EPA = A FQD [By AAS condition]

Therefore, EP = FQ [C.P.C.T]

( i i ) . Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same parallels EQ and AD

Therefore, ar {APEA) = ar ( A QFD) (1)

Since, triangles PEA and PFD stand on the same base PF and between same parallels PF and AD

Therefore, ar ( A PFA) = ar ( A PFD) (2)

Divide the equation 1 by equation 2

ar(A P E A ) ar(AQFD)ar(A P F A ) ~ ar(A PFD )

(iii). From part (i), A EPA = A FQD Then, ar (APEA)= ar(AQFD).

Q27. In figure, ABCD is a parallelogram. O is any point on AC. PQ IIAB and LM II AD. Provetha t:ar ( | |9,71 DLOP) — ar (||9171BMOQ).

Since a diagonal of a parallelogram divides it into two triangles of equal area

Therefore, ar (AADC) = ar ( A ABC)

=> ar (AAPO) + ar (| |̂ D L O P ) + ar (A OLC)

=>■ ar (AAOM) + ar (| |91,1 BMOQ) + ar (A OQC) (1)

Since AO and Oc are diagonals of parallelograms AM OP and OQCL respectively.

ar (AAPO) = ar (AAM O) (2)

And ar ( A OLC) = ar ( A OQC) (3)

Subtracting 2 and 3 from 1, we get

ar (||s™DLOP) = ar (W^BMOQ).

Q28. In a triangle ABC, If Land M are points on AB and A C respectively such that LM IIBC. Provethat:

(0 . ar (ALCM) = ar (ALBM).

00. ar (ALBC) = ar (AMBC).

OH) . ar ( A ABM) = ar (,AACL).

Ov) . ar (ALOB) = ar (A MOC).

Solution:

(i) . Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC.

:.a r (ALMB) = ar (ALMC) (1)

( i i ) . We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC.

ar (ALBC) = ar (AMBC) (2)

(i ii ) . We have,

ar ( A LMB) = ar ( A LMC) [From 1 ]

=> ar (AALM) + ar (ALMB) = ar (AALM) + ar (ALMC)

=> ar (AABM) = ar (AACL)

( iv ) . We have,

ar ( A LBC) = ar ( A MBC) [From 1 ]

=> ar (ALBC) - ar (ABOC) = ar (AMBC) - ar (ABOC)

=> ar (ALOB) = ar (A MOC).

Q29. In figure, D and E are two points on BC such that BD = D E=EC . Show that ar(AABD) = ar(AADE) = ar(AAEC).

Solution:

Draw a line I through A parallel to BC.

Given that, BD = DE = EC

We observed that the triangles ABD and AEC are on the equal bases and between the same parallels I and BC. Therefore, their areas are equal.

Hence, ar ( A ABD) = ar ( A ADE) = ar ( A AEC).

Q 30. In figure, ABC is a right angled triangle at A BCED, ACFG and ABM N are squares on die sides BC, CA and AB respectively: Line segment A X -L D E meets B CatY. Show that

( i ) . AM BC = AABD

00. ar (BYXD ) = 2ar {AMBC)

Oil) . ar {BYXD) = ar {ABMN)

(!v ). AFCB = A ACE

(v ). ar {CYXE) = 2ar {AFCB)

(v t). ar {CYXE) = ar {ACFG)

(v ll). ar {BCED) = ar {ABMN) + ar {ACFG)

( i ) . In A M B C and A A B D , we have

MB = AB

BC = BD

And Z MBC = Z ABD [since ,Z M B C and Z A B C are obtained by adding Z A B C to a right angle.]

So, by SAS congruence criterion, we have

A MBC =* A ABD

=> ar {AMBC) = ar (AA B D ) ..........(1)

( i i ) . Clearly, triangle ABC and rectangle BYXD are on the same base BD and between the same parallels AX and BD .

ar (A ABD) = \ar {rect BYXD)

=>• ar {rect BYXD) — 2ar {AABD)

=> ar {rect BYXD) = 2 o r {A M B C ) .................... (2 ) [From equ .1]

(i ii ) . Since triangles MBC and square MBAN are on the same base Mb and between the same parallels MB and NC.

2 a r ( A MBC) = ar {M B A N ) .................... (3 )

From equ. 2 and 3, we have

ar {sq. MBAN) = ar {rectBYXD)

( iv ) . In triangles FOB and ACE, we have

FC = AC

CB = CE

A n d , Z F C B = Z ACE [since ,Z FC B and Z A C E are obtained by adding Z A C B to a right angle.]

So, by SAS congruence criterion, we have

AFCB S A ACE

( v ) . We have.

A FCB =* A ACE

=> or (AFCB) = or {AACE)

Clearly, triangle ACE and rectangle CYXE are on the same base CE and between same parallels CE and AX.

2or (AACE) = or (CYXE)

=> 2or (AFCB) = or (AC Y X E )............(4)

(v l ) . Clearly, triangle FCb and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2or (AFCB) = or (FC A G )..........(5)


or (CYXE) = or (ACFG)

(v ii). Applying Pythagoras theorem In triangle ACB, we have

BC2 = AB2 + AC2

=>■ BC x BD = AB x M B + AC x FC

=> or (BCED) = or (ABMN) + or (ACFG)

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RD Sharma Solutions Class 9 Areas Of Parallelograms And ... · RD Sharma Solutions Class 9 Areas Of Parallelograms And Triangles Ex 15.1 RD Sharma Solutions Class 9 Chapter 15 Ex