RS Aggarwal Class 9 Solutions Area

RS Aggarwal Class 9 Solutions Chapter 10

Area Exercise 10.1

Q.1: In the adjoining figure, show that ABCD is parallelogram .Calculate the area parallelogram ABCD.

C:\Users\user\Desktop\12.PNG

Sol:

C:\Users\user\Desktop\12.PNG

AreaofΔABD=12×base×height =(12×5×7)cm2=352cm2 AreaofΔABD=(12×5×7)cm2=352cm2

Since the diagonal BD divides ABCD into two triangles of equal area.

Therefore, Areaofparallelogram=AreaofΔABD+AreaofΔCBD

=(352+352)cm2=702cm2=35cm2

Therefore, Area of parallelogram = 35 cm2

 

Q.2: In a parallelogram ABCD, it is being given that AB = 10cm and the altitudes corresponding to the sides AB and AD are DL = 6cm and BM = 8cm, respectively. Find AD.

Sol:

 

C:\Users\user\Desktop\a1.PNG

 

Since, ABCD is a parallelogram and DL Is perpendicular to AB

So, its area =AB x DL = (10 x 6) cm2 = 60cm2

Also in parallelogram ABCD, BMAD

Therefore, Area of parallelogram ABCD= AD x BM

60 = AD x 8cm

AD = 608=7.5cm

Therefore, AD = 7.5 cm

 

Q.3: Find the area of a rhombus, the lengths whose diagonals are 16cm and 24cm respectively.

Sol:

ABCD is a rhombus in which diagonal Ac = 24cm and BD = 16cm

These diagonals intersect at O:

 

C:\Users\user\Desktop\b.PNG

 

Since, diagonals of a rhombus are perpendicular to each other. So, in ΔACD,

OD is its altitude and AC is its base

So, area of ΔACD=1/2 x AC x OD

= 1/2 x 24 x BD/2 = (12×24×8)cm2[Since,BD=16cm] = 96cm2

i.e. Area of ΔABC=12×AC×OB

= (12×24×8) cm2=96cm2

Now, the Area of rhombus = Area of ΔACD + Area of ΔABC = (96 + 96) cm2 = 192 cm2

 

Q.4: Find the area of a rhombus, the lengths of whose diagonals are 16 cm and 24 cm respectively.

Sol:

ABCD is a trapezium in which, AB || CD

AB = 9cm and CD = 6cm

CE is a perpendicular drawn to AB through C and CE = 8cm

 

C:\Users\user\Desktop\c.PNG

 

Area of trapezium = 1/2 (sum of parallel sides) x distance between them = [1/2 (9+6) x 8] cm= [1/2 x 15 x 8] cm2 = 60 cm2

Therefore, the Area of trapezium = 60 cm2

 

Q.5: Calculate the area of quad ABCD, given in Fig.

 

C:\Users\user\Desktop\d.PNG

 

Sol:

ABCD is a quadrilateral:

 

C:\Users\user\Desktop\d.PNG

 

Now in right angled ΔDBC,

DB2 = DC2 – CB2 = 172 – 82 = 289 – 64 = 225 cm2

Therefore, DB=225=15cm

So, area of ΔDBC = (1/2 x 15 x 8) cm2 = 60  cm2

Again, in right angled ΔDAB,

AB2 = DB2 – AD2 = 152 – 92 = 225 – 81 = 144cm2

Therefore, AB=144=12cm

Therefore, the area of ΔDAB = (1/2 x 12 x 9) cm2 = 54  cm2

So, area of quadrilateral ABCD = Area of ΔDBC +Area of ΔDAB = (60 + 54) cm2 =114 cm2

Therefore, the area of quadrilateral ABCD = 114 cm2

 

Q.6: Calculate the area of trap PQRS, given in Fig. (ii)

 

C:\Users\user\Desktop\s.PNG

 

Sol:

 

C:\Users\user\Desktop\s.PNG

 

RTPQ

In right angled ΔRTQ

RT2 = RQ2 – TQ2 = 172 – 82 = 289 – 64 = 225 cm2

Therefore, RT=225=15cm

Therefore, Area of trapezium = 1/2(sum of parallel sides) x distance between them

= 1/2 x (PQ + SR) x RT =1/2 x (16 +8) x 15 = (1/2 x 24 x 15) cm2 = 180 cm2

Therefore, the Area of trapezium = 180 cm2

 

Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If ALBD and CMBD , show that

Area (quad. ABCD) = 1/2 * BD * (AL + CM)

 

C:\Users\user\Desktop\f.PNG

 

Sol:

Given: ABCD is a quadrilateral and BD is one of its diagonals

ALBDandCMBD

To prove: area (quad. ABCD) = 1/2 x BD x (AL + CM)

Proof:

 

C:\Users\user\Desktop\f.PNG

 

Area of ΔBAD=12×BD×AL

Area of ΔCBD=12×BD×CM

Therefore, Area of quadrilateral ABCD= Area of ΔABD +Area of ΔCBD = 1/2 x BD x AL + 1/2 x BD x CM

Therefore, the Area of quadrilateral ABCD = 1/2 x BD [AL + CM]

 

Q.8: In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14cm. If ALBD and CMBD such that AL = 8cm and CM = 6cm, find the area of quad ABCD.

 

C:\Users\user\Desktop\r.PNG

 

Sol:

 

C:\Users\user\Desktop\r.PNG

 

Area of ΔBAD=12×BD×AL = (1/2 x 14 x 8) cm2 = 56cm2

Area of ΔCBD=12×BD×CM = (1/2 x 14 x 6) cm2 = 42cm2

Therefore, the Area of quadrilateral ABCD= Area of ΔABD +Area of ΔCBD = (56 + 42) cm2 = 98 cm2

 

Q.9: In the adjoining figure, ABCD is a trapezium in which ABBD and its diagonals AC and BD intersect at O.

Prove that area of triangle AOD = area of the triangle of BOC.

 

C:\Users\user\Desktop\q.PNG

 

Sol:

 

C:\Users\user\Desktop\q.PNG

 

Consider ΔADCandΔDCB.. We find they have the same base CD and lie between two parallel lines DC and AB

Triangles on the same base and between the same parallels are equal in area.

So ΔCDAandΔCDB. are equal in area.

Therefore, area(ΔCDA)=area(ΔCDB)

Now, area(ΔAOD)=area(ΔADC)area(ΔOCD)

And, area(ΔBOC)=area(ΔCDB)area(ΔOCD)

= area(ΔADC)area(ΔOCD)

Therefore, Area (ΔAOD)=area(ΔBOC)

 

Q.10: In the adjoining figure, DEBC. Prove that:

 

C:\Users\user\Desktop\2a.PNG

 

(i) Area of triangle ACD = area ABE

(ii) Area of triangle OCE = area of triangle OBD

Sol:

 

C:\Users\user\Desktop\2a.PNG

 

(i) ΔDBEandΔDCE have the same base DE and lie between parallel lines BC and DE.

So, area (ΔDBE)=area(ΔDCE) . . . . . . . (1)

Adding area (ΔDBE) on both sides ,we get

ar(ΔDBE)+ar(ΔADE)=ar(ΔDCE)+ar(ΔADE)

Therefore, ar(ΔABE)=ar(ΔACD)

 

(ii) Since ar(ΔDBE)=ar(ΔDCE) [From (1)]

Subtracting ar(ΔODE) from both sides we get

ar(ΔDBE)ar(ΔODE)=ar(ΔDCE)ar(ΔODE)

Therefore, ar(ΔOBD)=ar(ΔOCE)

 

Q.11: In the adjoining figure, D and E are points on the sides AB and AC of triangle ABC such that area of triangle BCE = area of triangle BCD.

 

C:\Users\user\Desktop\g.PNG

 

Show that: DEBC

Sol:

 

C:\Users\user\Desktop\g.PNG

 

Given: A ΔABC in which points D and E lie on AB and AC,

Such that ar(ΔBCE)=ar(ΔBCD)

To prove: DEAC

Proof:

As ΔBCEandΔBCD have same base BC, and are equal in area, they have same altitudes.

This means that they lie between two parallel lines. Therefore, DEAC

 

Question 12: In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that:

 

C:\Users\user\Desktop\n.PNG

 

(i) ar.( OAB )+ar.( OCD)= 12(gmABCD)

(ii) ar.( OAD)+ ar.( OBC)=

12(gmABCD)

Sol:

 

C:\Users\user\Desktop\n.PNG

 

Given: A parallelogram ABCD in which O is a point inside it

To prove: (i) ar(ΔOAB)+ar(ΔOCD)=12ar(gmABCD)

(ii) ar(ΔOAD)+ar(ΔOBC)=12ar(gmABCD)

Construction: Though O draw PQABandRSAD

Proof: (i) ΔAOB and parallelogram ABQP have same base AB and lie between parallel lines AB and PQ.

If a triangle and a parallelogram are on the same base, and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Therefore, ar(ΔAOB)=12ar(gmABQP)

Similarly, ar(ΔCOD)=12ar(gmPQCD)

So, ar(ΔAOB)+ar(ΔCOD)

= 12ar(gmABQP)+12ar(gmPQCD)

= 12[ar(gmABQP)+ar(gmPQCD)]

= 12[argmABCD]

 

(ii) ΔAOD and gm ASRD have the same base AD and lie between same parallel lines AD and RS.

Thererfore, ar(ΔAOD)=12ar(gmASRD)

Similarly, ar(ΔBOC)=12ar(gmRSBC)

Therefore, ar(Δaod)+ar(ΔBOC)=12[ar(gmASRD)+ar(gmRSBC)]

12[argmABCD]

Related Links
NCERT Books NCERT Solutions RS Aggarwal
Lakhmir Singh RD Sharma Solutions NCERT Solutions Class 6 to 12
More RS Aggarwal Solutions
RS Aggarwal Solutions Class 9 Solutions Chapter 11 Circle Exercise 11 1RS Aggarwal Solutions Class 9 Solutions Chapter 13 Volume And Surface Area Exercise 13 2
RS Aggarwal Solutions Class 9 Solutions Chapter 14 Statistics Exercise 14 3RS Aggarwal Solutions Class 9 Solutions Chapter 14 Statistics Exercise 14 7
RS Aggarwal Solutions Class 9 Solutions Chapter 2 Polynomials Exercise 2 1RS Aggarwal Solutions Class 9 Solutions Chapter 5 Congruence And Inequalities In Triangle Exercise 5 1
RS Aggarwal Solutions Class 9 Solutions Chapter 9 Quadrilaterals And Parallelograms Exercise 9 1RS Aggarwal Solutions Class 9 Solutions Circle Exercise 11 2