RS Aggarwal Class 9 Solutions Area

A quantity that measures the top shape, planar lamina and plane. The surface area is the area of a two dimensional surface on a three dimensional object. The unit of area is primarily concerned with the side of the length and they are primarily measured in square meters, square kilometers, square yards, square miles etc. Some of the areas of different shapes are:

  1. Quadrilateral Area
  2. Triangle Area
  3. Rectangular Area
  4. Polygon Area
  5. Area of a Circle
  6. Area of an Ellipse
  7. Surface Area
  8. Fractals
  9. Area of Calculus

Learn more about RS Aggarwal Class 9 Solutions Chapter 10 Area below:

RS Aggarwal Class 9 Solutions Chapter 10

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





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

Therefore, Areaofparallelogram=AreaofΔABD+AreaofΔCBD


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.





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.


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

These diagonals intersect at O:




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.


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




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.





ABCD is a quadrilateral:




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)









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)





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


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





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.








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.








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:




(i) Area of triangle ACD = area ABE

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





(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


Therefore, ar(ΔABE)=ar(ΔACD)


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

Subtracting ar(ΔODE) from both sides we get


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.




Show that: DEBC





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

Such that ar(ΔBCE)=ar(ΔBCD)

To prove: DEAC


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:




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

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






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)]


Practise This Question

Alpha particle scattering experiment gave totally unexpected results. Which of the following observations were made?