# RS Aggarwal Class 9 Solutions Chapter 10 - Area

## RS Aggarwal Class 9 Chapter 10 - Area Solutions Free PDF

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.

Sol:

AreaofΔABD=12×base×height$Area\:of\:\Delta ABD=\frac{1}{2}\times base\times height$
=(12×5×7)cm2=352cm2$=\left ( \frac{1}{2}\times 5\times 7 \right )cm^{2}=\frac{35}{2}cm^{2}$
AreaofΔABD=(12×5×7)cm2=352cm2$Area\:of\:\Delta ABD=\left ( \frac{1}{2}\times 5\times 7 \right )cm^{2}=\frac{35}{2}cm^{2}$

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

Therefore, Areaofparallelogram=AreaofΔABD+AreaofΔCBD$Area\:of\:parallelogram=Area\:of\:\Delta ABD +Area\:of\:\Delta CBD$

=(352+352)cm2=702cm2=35cm2$=\left ( \frac{35}{2}+\frac{35}{2} \right )cm^{2}=\frac{70}{2}cm^{2}=35cm^{2}$

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:

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$BM\perp AD$

Therefore, Area of parallelogram ABCD= AD x BM

60 = AD x 8cm

AD = 608=7.5cm$\frac{60}{8}=7.5cm$

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:

Since, diagonals of a rhombus are perpendicular to each other. So, in ΔACD,$\Delta ACD,$

OD is its altitude and AC is its base

So, area of ΔACD$\Delta ACD$=1/2 x AC x OD

= 1/2 x 24 x BD/2 = (12×24×8)cm2[Since,BD=16cm]$\left ( \frac{1}{2} \times 24\times 8\right )cm^{2}\:\:\:\:\left [ Since, BD=16cm \right ]$ = 96cm2

i.e. Area of ΔABC=12×AC×OB$\Delta ABC=\frac{1}{2}\times AC\times OB$

= (12×24×8)$\left ( \frac{1}{2}\times 24\times 8 \right )$ cm2=96cm2$cm^{2}=96cm^{2}$

Now, the Area of rhombus = Area of ΔACD$\Delta ACD$ + Area of ΔABC$\Delta ABC$ = (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

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.

Sol:

ABCD is a quadrilateral:

Now in right angled ΔDBC,$\Delta DBC,$

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

Therefore, DB=225=15cm$DB = \sqrt{225}=15cm$

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

Again, in right angled ΔDAB,$\Delta DAB,$

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

Therefore, AB=144=12cm$AB = \sqrt{144}=12cm$

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

So, area of quadrilateral ABCD = Area of ΔDBC$\Delta DBC$ +Area of ΔDAB$\Delta DAB$ = (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)

Sol:

RTPQ$RT\perp PQ$

In right angled ΔRTQ$\Delta RTQ$

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

Therefore, RT=225=15cm$RT = \sqrt{225}=15cm$

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$AL\perp BD$ and CMBD$CM\perp BD$ , show that

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

Sol:

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

ALBDandCMBD$AL\perp BD\:and\:CM\perp BD$

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

Proof:

Area of ΔBAD=12×BD×AL$\Delta BAD=\frac{1}{2}\times BD\times AL$

Area of ΔCBD=12×BD×CM$\Delta CBD=\frac{1}{2}\times BD\times CM$

Therefore, Area of quadrilateral ABCD= Area of ΔABD$\Delta ABD$ +Area of ΔCBD$\Delta CBD$ = 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$AL\perp BD$ and CMBD$CM\perp BD$ such that AL = 8cm and CM = 6cm, find the area of quad ABCD.

Sol:

Area of ΔBAD=12×BD×AL$\Delta BAD=\frac{1}{2}\times BD\times AL$ = (1/2 x 14 x 8) cm2 = 56cm2

Area of ΔCBD=12×BD×CM$\Delta CBD=\frac{1}{2}\times BD\times CM$ = (1/2 x 14 x 6) cm2 = 42cm2

Therefore, the Area of quadrilateral ABCD= Area of ΔABD$\Delta ABD$ +Area of ΔCBD$\Delta CBD$ = (56 + 42) cm2 = 98 cm2

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

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

Sol:

Consider ΔADCandΔDCB.$\Delta ADC\:and\:\Delta DCB.$. 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.$\Delta CDA\:and\:\Delta CDB.$ are equal in area.

Therefore, area(ΔCDA)=area(ΔCDB)$area\left ( \Delta CDA \right )=area\left ( \Delta CDB \right )$

Now, area(ΔAOD)=area(ΔADC)area(ΔOCD)$area\left ( \Delta AOD \right )=area\left ( \Delta ADC \right )-area\left ( \Delta OCD \right )$

And, area(ΔBOC)=area(ΔCDB)area(ΔOCD)$area\left ( \Delta BOC \right )=area\left ( \Delta CDB \right )-area\left ( \Delta OCD \right )$

= area(ΔADC)area(ΔOCD)$area\left ( \Delta ADC \right )-area\left ( \Delta OCD \right )$

Therefore, Area (ΔAOD)=area(ΔBOC)$\left ( \Delta AOD \right )=area\left ( \Delta BOC \right )$

Q.10: In the adjoining figure, DEBC$DE\parallel BC$. Prove that:

(i) Area of triangle ACD = area ABE

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

Sol:

(i) ΔDBEandΔDCE$\Delta DBE \:and\:\Delta DCE$ have the same base DE and lie between parallel lines BC and DE.

So, area (ΔDBE)=area(ΔDCE)$\left ( \Delta DBE\right )=area\left ( \Delta DCE \right )$ . . . . . . . (1)

Adding area (ΔDBE)$\left ( \Delta DBE \right )$ on both sides ,we get

ar(ΔDBE)+ar(ΔADE)=ar(ΔDCE)+ar(ΔADE)$ar\left ( \Delta DBE \right )+ar\left ( \Delta ADE \right )=ar\left ( \Delta DCE \right )+ar\left ( \Delta ADE \right )$

Therefore, ar(ΔABE)=ar(ΔACD)$ar\left ( \Delta ABE \right )=ar\left ( \Delta ACD \right )$

(ii) Since ar(ΔDBE)=ar(ΔDCE)$ar\left ( \Delta DBE \right )=ar\left ( \Delta DCE \right )$ [From (1)]

Subtracting ar(ΔODE)$ar\left ( \Delta ODE \right )$ from both sides we get

ar(ΔDBE)ar(ΔODE)=ar(ΔDCE)ar(ΔODE)$ar\left ( \Delta DBE \right )-ar\left ( \Delta ODE \right )=ar\left ( \Delta DCE \right )-ar\left ( \Delta ODE \right )$

Therefore, ar(ΔOBD)=ar(ΔOCE)$ar\left ( \Delta OBD \right )=ar\left ( \Delta OCE \right )$

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$DE\parallel BC$

Sol:

Given: A ΔABC$\Delta ABC$ in which points D and E lie on AB and AC,

Such that ar(ΔBCE)=ar(ΔBCD)$ar\left ( \Delta BCE \right )=ar\left ( \Delta BCD \right )$

To prove: DEAC$DE\parallel AC$

Proof:

As ΔBCEandΔBCD$\Delta BCE\:and\:\Delta BCD$ have same base BC, and are equal in area, they have same altitudes.

This means that they lie between two parallel lines. Therefore, DEAC$DE\parallel AC$

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

(i) ar.( OAB$\triangle OAB$ )+ar.( OCD$\triangle OCD$)= 12(gmABCD)$\frac{1}{2}(\parallel gm\;ABCD)$

(ii) ar.( OAD$\triangle OAD$)+ ar.( OBC$\triangle OBC$)=

12(gmABCD)$\frac{1}{2}(\parallel gm\;ABCD)$

Sol:

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

To prove: (i) ar(ΔOAB)+ar(ΔOCD)=12ar(gmABCD)$ar\left ( \Delta OAB \right )+ar\left ( \Delta OCD \right )=\frac{1}{2}ar\left ( \parallel gmABCD \right )$

(ii) ar(ΔOAD)+ar(ΔOBC)=12ar(gmABCD)$ar\left ( \Delta OAD \right )+ar\left ( \Delta OBC \right )=\frac{1}{2}ar\left ( \parallel gmABCD \right )$

Construction: Though O draw PQABandRSAD$PQ\parallel AB\:and\:RS\parallel AD$

Proof: (i) ΔAOB$\Delta AOB$ 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)$ar\left ( \Delta AOB \right )=\frac{1}{2}ar\left ( \parallel gm\:ABQP \right )$

Similarly, ar(ΔCOD)=12ar(gmPQCD)$ar\left ( \Delta COD \right )=\frac{1}{2}ar\left ( \parallel gm\:PQCD \right )$

So, ar(ΔAOB)+ar(ΔCOD)$ar\left ( \Delta AOB \right )+ar\left ( \Delta COD \right )$

= 12ar(gmABQP)+12ar(gmPQCD)$\frac{1}{2}ar\left ( \parallel gm\:ABQP \right )+\frac{1}{2}ar\left ( \parallel gm\:PQCD \right )$

= 12[ar(gmABQP)+ar(gmPQCD)]$\frac{1}{2}\left [ ar\left ( \parallel gm\:ABQP \right )+ar\left ( \parallel gm\:PQCD \right ) \right ]$

= 12[argmABCD]$\frac{1}{2}\left [ ar\parallel gm\:ABCD \right ]$

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

Thererfore, ar(ΔAOD)=12ar(gmASRD)$ar\left ( \Delta AOD \right )=\frac{1}{2}ar\left ( \parallel gm\:ASRD \right )$

Similarly, ar(ΔBOC)=12ar(gmRSBC)$ar\left ( \Delta BOC \right )=\frac{1}{2}ar\left ( \parallel gm\:RSBC \right )$

Therefore, ar(Δaod)+ar(ΔBOC)=12[ar(gmASRD)+ar(gmRSBC)]$ar\left ( \Delta aod \right )+ar\left ( \Delta BOC \right ) = \frac{1}{2}\left [ ar\left ( \parallel gm\:ASRD \right )+ar\left ( \parallel gm\:RSBC \right ) \right ]$

12[argmABCD]$\frac{1}{2}\left [ ar\parallel gm\:ABCD \right ]$

#### Practise This Question

POQ and QOR from linear pair of angles. The sum of POQ and QOR is ___________.