# 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:

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

## RS Aggarwal Class 9 Solutions Chapter 10

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

Sol:

AreaofÎ”ABD=12Ã—baseÃ—height$Area\:of\:\Delta ABD=\frac{1}{2}\times base\times height$
=(12Ã—5Ã—7)cm2=352cm2$=\left ( \frac{1}{2}\times 5\times 7 \right )cm^{2}=\frac{35}{2}cm^{2}$
AreaofÎ”ABD=(12Ã—5Ã—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=AreaofÎ”ABD+AreaofÎ”CBD$Area\:of\:parallelogram=Area\:of\:\Delta ABD +Area\:of\:\Delta CBD$

=(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, BMâŠ¥AD$BM\perp AD$

Therefore, Area of parallelogram ABCD= AD x BM

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

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 Î”ACD,$\Delta ACD,$

OD is its altitude and AC is its base

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

= 1/2 x 24 x BD/2 = (12Ã—24Ã—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 Î”ABC=12Ã—ACÃ—OB$\Delta ABC=\frac{1}{2}\times AC\times OB$

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

Now, the Area of rhombus = Area of Î”ACD$\Delta ACD$ + Area of Î”ABC$\Delta ABC$ = (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] cm2Â = [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:

Now in right angled Î”DBC,$\Delta DBC,$

DB2 = DC2 â€“ CB2 = 172 â€“ 82 = 289 â€“ 64 = 225 cm2

Therefore,Â DB=225âˆ’âˆ’âˆ’âˆš=15cm$DB = \sqrt{225}=15cm$

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

Again, in right angled Î”DAB,$\Delta DAB,$

AB2 = DB2 â€“ AD2 = 152 â€“ 92 = 225 â€“ 81 = 144cm2

Therefore,Â AB=144âˆ’âˆ’âˆ’âˆš=12cm$AB = \sqrt{144}=12cm$

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

So, area of quadrilateral ABCD = Area of Î”DBC$\Delta DBC$ +Area of Î”DAB$\Delta DAB$ = (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:

RTâŠ¥PQ$RT\perp PQ$

In right angled Î”RTQ$\Delta RTQ$

RT2 = RQ2 â€“ TQ2 = 172 â€“ 82 = 289 â€“ 64 = 225 cm2

Therefore, RT=225âˆ’âˆ’âˆ’âˆš=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, theÂ Area of trapezium = 180 cm2

Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If ALâŠ¥BD$AL\perp BD$ and CMâŠ¥BD$CM\perp BD$ , show that

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

Sol:

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

ALâŠ¥BDandCMâŠ¥BD$AL\perp BD\:and\:CM\perp BD$

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

Proof:

Area of Î”BAD=12Ã—BDÃ—AL$\Delta BAD=\frac{1}{2}\times BD\times AL$

Area of Î”CBD=12Ã—BDÃ—CM$\Delta CBD=\frac{1}{2}\times BD\times CM$

Therefore, Area of quadrilateral ABCD= Area of Î”ABD$\Delta ABD$ +Area of Î”CBD$\Delta CBD$ = 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 ALâŠ¥BD$AL\perp BD$ and CMâŠ¥BD$CM\perp BD$ such that AL = 8cm and CM = 6cm, find the area of quad ABCD.

Sol:

Area of Î”BAD=12Ã—BDÃ—AL$\Delta BAD=\frac{1}{2}\times BD\times AL$ = (1/2 x 14 x 8) cm2 = 56cm2

Area of Î”CBD=12Ã—BDÃ—CM$\Delta CBD=\frac{1}{2}\times BD\times CM$ = (1/2 x 14 x 6) cm2 = 42cm2

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

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

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

Sol:

Consider Î”ADCandÎ”DCB.$\Delta ADC\:and\:\Delta DCB.$. 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 Î”CDAandÎ”CDB.$\Delta CDA\:and\:\Delta CDB.$ are equal in area.

Therefore, area(Î”CDA)=area(Î”CDB)$area\left ( \Delta CDA \right )=area\left ( \Delta CDB \right )$

Now, area(Î”AOD)=area(Î”ADC)âˆ’area(Î”OCD)$area\left ( \Delta AOD \right )=area\left ( \Delta ADC \right )-area\left ( \Delta OCD \right )$

And, area(Î”BOC)=area(Î”CDB)âˆ’area(Î”OCD)$area\left ( \Delta BOC \right )=area\left ( \Delta CDB \right )-area\left ( \Delta OCD \right )$

= area(Î”ADC)âˆ’area(Î”OCD)$area\left ( \Delta ADC \right )-area\left ( \Delta OCD \right )$

Therefore, Area (Î”AOD)=area(Î”BOC)$\left ( \Delta AOD \right )=area\left ( \Delta BOC \right )$

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

(i) Area of triangle ACD = area ABE

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

Sol:

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

So, area (Î”DBE)=area(Î”DCE)$\left ( \Delta DBE\right )=area\left ( \Delta DCE \right )$ . . . . . . . (1)

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

ar(Î”DBE)+ar(Î”ADE)=ar(Î”DCE)+ar(Î”ADE)$ar\left ( \Delta DBE \right )+ar\left ( \Delta ADE \right )=ar\left ( \Delta DCE \right )+ar\left ( \Delta ADE \right )$

Therefore, ar(Î”ABE)=ar(Î”ACD)$ar\left ( \Delta ABE \right )=ar\left ( \Delta ACD \right )$

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

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

ar(Î”DBE)âˆ’ar(Î”ODE)=ar(Î”DCE)âˆ’ar(Î”ODE)$ar\left ( \Delta DBE \right )-ar\left ( \Delta ODE \right )=ar\left ( \Delta DCE \right )-ar\left ( \Delta ODE \right )$

Therefore, ar(Î”OBD)=ar(Î”OCE)$ar\left ( \Delta OBD \right )=ar\left ( \Delta OCE \right )$

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: DEâˆ¥BC$DE\parallel BC$

Sol:

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

Such that ar(Î”BCE)=ar(Î”BCD)$ar\left ( \Delta BCE \right )=ar\left ( \Delta BCD \right )$

To prove: DEâˆ¥AC$DE\parallel AC$

Proof:

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

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

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

(i) ar.( â–³OAB$\triangle OAB$ )+ar.( â–³OCD$\triangle OCD$)= 12(âˆ¥gmABCD)$\frac{1}{2}(\parallel gm\;ABCD)$

(ii) ar.( â–³OAD$\triangle OAD$)+ ar.( â–³OBC$\triangle OBC$)=

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

Sol:

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

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

(ii) ar(Î”OAD)+ar(Î”OBC)=12ar(âˆ¥gmABCD)$ar\left ( \Delta OAD \right )+ar\left ( \Delta OBC \right )=\frac{1}{2}ar\left ( \parallel gmABCD \right )$

Construction: Though O draw PQâˆ¥ABandRSâˆ¥AD$PQ\parallel AB\:and\:RS\parallel AD$

Proof: (i) Î”AOB$\Delta AOB$ 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(Î”AOB)=12ar(âˆ¥gmABQP)$ar\left ( \Delta AOB \right )=\frac{1}{2}ar\left ( \parallel gm\:ABQP \right )$

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

So, ar(Î”AOB)+ar(Î”COD)$ar\left ( \Delta AOB \right )+ar\left ( \Delta COD \right )$

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

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

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

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

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

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

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

=Â 12[arâˆ¥gmABCD]$\frac{1}{2}\left [ ar\parallel gm\:ABCD \right ]$

#### Practise This Question

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