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:

- Quadrilateral Area
- Triangle Area
- Rectangular Area
- Polygon Area
- Area of a Circle
- Area of an Ellipse
- Surface Area
- Fractals
- 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:**

AreaofÎ”ABD=12Ã—baseÃ—height

=(12Ã—5Ã—7)cm2=352cm2

AreaofÎ”ABD=(12Ã—5Ã—7)cm2=352cm2

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

Therefore, Areaofparallelogram=AreaofÎ”ABD+AreaofÎ”CBD

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

Therefore, Area of parallelogram = 35 cm^{2}

**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) cm^{2} = 60cm^{2}

Also in parallelogram ABCD, BMâŠ¥AD

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:

Since, diagonals of a rhombus are perpendicular to each other. So, in Î”ACD,

OD is its altitude and AC is its base

So, area of Î”ACD=1/2 x AC x OD

= 1/2 x 24 x BD/2 = (12Ã—24Ã—8)cm2[Since,BD=16cm] = 96cm^{2}

i.e. Area of Î”ABC=12Ã—ACÃ—OB

= (12Ã—24Ã—8)cm2=96cm2

Now, the Area of rhombus = Area of Î”ACD + Area of Î”ABC = (96 + 96) cm^{2} = 192 cm^{2}

**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^{2Â }= [1/2 x 15 x 8] cm^{2} = 60 cm^{2}

Therefore, the Area of trapezium = 60 cm^{2}

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

**Sol:**

ABCD is a quadrilateral:

Now in right angled Î”DBC,

DB^{2 }= DC^{2} â€“ CB^{2} = 17^{2} â€“ 8^{2} = 289 â€“ 64 = 225 cm^{2}

Therefore,Â DB=225âˆ’âˆ’âˆ’âˆš=15cm

So, area of Î”DBC = (1/2 x 15 x 8) cm^{2} = 60 Â cm^{2}

Again, in right angled Î”DAB,

AB^{2} = DB^{2} â€“ AD^{2} = 15^{2} â€“ 9^{2} = 225 â€“ 81 = 144cm^{2}

Therefore,Â AB=144âˆ’âˆ’âˆ’âˆš=12cm

Therefore, the area of Î”DAB = (1/2 x 12 x 9) cm^{2} = 54 Â cm^{2}

So, area of quadrilateral ABCD = Area of Î”DBC +Area of Î”DAB = (60 + 54) cm^{2} =114 cm^{2}

Therefore, the area of quadrilateral ABCD = 114 cm^{2}

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

**Sol:**

RTâŠ¥PQ

In right angled Î”RTQ

RT^{2} = RQ^{2} â€“ TQ^{2} = 17^{2} â€“ 8^{2} = 289 â€“ 64 = 225 cm^{2}

Therefore, RT=225âˆ’âˆ’âˆ’âˆš=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) cm^{2} = 180 cm^{2}

Therefore, theÂ Area of trapezium = 180 cm^{2}

**Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If ALâŠ¥BD and CMâŠ¥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

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

**Proof:**

Area of Î”BAD=12Ã—BDÃ—AL

Area of Î”CBD=12Ã—BDÃ—CM

Therefore, Area of quadrilateral ABCD= Area of Î”ABD +Area of Î”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 and CMâŠ¥BD such that AL = 8cm and CM = 6cm, find the area of quad ABCD.**

**Sol:**

Area of Î”BAD=12Ã—BDÃ—AL = (1/2 x 14 x 8) cm^{2} = 56cm^{2}

Area of Î”CBD=12Ã—BDÃ—CM = (1/2 x 14 x 6) cm^{2} = 42cm^{2}

Therefore,Â the Area of quadrilateral ABCD= Area of Î”ABD +Area of Î”CBD = (56 + 42) cm^{2} = 98 cm^{2}

**Q.9: In the adjoining figure, ABCD is a trapezium in which ABâˆ¥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.. 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. are equal in area.

Therefore, area(Î”CDA)=area(Î”CDB)

Now, area(Î”AOD)=area(Î”ADC)âˆ’area(Î”OCD)

And, area(Î”BOC)=area(Î”CDB)âˆ’area(Î”OCD)

= area(Î”ADC)âˆ’area(Î”OCD)

Therefore, Area (Î”AOD)=area(Î”BOC)

**Q.10: In the adjoining figure, DEâˆ¥BC. Prove that:**

**(i) Area of triangle ACD = area ABE**

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

**Sol:**

**(i)** Î”DBEandÎ”DCE have the same base DE and lie between parallel lines BC and DE.

So, area (Î”DBE)=area(Î”DCE) . . . . . . . (1)

Adding area (Î”DBE) on both sides ,we get

ar(Î”DBE)+ar(Î”ADE)=ar(Î”DCE)+ar(Î”ADE)

Therefore, ar(Î”ABE)=ar(Î”ACD)

**(ii)** Since ar(Î”DBE)=ar(Î”DCE) [From (1)]

Subtracting ar(Î”ODE) from both sides we get

ar(Î”DBE)âˆ’ar(Î”ODE)=ar(Î”DCE)âˆ’ar(Î”ODE)

Therefore, ar(Î”OBD)=ar(Î”OCE)

**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**

**Sol:**

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

Such that ar(Î”BCE)=ar(Î”BCD)

**To prove:** DEâˆ¥AC

**Proof:**

As Î”BCEandÎ”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

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

**(i) ar.(** **â–³OAB )+ar.(** **â–³OCD)=** **12(âˆ¥gmABCD)**

**(ii) ar.( â–³OAD)+ ar.( â–³OBC)=**

**12(âˆ¥gmABCD)**

**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)

**(ii)** ar(Î”OAD)+ar(Î”OBC)=12ar(âˆ¥gmABCD)

**Construction:** Though O draw PQâˆ¥ABandRSâˆ¥AD

**Proof: (i)** Î”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)

Similarly, ar(Î”COD)=12ar(âˆ¥gmPQCD)

So, ar(Î”AOB)+ar(Î”COD)

= 12ar(âˆ¥gmABQP)+12ar(âˆ¥gmPQCD)

= 12[ar(âˆ¥gmABQP)+ar(âˆ¥gmPQCD)]

= 12[arâˆ¥gmABCD]

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

Thererfore, ar(Î”AOD)=12ar(âˆ¥gmASRD)

Similarly, ar(Î”BOC)=12ar(âˆ¥gmRSBC)

Therefore, ar(Î”aod)+ar(Î”BOC)=12[ar(âˆ¥gmASRD)+ar(âˆ¥gmRSBC)]

=Â 12[arâˆ¥gmABCD]