## 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

=(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 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, BM⊥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 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] = 96cm^{2}

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) 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^{2 }= [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 ΔDBC,

DB^{2 }= DC^{2} – CB^{2} = 17^{2} – 8^{2} = 289 – 64 = 225 cm^{2}

Therefore, DB=225−−−√=15cm

So, area of ΔDBC = (1/2 x 15 x 8) cm^{2} = 60 cm^{2}

Again, in right angled ΔDAB,

AB^{2} = DB^{2} – AD^{2} = 15^{2} – 9^{2} = 225 – 81 = 144cm^{2}

Therefore, AB=144−−−√=12cm

Therefore, the area of ΔDAB = (1/2 x 12 x 9) cm^{2} = 54 cm^{2}

So, area of quadrilateral ABCD = Area of ΔDBC +Area of ΔDAB = (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:**

RT⊥PQ

In right angled ΔRTQ

RT^{2} = RQ^{2} – TQ^{2} = 17^{2} – 8^{2} = 289 – 64 = 225 cm^{2}

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) cm^{2} = 180 cm^{2}

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

**Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If AL⊥BD and CM⊥BD , show that**

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

**Sol:**

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

AL⊥BDandCM⊥BD

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

**Proof:**

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 AL⊥BD and CM⊥BD such that AL = 8cm and CM = 6cm, find the area of quad ABCD.**

**Sol:**

Area of ΔBAD=12×BD×AL = (1/2 x 14 x 8) cm^{2} = 56cm^{2}

Area of ΔCBD=12×BD×CM = (1/2 x 14 x 6) cm^{2} = 42cm^{2}

Therefore, the Area of quadrilateral ABCD= Area of ΔABD +Area of ΔCBD = (56 + 42) cm^{2} = 98 cm^{2}

**Q.9: In the adjoining figure, ABCD is a trapezium in which AB∥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.. 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, DE∥BC. Prove that:**

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

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

**Sol:**

**(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.**

**Show that: DE∥BC**

**Sol:**

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

Such that ar(ΔBCE)=ar(ΔBCD)

**To prove:** DE∥AC

**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, DE∥AC

**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)=**

**12(∥gmABCD)**

**Sol:**

**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 PQ∥ABandRS∥AD

**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[ar∥gmABCD]

**(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[ar∥gmABCD]