RD Sharma 2014 Solutions for Class 7 Math Chapter 20 Mensuration I are provided here with simple step-by-step explanations. These solutions for Mensuration I are extremely popular among class 7 students for Math Mensuration I Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 20 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

#### Page No 20.13:

#### Question 1:

A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by a path which is 2 m wide. Calculate the cost of levelling the path at the rate of Rs 8.25 per square metre.

#### Answer:

We have,

Length *AB* = 40 m and breadth *BC* = 25 m

∴ Area of lawn *ABCD* = 40 m x 25 m = 1000 m^{2}

Length *PQ *= (40 + 2 + 2 ) m = 44 m

Breadth *QR* = ( 25 + 2 + 2 ) m = 29 m

∴ Area of *PQRS* = 44 m x 29 m = 1276 m^{2}

Now,

Area of the path = Area of *PQRS* − Area of the lawn *ABCD*

= 1276 m^{2}^{ }− 1000 m^{2}

= 276 m^{2}

Rate of levelling the path = Rs. 8.25 per m^{2}

∴ Cost of levelling the path = Rs. ( 8.25 x 276)

= Rs. 2277

#### Page No 20.13:

#### Question 2:

One metre wide path is built inside a square park of side 30 m along its sides. The remaining part of the park is covered by grass. If the total cost of covering by grass is Rs 1176, find the rate per square metre at which the park is covered by the grass.

#### Answer:

We have,

The side of the square garden (*a*) = 30 m

∴ Area of the square garden including the path = *a*^{2} = (30)^{2} = 900 m^{2}

From the figure, it can be observed that the side of the square garden, when the path is not included, is 28 m.

Area of the square garden not including the path = (28)^{2} = 784 m^{2}

Total cost of covering the park with grass = Area of the park covering with green grass x Rate per square metre

1176 = 784 x Rate per square metre

∴ Rate per square metre at which the park is covered with grass = Rs. (1176 ÷ 784 )

= Rs. 1.50

#### Page No 20.13:

#### Question 3:

Through a rectangular field of sides 90 m × 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the field. If the which of the road is 3 m, find the total area cobered by the two roads.

#### Answer:

We have,

Length of the rectangular field = 90 m and breadth of the rectangular field = 60 m

∴ Area of the rectangular field = 90 m x 60 m = 5400 m^{2}

Area of the road *PQRS* = 90 m x 3 m = 270 m^{2}

Area of the road *ABCD* = 60 m x 3 m = 180 m^{2}

Clearly, area of *KLMN* is common to the two roads.

Thus, area of KLMN = 3 m x 3 m = 9 m^{2}

Hence,

Area of the roads = Area (*PQRS*) + Area (*ABCD*) − Area (*KLMN*)

= (270 + 180 )m^{2} − 9 m^{2} = 441 m^{2}

#### Page No 20.13:

#### Question 4:

From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.

#### Answer:

We have,

Length of the rectangular sheet = 100 cm

Breadth of the rectangular sheet = 80 cm

Area of the rectangular sheet of tin = 100 cm x 80 cm = 8000 cm^{2}

Side of the square at the corner of the sheet = 10 cm

Area of one square at the corner of the sheet = (10 cm)^{2} = 100 cm^{2}

∴ Area of 4 squares at the corner of the sheet = 4 x 100 cm^{2}^{ }= 400 cm^{2}

Hence,

Area of the remaining sheet of tin =Area of the rectangular sheet − Area of the 4 squares

Area of the remaining sheet of tin = (8000^{ }− 400) cm^{2}

= 7600 cm^{2}

#### Page No 20.14:

#### Question 5:

A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a margin of 1.5 cm along each of its sides. Fund the total area of the margin.

#### Answer:

We have,

Length of the cardboard = 8 cm and breadth of the cardboard = 5 cm

∴ Area of the cardboard including the margin = 8 cm x 5 cm = 40 cm^{2}

From the figure, it can be observed that,

New length of the painting when the margin is not included = 8 cm − (1.5 cm + 1.5 cm) = (8 − 3) cm = 5 cm

New breadth of the painting when the margin is not included = 5 cm − (1.5 cm + 1.5 cm) = (5 − 3) cm = 2 cm

∴ Area of the painting not including the margin = 5 cm x 2 cm = 10 cm^{2}

Hence,

Area of the margin = Area of the cardboard including the margin − Area of the painting

= (40 − 10) cm^{2}

= 30 cm^{2}

#### Page No 20.14:

#### Question 6:

Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to make a garden 10 m long and 4 m broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining part of the field, he wants to apply mansures. Find the cost of applying the manures at the rate of Rs 300 per are.

#### Answer:

Length of the rectangular field = 80 m

Breadth of the rectangular field = 60 m

∴ Area of the rectangular field = 80 m x 60

= 4800 m^{2}

Again,

Area of the garden = 10 m x 4 m = 40 m^{2}

Area of one flower bed = 4 m x 1.5 m = 6 m^{2}

Thus,

Area of two flower beds = 2 x 6 m^{2} = 12 m^{2}

Remaining area of the field for applying manure = Area of the rectangular field − (Area of the garden + Area of the two flower beds)

Remaining area of the field for applying manure = 4800 m^{2} − (40 + 12 ) m^{2}

= (4800 − 52 )m^{2}

= 4748 m^{2}

Since 100 m^{2} = 1 are

∴ 4748 m^{2} = 47.48 ares

So, cost of applying manure at the rate of Rs. 300 per are will be Rs. (300 x 47.48) = Rs. 14244

#### Page No 20.14:

#### Question 7:

Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip 30 cm wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.

#### Answer:

We have,

Side of the flower bed = 2 m 80 cm = 2.80 m [ Since 100 cm = 1 m]

∴ Area of the square flower bed = (Side)^{2 }= (2.80 m )^{2} = 7.84 m^{2}

Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm

= (2.80 + 0.3 + 0.3) m = 3.4 m

Area of the enlarged flower bed with the digging strip = (Side )^{2} = (3.4 )^{2} = 11.56 m^{2}

Thus,

Increase in the area of the flower bed = 11.56 m^{2} − 7.84 m^{2}

= 3.72 m^{2}

#### Page No 20.14:

#### Question 8:

A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m^{2}, find the width of the varandah.

#### Answer:

Let the width of the verandah be *x* m.

Length of the room AB = 5 m and BC = 4 m

∴ Area of the room = 5 m x 4 m = 20 m^{2}

Length of the verandah* PQ* = (5 + *x* + *x*) = (5 + 2*x*) m

Breadth of the verandah *QR *= ( 4 + *x* + *x*) = (4 + 2*x*) m

Area of verandah *PQRS *= (5 + 2*x*) x (4 + 2*x*) = (4*x*^{2} + 18*x** *+ 20 ) m^{2}

∴ Area of verandah = Area of *PQRS* − Area of *ABCD*

⇒ 22 = 4*x*^{2} + 18*x** *+ 20 − 20

⇒ 22 = 4*x*^{2} + 18*x*

⇒ 11 = 2*x*^{2} + 9*x*

⇒ 2*x*^{2} + 9*x* − 11 = 0

⇒ 2*x*^{2} + 11*x* − 2*x* − 11 = 0

⇒ *x*(2*x** *+ 11)* *− 1(2*x* + 11) = 0

⇒ (*x* − 1) (2*x* +11) = 0

When *x* − 1 = 0,* x* = 1

When 2*x* + 11 = 0, *x* = $-\frac{11}{2}$

The width cannot be a negative value.

So, width of the verandah = *x* = 1 m.

#### Page No 20.14:

#### Question 9:

A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m^{2}, find the area of the lawn.

#### Answer:

We have,

Let* ABCD *be the square lawn and* PQRS* be the outer boundary of the square path.

Let side of the lawn *AB* be *x *m.

Area of the square lawn = *x*^{2}

Length *PQ* = (*x* m + 2 m + 2 m) = (*x* + 4) m

∴ Area of* PQRS* = (*x* + 4)^{2} = *(**x*^{2} + 8*x** *+ 16) m^{2}

Now,

Area of the path = Area* *of* PQRS −* Area of the square lawn

⇒ 136 = *x*^{2} + 8*x** *+ 16^{ }*−* *x*^{2}

⇒ 136 = 8*x** *+ 16

⇒ 136 *− *16 = 8*x** *

⇒ 120* = *8*x** *

∴* x = *120* *÷ 8 = 15

∴ Side of the lawn = 15 m

Hence,

Area of the lawn = (Side)^{2} = (15 m)^{2} = 225 m^{2}

#### Page No 20.14:

#### Question 10:

A poster of size 10 cm by 8 cm is pasted on a sheet of cardboard such that there is a margin of width 1.75 cm along each side of the poster. Find (i) the total area of the margin (ii) the cost of the cardboard used at the rate of Re 0.60 per cm^{2}.

#### Answer:

We have,

Length of the poster = 10 cm and breadth of the poster = 8 cm

∴ Area of the poster = Length x Breadth = 10 cm x 8 cm = 80 cm^{2}

From the figure, it can be observed that,

Length of the cardboard when the margin is included = 10 cm + 1.75 cm + 1.75 cm = 13.5 cm

Breadth of the cardboard when the margin is included = 8 cm + 1.75 cm + 1.75 cm = 11.5 cm

∴ Area of the cardboard = Length x Breadth = 13.5 cm x 11.5 cm = 155.25 cm^{2}

Hence,

**(i)** Area of the margin = Area of cardboard including the margin − Area of the poster

= 155.25 cm^{2}^{ }− 80 cm^{2}

= 75.25 cm^{2}

**(ii) **Cost of the cardboard = Area of the cardboard x Rate of the cardboard Rs. 0.60 per cm^{2}

= Rs. (155.25 x 0.60)

= Rs. 93.15

#### Page No 20.14:

#### Question 11:

A rectangulr field is 50 m by 40 m. It has two roads through its centre, running parallel to its sides. The width of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the erea of the remaining portion of the field.

#### Answer:

Let* ABCD* be the rectangular field and *KLMN *and *PQRS* the two rectangular roads with width 1.8 m and 2.5 m, respectively.

Length of the rectangular field *CD* = 50 cm and breadth of the rectangular field *BC* = 40 m

∴ Area of the rectangular field *ABCD* = 50 m x 40 m = 2000 m^{2}

Area of the road *KLMN *= 40 m x 2.5 m = 100 m^{2}

Area of the road *PQRS* = 50 m x 1.8 m = 90 m^{2}

Clearly area of *EFGH* is common to the two roads.

Thus, Area of *EFGH* = 2.5 m x 1.8 m = 4.5 m^{2}

Hence,

Area of the roads = Area (*KLMN*) + Area (*PQRS*) − Area (*EFGH)*

= (100 m^{2} + 90 m^{2}) − 4.5 m^{2} = 185.5 m^{2}

Area of the remaining portion of the field = Area of the rectangular field *ABCD* − Area of the roads

= (2000 − 185.5) m^{2}

^{ }= 1814.5 m^{2}

#### Page No 20.14:

#### Question 12:

There is a rectangular field of size 94 m × 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.

#### Answer:

Let* ABCD *be the rectangular field.

Here,

Two roads which are parallel to the breadth of the field *KLMN* and *EFGH* with width 2 m each.

One road which is parallel to the length of the field *PQRS* with width 2 m.

Length of the rectangular field* AB* = 94 m and breadth of the rectangular field* BC* = 32 m

∴ Area of the rectangular field = Length x Breadth = 94 m x 32 m = 3008 m^{2}

Area of the road* KLMN* = 32 m x 2 m = 64 m^{2}

Area of the road* EFGH *= 32 m x 2 m = 64 m^{2}

Area of the road *PQRS *= 94 m x 2 m = 188 m^{2}

Clearly area of *TUVI *and *WXYZ* is common to these three roads.

Thus,

Area of* **TUVI *= 2 m x 2 m = 4 m^{2}

Area of *WXYZ* = 2 m x 2 m = 4 m^{2}

Hence,

**(i)** Area of the field covered by the three roads:

= Area (*KLMN*) + Area (*EFGH)* + Area (*PQRS*) − {Area (*TUVI *) + Area (*WXYZ*)}

= [ 64+ 64 + 188 − (4 + 4 ^{ })] m^{2}

= 316 m^{2} − 8 m^{2}

= 308 m^{2}

**(ii) **Area of the field not covered by the roads:

= Area of the rectangular field* ABCD* − Area of the field covered by the three roads

= 3008 m^{2} − 308 m^{2}

= 2700 m^{2}

#### Page No 20.14:

#### Question 13:

A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncoverd. If the width of the carpet is 82 cm, find the cost at the rate of Rs 18 per metre.

#### Answer:

We have,

Length of the hall *PQ *= 22 m and breadth of the hall *QR* = 15.5 m

∴ Area of the school hall *PQRS* = 22 m x 15.5 m = 341 m^{2}

Length of the carpet *AB* = 22 m − ( 0.75 m + 0.75 m) = 20.5 m [ Since 100 cm = 1 m]

Breadth of the carpet *BC* = 15.5 m − ( 0.75 m + 0.75 m) = 14 m

∴ Area of the carpet *ABCD *= 20.5 m x 14 m = 287 m^{2}

Area of the strip = Area of the school hall *PQRS* − Area of the carpet *ABCD*

= 341 m^{2} − 287 m^{2}

= 54 m^{2}

Again,

Area of the 1 m length of carpet = 1 m x 0.82 m = 0.82 m^{2}

Thus,

Length of the carpet^{ }whose area is 287 m^{2}^{ }= 287 m^{2}^{ }÷ 0.82 m^{2}^{ }= 350 m

Cost of the 350 m long carpet = Rs. 18 x 350 = Rs. 6300

#### Page No 20.14:

#### Question 14:

Two cross roads, each of width 5 m, run at right angles through the centre of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m^{2}.

#### Answer:

Let* ABCD *be the rectangular park then *EFGH* and* IJKL* the two rectangular roads with width 5 m.

Length of the rectangular park *AD* = 70 cm

Breadth of the rectangular park *CD* = 45 m

∴ Area of the rectangular park = Length x Breadth = 70 m x 45 m = 3150 m^{2}

Area of the road *EFGH* = 70 m x 5 m = 350 m^{2}

Area of the road* JKIL *= 45 m x 5 m = 225 m^{2}

Clearly area of *MNOP* is common to the two roads.

Thus, Area of* MNOP* = 5 m x 5 m = 25 m^{2}

Hence,

Area of the roads = Area (*EFGH*) + Area (*JKIL*) − Area (*MNOP*)

= (350 + 225 ) m^{2}− 25 m^{2} = 550 m^{2}

Again, it is given that the cost of constructing the roads = Rs. 105 per m^{2}

Therefore,

Cost of constructing 550 m^{2} area of the roads = Rs. (105 × 550)

= Rs. 57750.

#### Page No 20.14:

#### Question 15:

The length and breadth of a rectangular park are in the ratio 5 : 2. A 2.5 m wide path running all around the outside the park has an area 305 m^{2}. Find the dimensiions of the park.

#### Answer:

We have,

Area of the path = 305 m^{2}

Let the length of the park be 5*x** *m and the breadth of the park be 2*x** *m

Thus,

Area of the rectangular park = 5*x* x 2*x* = 10*x ^{2}* m

^{2}

Width of the path = 2.5 m

Outer length

*PQ*= 5

*x*

*m*

*+*2.5 m + 2.5 m

*=*(5

*x*

*+*5) m

Outer breadth

*QR*= 2

*x*

*+*2.5 m + 2.5 m

*=*(2

*x*

*+*5) m

Area of

*PQRS*= (5

*x*

*+*5) m x (2

*x*

*+*5) m = (10

*x*+ 25

^{2}*x*+ 10

*x*+ 25) m

^{2}= (10

*x*

*+ 35*

^{2}*x*+ 25) m

^{2}

∴ Area of the path = [(10

*x*

*+ 35*

^{2}*x*

*+*25) − 10

*x*

*] m*

^{2}^{2}

⇒ 305 = 35

*x*+ 25

⇒ 305 − 25 = 35

*x*

⇒ 280 = 35

*x*

⇒

*x*= 280 ÷ 35 = 8

Therefore,

Length of the park = 5

*x*

*=*5 x 8 = 40 m

Breadth of the park = 2

*x*= 2 x 8 = 16 m

#### Page No 20.14:

#### Question 16:

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m^{2}, find the area of the lawn.

#### Answer:

Let the side of the lawn be *x *m*.
*

Given that width of the path = 2.5 m

Side of the lawn including the path = (

*x*+ 2.5 + 2.5) m = (

*x*+ 5 ) m

So, area of lawn = (Area of the lawn including the path) − (Area of the path)

We know that the area of a square = (Side)

^{2}

∴ Area of lawn (

*x*

^{2}

^{ }) = (

*x*+ 5)

^{2}− 165

⇒

*x*

^{2}

^{ }= (

*x*

^{2}+ 10

*x*

*+ 25) − 165*

⇒ 165

^{ }= 10

*x*

*+ 25*

⇒ 165 − 25

^{ }= 10

*x*

⇒ 140 = 10

*x*

Therefore

*x*= 140 ÷ 10 = 14

Thus the side of the lawn = 14 m

Hence,

The area of the lawn = (14 m)

^{2}= 196 m

^{2}

#### Page No 20.20:

#### Question 1:

Find the area of a parallogram with base 8 cm and altitude 4.5 cm.

#### Answer:

We have,

Base = 8 cm and altitude = 4.5 cm

Thus,

Area of the parallelogram = Base x Altitude

= 8 cm x 4.5 cm

= 36 cm^{2}

#### Page No 20.20:

#### Question 2:

Find the area in square metres of the parallelogram whose base and altitudes are as under:

(i) Base = 15 dm, altitude = 6.4 dm

(ii) Base = 1 m 40 cm, altitude = 60 cm

#### Answer:

We have,

(i) Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m [Since 100 cm = 1 m]

Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m

Thus,

Area of the parallelogram = Base x Altitude

= 1.5 m x 0.64 m

= 0.96 m^{2}

(ii) Base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]

Altitude = 60 cm = 0.6 m

Thus,

Area of the parallelogram = Base x Altitude

= 1.4 m x 0.6 m

= 0.84 m^{2}

#### Page No 20.20:

#### Question 3:

Find the altitude of a parallelogram whose area is 54 dm^{2} and base is 12 dm.

#### Answer:

We have,

Area of the given parallelogram = 54 dm^{2}

Base of the given parallelogram = 12 dm

∴ Altitude of the given parallelogram = $\frac{\mathrm{Area}}{\mathrm{Base}}=\frac{54}{12}\mathrm{dm}=4.5\mathrm{dm}$

#### Page No 20.20:

#### Question 4:

The area of a rhombus is 28 m^{2}. If its perimeter be 28 m, find its altitude.

#### Answer:

We have,

Perimeter of a rhombus = 28 m

∴ 4(Side) = 28 m [Since perimeter = 4(Side)]

⇒ Side = $\frac{28\mathrm{m}}{4}=7\mathrm{m}$

Now,

Area of the rhombus = 28 m^{2}

⇒ (Side x Altitude) = 28 m^{2}

⇒ (7 m x Altitude) = 28 m^{2}

⇒ Altitude = ^{$\frac{28{\mathrm{m}}^{2}}{7\mathrm{m}}=4\mathrm{m}$}

#### Page No 20.20:

#### Question 5:

In Fig. 20, *ABCD* is a parallelogram, *DL* ⊥ *AB* and *DM* ⊥ *BC*. If AB = 18 cm, *BC* = 12 cm and *DM* = 9.3 cm, find *DL*.

#### Answer:

We have,

Taking *BC* as the base,

*BC* = 12 cm and altitude* DM* = 9.3 cm

∴ Area of parallelogram *ABCD* = Base x Altitude

= (12 cm x 9.3 cm) = 111.6 cm^{2} ......... (i)

Now,

Taking *AB* as the base, we have,

Area of the parallelogram *ABCD* = Base x Altitude = (18 cm x *DL*).................(ii)

From (i) and (ii), we have

18 cm x *DL* = 111.6 cm^{2}

⇒ *DL* = $\frac{111.6{\mathrm{cm}}^{2}}{18\mathrm{cm}}=6.2\mathrm{cm}$

#### Page No 20.20:

#### Question 6:

The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.

#### Answer:

We have,

*ABCD* is a parallelogram with the longer side* AB* = 54 cm and corresponding altitude *AE* = 16 cm.

The shorter side is *BC* and the corresponding altitude is* CF *= 24 cm.

Area of a parallelogram = base × height.We have two altitudes and two corresponding bases. So,

$\frac{1}{2}\times BC\times CF=\frac{1}{2}\times AB\times AE$

⇒* BC* x *CF* = *AB *x *AE*

⇒ *BC* x 24 = 54 x 16

⇒ BC = $\frac{54\times 16}{24}=36\mathrm{cm}$

Hence, the length of the shorter side *BC *= *AD* = 36 cm.

#### Page No 20.20:

#### Question 7:

In Fig. 21, *ABCD* is a parallelogram, *DL* ⊥ *AB*. If *AB* = 20 cm, *AD* = 13 cm and area of the parallelogram is 100 cm^{2}, find *AL*.

#### Answer:

We have,

*ABCD* is a parallelogram with base *AB* = 20 cm and corresponding altitude *DL.*

It is given that the area of the parallelogram *ABCD* = 100 cm^{2}

Now,

Area of a parallelogram = Base x Height

100 cm^{2} = *AB* x *DL*

100 cm^{2} = 20 cm x *DL*

∴* DL* = $\frac{100{\mathrm{cm}}^{2}}{20\mathrm{cm}}=5\mathrm{cm}$

Again by Pythagoras theorem, we have,

(*AD*)^{2} = (*AL*)^{2 }+ (*DL*)^{2}

⇒ (13)^{2} = (*AL*)^{2} + (5)^{2}

⇒ (*AL*)^{2 }= (13)^{2 }- (5)^{2
} = 169 − 25 = 144

⇒ (*AL*)^{2} = (12)^{2}

⇒ *AL* = 12 cm

Hence. length of *AL* is 12 cm.

#### Page No 20.20:

#### Question 8:

In Fig. 21, if *AB* = 35 cm, *AD* = 20 cm and area of the parallelogram is 560 cm^{2}, find *LB*.

#### Answer:

We have,

*ABCD *is a parallelogram with base *AB *= 35 cm and corresponding altitude *DL*. The adjacent side of the parallelogram *AD* = 20 cm.

It is given that the area of the parallelogram* ABCD* = 560 cm^{2}

Now,

Area of the parallelogram = Base x Height

560 cm^{2} = *AB* x *DL*

560 cm^{2} = 35 cm x *DL*

∴ *DL* = $\frac{560{\mathrm{cm}}^{2}}{35\mathrm{cm}}=16\mathrm{cm}$

Again by Pythagoras theorem, we have,

(*AD*)^{2} = (*AL*)^{2 }+ (*DL)*^{2}

⇒ (20)^{2} = (*AL*)^{2} + (16)^{2}

⇒ (*AL*)^{2 }= (20)^{2 }− (16)^{2
} = 400 − 256 = 144

⇒ (*AL)*^{2} = (12)^{2}

⇒ *AL *= 12 cm

From the figure,

*AB* =* AL* + *LB*

35 cm = 12 cm +* LB*

∴* LB* = 35 cm − 12 cm

= 23 cm

Hence, length of* LB* is 23 cm.

#### Page No 20.20:

#### Question 9:

The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides.

#### Answer:

We have,

*ABCD* is a parallelogram with side *AB* = 10 m and corresponding altitude *AE* = 4 m.

The adjacent side *AD* = 8 m and the corresponding altitude is* CF.*

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases. So,

*AD* x *CF* = *AB* x *AE*

⇒ 8 m x *CF* = 10 m x 4 m

⇒* CF* = $\frac{10\times 4}{8}=5\mathrm{m}$

Hence, the distance between the shorter sides is 5 m.

#### Page No 20.20:

#### Question 10:

The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm^{2}, find the base and height.

#### Answer:

Let the height of the parallelogram be *x* cm.

Then the base of the parallelogram is *2x* cm.

It is given that the area of the parallelogram = 512 cm^{2}

So,

Area of a parallelogram = Base x Height

512 cm^{2} = 2*x* x *x*

512 cm^{2} = *2x ^{2}*

⇒

*x*= $\frac{512{\mathrm{cm}}^{2}}{2}=256{\mathrm{cm}}^{2}$

^{2}⇒

*x*(16 cm)

^{2}^{ }=^{2}

⇒

*x*= 16 cm

Hence, base =

*2x*

*=*2 x 16 = 32 cm and height =

*x*= 16 cm.

#### Page No 20.20:

#### Question 11:

Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.

#### Answer:

Let *ABCD* be the rhombus where diagonals intersect at *O*.

Then* AB* = 15 cm and* AC* = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore, Δ *AOB* is a right-angled triangle, right angled at *O* such that

*OA* = $\frac{1}{2}AC$ = 12 cm and* AB* = 15 cm.

By Pythagoras theorem, we have,

(*AB*)^{2} = *(OA*)^{2} + (*OB*)^{2}

⇒ (15)^{2} = (12)^{2} + (*OB*)^{2}

⇒ (*OB*)^{2 }= (15)^{2 }− (12)^{2}

⇒ *(OB*)^{2 }= 225 − 144 = 81

⇒ (*OB*)^{2} = (9)^{2}

⇒ *OB* = 9 cm

∴ *BD* = 2 x *OB* = 2 x 9 cm = 18 cm

Hence,

Area of the rhombus* ABCD* = $\left(\frac{1}{2}\times AC\times BD\right)=\left(\frac{1}{2}\times 24\times 18\right)=216{\mathrm{cm}}^{2}$

#### Page No 20.20:

#### Question 12:

Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

#### Answer:

Let* ABCD *be the rhombus whose diagonals intersect at *O*.

Then *AB* = 20 cm and *AC* = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore Δ *AOB* is a right-angled triangle, right angled at *O* such that

*OA* = $\frac{1}{2}AC$ = 12 cm and *AB* = 20 cm

By Pythagoras theorem, we have,

(*AB*)^{2} = (*OA*)^{2} + (*OB*)^{2}

⇒ (20)^{2} = (12)^{2} + (*OB*)^{2}

⇒ (*OB*)^{2 }= (20)^{2 }− (12)^{2}

⇒ (*OB*)^{2 }= 400 − 144 = 256

⇒ (*OB*)^{2} = (16)^{2}

⇒* OB* = 16 cm

∴ *BD* = 2 x *OB* = 2 x 16 cm = 32 cm

Hence,

Area of the rhombus *ABCD* = $\left(\frac{1}{2}\times AC\times BD\right)=\left(\frac{1}{2}\times 24\times 32\right)=384{\mathrm{cm}}^{2}$

#### Page No 20.20:

#### Question 13:

The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?

#### Answer:

We have,

Side of a square = 4 m and one diagonal of a square = 2 m

Area of the rhombus = Area of the square of side 4 m

⇒$\left(\frac{1}{2}\times AC\times BD\right)={\left(4\mathrm{m}\right)}^{2}$

⇒ $\left(\frac{1}{2}\times AC\times 2\mathrm{m}\right)=16{\mathrm{m}}^{2}$

⇒ *AC* = 16 m

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

⇒ $AO=\frac{1}{2}AC=8\mathrm{m}$ and $BO=\frac{1}{2}BD=1\mathrm{m}$

By Pythagoras theorem, we have:

*AO*^{2} + *BO*^{2} = *AB*^{2}

⇒* **AB*^{2} = (8 m)^{2} + (1m)^{2} = 64 m^{2} + 1 m^{2} = 65 m^{2}

⇒ Side of a rhombus = *AB* = $\sqrt{65}$ m.

Let* DX *be the altitude.

Area of the rhombus = *AB* × *DX*

16 m^{2} = $\sqrt{65}$ m x *DX*

∴ *DX* = $\frac{16}{\sqrt{65}}\mathrm{m}$

Hence, the altitude of the rhombus will be $\frac{16}{\sqrt{65}}\mathrm{m}$.

#### Page No 20.21:

#### Question 14:

Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.

#### Answer:

We have,

*ABCD* is a parallelogram with longer side *AB *= 25 cm and altitude* AE* = 10 cm.

As ABCD is a parallelogram .hence *AB=CD* (opposite sides of parallelogram are equal)

The shorter side is* AD* = 20 cm and the corresponding altitude is* CF*.

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

So,

⇒ *AD* x* CF* = *CD* x *AE*

⇒ 20 x* CF =* 25 x 10

∴ *CF* = $\frac{25\times 10}{20}=12.5\mathrm{cm}$

Hence, the altitude corresponding to the other pair of the side *AD* is 12.5 cm.

#### Page No 20.21:

#### Question 15:

The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.

#### Answer:

We have,

*ABCD* is a parallelogram with side *AB* = *CD* = 10 cm (Opposite sides of parallelogram are equal) and corresponding altitude *AM* = 12 cm.

The other side is *AD* and the corresponding altitude is* CN* = 8 cm

Area of a parallelogram = Base × Height

We have two altitudes and two corresponding bases.

So,

⇒ *AD* x *CN* =* **CD *x *AM*

⇒* AD* x 8 = 10 x 12

⇒ *AD* = $\frac{10\times 12}{8}=15\mathrm{cm}$

Hence, the length of the other pair of the parallel side = 15 cm.

#### Page No 20.21:

#### Question 16:

A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm^{2}.

#### Answer:

We have,

Altitude of a tile = 3 cm

Base of a tile = 5 cm

Area of one tile = Altitude x Base = 5 cm x 3 cm = 15 cm^{2}

Area of 280 tiles = 280 x 15 cm^{2} = 4200 cm^{2}

Rate of polishing the tiles at 50 paise per cm^{2} = Rs. 0.5 per cm^{2}

Thus,

Total cost of polishing the design = Rs. (4200 x 0.5) = Rs. 2100

#### Page No 20.26:

#### Question 1:

Find the area in square centimetres of a triangle whose base and altitude are as under:

(i) base = 18 cm, altitude = 3.5 cm

(ii) base = 8 dm, altitude = 15 cm

#### Answer:

We know that the area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

(i) Here, base = 18 cm and height = 3.5 cm

∴ Area of the triangle = $\left(\frac{1}{2}\times 18\times 3.5\right)=31.5{\mathrm{cm}}^{2}$

(ii) Here, base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]

and height = 3.5 cm

∴ Area of the triangle = $\left(\frac{1}{2}\times 80\times 15\right)=600{\mathrm{cm}}^{2}$

#### Page No 20.26:

#### Question 2:

Find the altitude of a triangle whose area is 42 cm^{2} and base is 12 cm.

#### Answer:

We have,

Altitude of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Base}}$

Here, base = 12 cm and area = 42 cm^{2}

∴ Altitude = $\frac{2\times 42}{12}=7\mathrm{cm}$

#### Page No 20.26:

#### Question 3:

The area of a triangle is 50 cm^{2}. If the altitude is 8 cm, what is its base?

#### Answer:

We have,

Base of a triangle = $\frac{2\times \mathrm{Area}}{\mathrm{Altitude}}$

Here, altitude = 8 cm and area = 50 cm^{2}

∴ Altitude =$\frac{2\times 50}{8}=12.5\mathrm{cm}$

2×4212= 7 cm

#### Page No 20.26:

#### Question 4:

Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.

#### Answer:

In a right-angled triangle, the sides containing the right angles are of lengths 20.8 m and 14.7 m.

Let the base be 20.8 m and the height be 14.7 m.

Then,

Area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}$

= $\frac{1}{2}\times 20.8\times 14.7=152.88{\mathrm{m}}^{2}$

#### Page No 20.26:

#### Question 5:

The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

#### Answer:

For the first triangle, we have,

Base = 15 cm and altitude = 7 cm

Thus, area of a triangle = $\frac{1}{2}\times \mathrm{Base}\times \mathrm{Altitude}$

= $\frac{1}{2}\times 15\times 7=52.5{\mathrm{cm}}^{2}$

It is given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 cm^{2}

One side of the second triangle = 10.5 cm

Therefore,

The other side of the second triangle = $\frac{2\times \mathrm{Area}}{\mathrm{One}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{triangle}}$

= $\frac{2\times 52.5}{10.5}=10\mathrm{cm}$

Hence, the other side of the second triangle will be 10 cm.

#### Page No 20.26:

#### Question 6:

A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?

#### Answer:

We have,

Length of the rectangular field = 48 m

Breadth of the rectangular field = 20 m

Area of the rectangular field = Length x Breadth = 48 m x 20 m = 960 m^{2}

Area of one right triangular flower bed = $\frac{1}{2}\times 12\mathrm{m}\times 5\mathrm{m}=30{\mathrm{m}}^{2}$

Therefore,

Required number of right triangular flower beds = $\frac{960{\mathrm{m}}^{2}}{30{\mathrm{m}}^{2}}=32$

#### Page No 20.26:

#### Question 7:

In Fig. 29, *ABCD* is a quadrilateral in which diagonal *AC* = 84 cm; *DL* ⊥ *AC*, *BM* ⊥ *AC*, *DL* = 16.5 cm and *BM* = 12 cm. Find the area of quadrilateral *ABCD*.

#### Answer:

We have,

*AC* = 84 cm, *DL* = 16.5 cm and *BM *= 12 cm

Area of Δ *ADC *= $$$\frac{1}{2}$x *AC* x* DL*

= $$$\frac{1}{2}$x 84 cm x 16.5 cm = 693 cm^{2}

Area of Δ *ABC* = $$$\frac{1}{2}$x *AC* x *BM*

= $$$\frac{1}{2}$x 84 cm x 12 cm = 504 cm^{2}

Hence,

Area of quadrilateral *ABCD* = Area of Δ* ADC* + Area of Δ* ABC*

= (693 + 504) cm^{2}

^{ }= 1197 cm^{2}

#### Page No 20.27:

#### Question 8:

Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals *AC* and BD measure 48 m and 32 m respectively and are perpendicular to each other.

#### Answer:

We have,

Diagonal *AC* = 48 cm and diagonal *BD* = 32 m

∴ Area of a quadrilateral = $\frac{1}{2}$x Product of diagonals

= $\frac{1}{2}$x *AC* x *BD*

= ($\frac{1}{2}$ x 48 x 32) m^{2} = (24 x 32) m^{2} = 768 m^{2}

#### Page No 20.27:

#### Question 9:

In Fig 31, *ABCD* is a rectangle with dimensions 32 m by 18 m. *ADE* is a triangle such that *EF* ⊥ *AD* and *EF* = 14 cm. Calculate the area of the shaded region.

#### Answer:

We have,

Area of the rectangle = *AB* x* BC*

= 32 m x 18 m

= 576 m^{2}

Area of the triangle = $\frac{1}{2}$x *AD* x *FE*

= $\frac{1}{2}$x BC x FE [Since AD = BC]

= $\frac{1}{2}$x 18 m x 14 m

= 9 m x 14 m = 126 m^{2}

∴ Area of the shaded region = Area of the rectangle − Area of the triangle

=(576 − 126) m^{2}

^{ }=^{ }450^{ }m^{2}

#### Page No 20.27:

#### Question 10:

In Fig. 32, *ABCD* is a rectangle of length *AB* = 40 cm and breadth *BC* = 25 cm. If *P*, *Q*, *R*, *S* be the mid-points of the sides *AB*, *BC*, *CD* and *DA* respectively, find the area of the shaded region.

#### Answer:

We have,

Join points *PR *and *SQ*.

These two lines bisect each other at point* O.*

Here, *AB* = *DC* = *SQ* = 40 cm and *AD = BC =RP* = 25 cm

Also* OP = OR *= $\frac{RP}{2}=\frac{25}{2}=12.5\mathrm{cm}$

From the figure we observed that,

Area of Δ *SPQ* = Area of Δ *SRQ*

Hence, area of the shaded region = 2 x (Area of Δ *SPQ*)

= 2 x ($\frac{1}{2}$ x* SQ* x *OP)*

= 2 x ($\frac{1}{2}$ x 40 cm x 12.5 cm)

= 500 cm^{2}

#### Page No 20.27:

#### Question 11:

Calculate the area of the quadrilateral *ABCD* as shown in Fig. 33, given that *BD* = 42 cm, *AC* = 28 cm, *OD* = 12 cm and *AC* ⊥ *BD*.

#### Answer:

We have,

*BD* = 42 cm, *AC* = 28 cm, *OD* = 12 cm

Area of Δ*ABC* = $\frac{1}{2}$ x *AC* x *OB*

= $\frac{1}{2}$ x* AC* x (*BD* − *OD*)

= $\frac{1}{2}$ x 28 cm x (42 cm − 12 cm) = $\frac{1}{2}$ x 28 cm x 30 cm = 14 cm x 30 cm = 420 cm^{2}

Area of Δ *ADC *= $\frac{1}{2}$ x *AC* x *OD*

= $\frac{1}{2}$ x 28 cm x 12 cm = 14 cm x 12 cm = 168 cm^{2}

Hence,

Area of the quadrilateral *ABCD* = Area of Δ* ABC* + Area of Δ *ADC*

= (420 + 168) cm^{2} = 588 cm^{2}

#### Page No 20.27:

#### Question 12:

Find the area of a figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.

#### Answer:

Let *x *cm be one of the equal sides of an isosceles triangle.

Given that the perimeter of the isosceles triangle = 18 cm

Then,

*x* + *x* + 8 = 18

⇒ 2*x* = (18 − 8) cm = 10 cm

⇒ *x *= 5 cm

Area of the figure formed = Area of the square + Area of the isosceles triangle

= (Side of square)^{2} + $\frac{1}{2}\times \mathrm{Base}\times \sqrt{{\left(\mathrm{Equal}\mathrm{side}\right)}^{2}-\frac{1}{4}\times {\left(\mathrm{Base}\right)}^{2}}$

= (8)^{2} + $\frac{1}{2}\times 8\times \sqrt{{\left(5\right)}^{2}-\frac{1}{4}{\left(8\right)}^{2}}$

= $64+4\times \sqrt{25-16}$

= $64+4\times \sqrt{9}$

= $64+4\times 3$

= 64 + 12 = 76 cm^{2}

#### Page No 20.27:

#### Question 13:

Find the area of Fig. 34 in the following ways:

(i) Sum of the areas of three triangles

(ii) Area of a rectangle − sum of the areas of five triangles

#### Answer:

We have,

(i) * P* is the midpoint of* AD*.

Thus* AP = PD* = 25 cm and *AB = CD* = 20 cm

From the figure, we observed that,

Area of Δ* APB* = Area of Δ *PDC*

Area of Δ APB = $\frac{1}{2}$x* AB *x *AP*

= $\frac{1}{2}$x 20 cm x 25 cm = 250 cm^{2}

Area of Δ *PDC* = Area of Δ *APB* = 250 cm^{2}

Area of Δ *RPQ *= $\frac{1}{2}$x Base x Height

= $\frac{1}{2}$x 25 cm x 10 cm = 125 cm^{2}

Hence,

Sum of the three triangles = (250 + 250 + 125) cm^{2}^{ }

= 625 cm^{2}

(ii) Area of the rectangle *ABCD* = 50 cm x 20 cm = 1000 cm^{2}

Thus,

Area of the rectangle − Sum of the areas of three triangles (There is a mistake in the question; it should be area of three triangles)

= (1000 − 625 ) cm^{2} = 375 cm^{2}

#### Page No 20.27:

#### Question 14:

Calculate the area of quadrilateral field *ABCD* as shown in Fig. 35, by dividing it into a rectangle and a triangle.

#### Answer:

We have,

Join *CE*, which intersect *AD* at point* E.*

Here,* AE = ED = BC *= 25 m and *EC = AB* = 30 m

Area of the rectangle *ABCE *=* AB *x *BC*

= 30 m x 25 m

= 750 m^{2}

Area of Δ *CED* = $\frac{1}{2}$x* EC x ED*

= $\frac{1}{2}$x 30 m x 25 m

= 375 m^{2}

Hence,

Area of the quadrilateral *ABCD* = (750 + 375) m^{2}

= 1125 m^{2}

#### Page No 20.28:

#### Question 15:

Calculate the area of the pentagon *ABCDE*, where *AB* = *AE* and with dimensions as shown in Fig. 36.

#### Answer:

Join *BE*.

Area of the rectangle *BCDE* =* CD* x *DE*

= 10 cm x 12 cm = 120 cm^{2}

Area of Δ*ABE* = $\frac{1}{2}$x *BE* x height of the triangle

= $\frac{1}{2}$x 10 cm x (20 − 12) cm

= $\frac{1}{2}$x 10 cm x 8 cm = 40 cm^{2}

Hence,

Area of the pentagon *ABCDE =* (120 + 40) cm^{2}^{ }= 160 cm^{2}

#### Page No 20.28:

#### Question 16:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332. 10, find its base and height.

#### Answer:

Let altitude of the triangular field be* h *m

Then base of the triangular field is* 3h *m.

Area of the triangular field = $\frac{1}{2}\times h\times 3h=\frac{3{h}^{2}}{2}{\mathrm{m}}^{2}$..........(i)

The rate of cultivating the field is Rs 24.60 per hectare.

Therefore,

Area of the triangular field = $\frac{332.10}{24.60}=13.5\mathrm{hectare}$

= 135000 m^{2} [Since 1 hectare = 10000 m^{2} ]..........(ii)

From equation (i) and (ii) we have,

$\frac{3{h}^{2}}{2}=135000{\mathrm{m}}^{2}$

3*h ^{2}* = 135000 x 2 = 270000 m

^{2}

*h*= $\frac{270000}{3}\mathrm{m}$

^{2}^{2}

^{2}

⇒

*h*= 300 m

Hence,

Height of the triangular field = 300 m and base of the triangular field = 3 x 300 m = 900 m

#### Page No 20.28:

#### Question 17:

A wall is 4.5 m long and 3 m high. It has two equal windows, each having form and dimensions as shown in Fig. 37. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m_{2}.

#### Answer:

We have,

Length of a wall = 4.5 m

Breadth of the wall =3 m

Area of the wall = Length x Breadth = 4.5 m x 3 m = 13.5 m^{2}

From the figure we observed that,

Area of the window = Area of the rectangle + Area of the triangle

= (0.8 m x 0.5 m) + ($\frac{1}{2}$ x 0.8 m x 0.2 m) [Since 1 m = 100 cm]

= 0.4 m^{2} + 0.08 m^{2} = 0.48 m^{2}

Area of two windows = 2 x 0.48 = 0.96 m^{2}

Area of the remaining wall (leaving windows ) = (13.5 − 0.96 )m^{2} = 12.54 m^{2}

Cost of painting the wall per m^{2} = Rs. 15

Hence, the cost of painting on the wall = Rs. (15 x 12.54) = Rs. 188.1

(In the book, the answer is given for one window, but we have 2 windows.)

#### Page No 20.8:

#### Question 1:

Find the area, in square metres, of a rectangle whose

(i) Length = 5.5 m, breadth = 2.4 m

(ii) Length = 180 cm, breadth = 150 cm

#### Answer:

We have,

(i) Length = 5.5 m, Breadth = 2.4 m

Therefore,

Area of rectangle = Length x Breadth

= 5.5 m x 2.4 m

= 13.2 m^{2}^{ }

(ii) Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m [ Since 100 cm = 1 m]

Therefore,

Area of rectangle = Length x Breadth

= 1.8 m x 1.5 m

= 2.7 m^{2}^{ }

#### Page No 20.8:

#### Question 2:

Find the area, in square centimetres, of a square whose side is

(i) 2.6 cm

(ii) 1.2 dm

#### Answer:

We have,

(i) Side of the square = 2.6 cm

Therefore, area of the square = (Side)^{2}

= (2.6 cm)^{2 } = 6.76 cm^{2}

(ii) Side of the square = 1.2 dm = 1.2 x 10 cm = 12 cm [ Since 1 dm = 10 cm]

Therefore, area of the square = (Side)^{2}

= (12 cm)^{2 } = 144 cm^{2}

#### Page No 20.8:

#### Question 3:

Find in square metres, the area of a square of side 16.5 dam.

#### Answer:

We have,

Side of the square = 16.5 dam = 16.5 x 10 m = 165 m [ Since 1 dam = 10 m ]

Area of the square = (Side)^{2} = (165 m)^{2}

= 27225 m^{2}

#### Page No 20.8:

#### Question 4:

Find the area of a rectangular feild in ares whose sides are:

(i) 200 m and 125 m

(ii) 75 m 5 dm and 125 m

#### Answer:

We have,

(i) Length of the rectangular field = 200 m

Breadth of the rectangular field = 125 m

Therefore,

Area of the rectangular field = Length x Breadth

= 200 m x 125 m

= 25000 m^{2} = 250 ares [Since 100 m^{2} = 1 are]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m = 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

Therefore,

Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m

= 9060 m^{2} = 90.6 ares [Since 100 m^{2} = 1 are]

#### Page No 20.8:

#### Question 5:

Find the area of a rectangular field in hectares whose sides are:

(i) 125 m and 400 m

(ii) 75 m 5 dm and 120 m

#### Answer:

We have,

(i) Length of the rectangular field = 125 m

Breadth of the rectangular field = 400 m

Therefore,

Area of the rectangular field = Length x Breadth

= 125 m x 400 m

= 50000 m^{2} = 5 hectares [Since 10000 m^{2} = 1 hectare]

(ii) Length of the rectangular field =75 m 5 dm = (75 + 0.5) m = 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

Therefore,

Area of the rectangular field = Length x Breadth

= 75.5 m x 120 m

= 9060 m^{2} = 0.906 hectares [Since 10000 m^{2} = 1 hectare]

#### Page No 20.8:

#### Question 6:

A door of dimensions 3 m × 2m is on the wall of dimension 10 m × 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.

#### Answer:

We have,

Length of the door = 3 m

Breadth of the door = 2 m

Side of the wall = 10 m

Area of the wall = Side x Side = 10 m x 10 m = 100 m^{2}

Area of the door = Length x Breadth = 3 m x 2 m = 6 m^{2}

Thus,

Required area of the wall for painting = Area of the wall − Area of the door = (100 − 6 ) m^{2}^{ }= 94 m^{2}

Rate of painting per square metre = Rs. 2.50

Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235

#### Page No 20.8:

#### Question 7:

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?

#### Answer:

We have,

Perimeter the of rectangle = 2(Length + Breadth)

= 2(40 cm + 22 cm) = 124 cm

It is given that the wire which was in the shape of a rectangle is now bent into a square.

Therefore, the perimeter of the square = Perimeter of the rectangle

=> Perimeter of the square = 124 cm

⇒ 4 x side = 124 cm

∴ Side = $\frac{124}{4}=31\mathrm{cm}$

Now,

Area of the rectangle = 40 cm x 22 cm = 880 cm^{2}

Area of the square = (Side)^{2} = (31 cm)^{2} = 961 cm^{2}

Therefore, the square-shaped wire encloses more area.

#### Page No 20.8:

#### Question 8:

How many square metres of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?

#### Answer:

We have,

Length of the glass pane = 25 cm

Breadth of the glass pane = 16 cm

Area of one glass pane = 25 cm x 16 cm = 400 cm^{2} = 0.04 m^{2}^{ } [ Since 1 m^{2} = 10000 cm^{2} ]

Thus,

Area of 12 such panes = 12 x 0.04 = 0.48 m^{2}

#### Page No 20.8:

#### Question 9:

A marble tile measures 10 cm × 12 cm. How many tiles will be required to cover a wall of size 3 m × 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

#### Answer:

We have,

Area of the wall = 3 m x 4 m = 12 m^{2}

Area of one marble tile = 10 cm x 12 cm = 120 cm^{2} = 0.012 m^{2} [ Since 1 m^{2} = 10000 cm^{2} ]

Thus,

Number of tiles = $\frac{\mathrm{Area}\mathrm{of}\mathrm{wall}}{\mathrm{Area}\mathrm{of}\mathrm{one}\mathrm{tile}}=\frac{12{\mathrm{m}}^{2}}{0.012{\mathrm{m}}^{2}}=1000$

Cost of one tile = Rs. 2

Total cost = Number of tiles x Cost of one tile

= Rs. (1000 x 2) = Rs. 2000

#### Page No 20.8:

#### Question 10:

A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?

#### Answer:

We have,

Length of the table top = 9 dm 5 cm = (9 x 10 + 5) cm = 95 cm [ Since 1 dm = 10 cm]

Breadth of the table top = 6 dm 5 cm = (6 x 10 + 5) cm = 65 cm

∴ Area of the table top = Length x Breadth = (95 cm x 65 cm) = 6175 cm^{2}

Rate of polishing per square centimetre = 20 paise = Rs. 0.20

Total cost = Rs. (6175 x 0.20) = Rs. 1235

#### Page No 20.9:

#### Question 11:

A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile.

#### Answer:

We have,

Length of the floor of the room = 9.68 m

Breadth of the floor of the room = 6.2 m

Area of the floor = 9.68 m x 6.2 m = 60.016 m^{2}

Length of the tile = 22 cm

Breadth of the tile = 10 cm

Area of one tile = 22 cm x 10 cm = 220 cm^{2} = 0.022 m^{2} [Since 1 m^{2} = 10000 cm^{2}]

Thus,

Number of tiles = $\frac{60.016{\mathrm{m}}^{2}}{0.022{\mathrm{m}}^{2}}=2728$

Cost of one tile = Rs. 2.50

Total cost = Number of tiles x Cost of one tile

= Rs. (2728 x 2.50) = Rs. 6820

#### Page No 20.9:

#### Question 12:

One side of a square field is 179 m. Find the cost of raising a lown on the field at the rate of Rs 1.50 per square metre.

#### Answer:

We have,

Side of the square field = 179 m

Area of the field = (Side)^{2} = (179 m)^{2} = 32041 m^{2}

Rate of raising a lawn on the field per square metre = Rs. 1.50

Thus,

Total cost of raising a lawn on the field = Rs.(32041 x 1.50) = Rs. 48061.50

#### Page No 20.9:

#### Question 13:

A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec?

#### Answer:

We have,

Length of the rectangular field = 290 m

Breadth of the rectangular field = 210 m

Perimeter of the rectangular field = 2(Length + Breadth)

= 2(290 + 210) = 1000 m

Distance covered by the girl = 2 x Perimeter of the rectangular field

= 2 x 1000 = 2000 m

The girl walks at the rate of 1.5 m/sec.

or,

Rate = 1.5 x 60 m/min = 90 m/min

Thus,

Required time to cover a distance of 2000 m = $\frac{2000\mathrm{m}}{90\mathrm{m}/\mathrm{min}}=22\frac{2}{9}\mathrm{min}$

Hence, the girl will take $22\frac{2}{9}\mathrm{min}$ to go two times around the field.

#### Page No 20.9:

#### Question 14:

A corridor of a school is 8 m long and 6 m wide. It is to be covered with convas sheets. If the available canvas sheets have the size 2 m × 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet.

#### Answer:

We have,

Length of the corridor = 8 m

Breadth of the corridor = 6 m

Area of the corridor of a school = Length x Breadth = (8 m x 6 m) = 48 m^{2}

Length of the canvas sheet = 2 m

Breadth of the canvas sheet = 1 m

Area of one canvas sheet = Length x Breadth = (2 m x 1 m) = 2 m^{2}

Thus,

Number of canvas sheets = $\frac{48{\mathrm{m}}^{2}}{2{\mathrm{m}}^{2}}=24$

Cost of one canvas sheet = Rs. 8

∴ Total cost of the canvas sheets = Rs. (24 x 8) = Rs. 192

#### Page No 20.9:

#### Question 15:

The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metres per second.

#### Answer:

We have,

Length of a playground = 62 m 60 cm = 62.6 m [ Since 10 cm = 0.1 m]

Breadth of a playground = 25 m 40 cm = 25.4 m

Area of a playground = Length x Breadth= 62.6 m x 25.4 m = 1590.04 m^{2}

Rate of turfing = Rs. 2.50/m^{2}

∴ Total cost of turfing = Rs. (1590.04 x 2.50) = Rs. 3975.10

Again,

Perimeter of a rectangular field = 2(Length + Breadth)

= 2(62.6 + 25.4) = 176 m

Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field

= 3 x 176 m = 528 m

The man walks at the rate of 2 m/sec.

or,

Rate = 2 x 60 m/min = 120 m/min

Thus,

Required time to cover a distance of 528 m = $\frac{528\mathrm{m}}{120\mathrm{m}/\mathrm{min}}=4.4\mathrm{min}$

= 4 minutes 24 seconds [ since 0.1 minutes = 6 seconds]

#### Page No 20.9:

#### Question 16:

A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

#### Answer:

We have,

Length of the lane = 180 m

Breadth of the lane = 5 m

Area of a lane = Length x Breadth = 180 m x 5 m = 900 m^{2}

Length of the brick = 20 cm

Breadth of the brick = 15 cm

Area of a brick = Length x Breadth = 20 cm x 15 cm = 300 cm^{2} = 0.03 m^{2} [Since 1 m^{2} = 10000 cm^{2} ]

Required number of bricks = $\frac{900{\mathrm{m}}^{2}}{0.03{\mathrm{m}}^{2}}=30000$

Cost of 1000 bricks = Rs. 750

∴ Total cost of 30,000 bricks = Rs. $\left(\frac{750\times 30,000}{1000}\right)=\mathrm{Rs}.22,500$

#### Page No 20.9:

#### Question 17:

How many envelopes can be made out of a sheet of paper 125 cm by 85 cm; supposing one envelope requires a piece of paper of size 17 cm by 5 cm?

#### Answer:

We have,

Length of the sheet of paper = 125 cm

Breadth of the sheet of paper = 85 cm

Area of a sheet of paper = Length x Breadth = 125 cm x 85 cm = 10,625 cm^{2}

Length of sheet required for an envelope = 17 cm

Breadth of sheet required for an envelope = 5 cm

Area of the sheet required for one envelope = Length x Breadth = 17 cm x 5 cm = 85 cm^{2}

Thus,

Required number of envelopes = $\frac{10,625{\mathrm{cm}}^{2}}{85{\mathrm{cm}}^{2}}=125$

#### Page No 20.9:

#### Question 18:

The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.

#### Answer:

We have,

Length of the diaper = 50 cm

Breadth of the diaper = 17 cm

Area of cloth to make 1 diaper = Length x Breadth = 50 cm x 17 cm = 850 cm^{2}

Thus,

Area of 25 such diapers = (25 x 850) cm^{2} = 21,250 cm^{2}

Area of total cloth = Area of 25 diapers

= 21,250 cm^{2}

It is given that width of a cloth = 170 cm

∴ Length of the cloth = ^{$\frac{\mathrm{Area}\mathrm{of}\mathrm{cloth}}{\mathrm{Width}\mathrm{of}\mathrm{a}\mathrm{cloth}}=\frac{21,250{\mathrm{cm}}^{2}}{170\mathrm{cm}}=125\mathrm{cm}$}

Hence, length of the cloth will be 125 cm.

#### Page No 20.9:

#### Question 19:

The carpet for a room 6.6 m by 5.6 m costs Rs 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per metro.

#### Answer:

We have,

Length of a room = 6.6 m

Breadth of a room = 5.6 m

Area of a room = Length x Breadth = 6.6 m x 5.6 m = 36.96 m^{2}

Width of a carpet = 70 cm = 0.7 m [Since 1 m = 100 cm]

Length of a carpet =$\frac{\mathrm{Area}\mathrm{of}\mathrm{a}\mathrm{room}}{\mathrm{Width}\mathrm{of}\mathrm{a}\mathrm{carpet}}=\frac{36.96{\mathrm{m}}^{2}}{0.7\mathrm{m}}=52.8\mathrm{m}$

Cost of 52.8 m long roll of carpet = Rs. 3960

Therefore,

Cost of 1 m long roll of carpet = Rs. $\frac{3960}{52.8}=\mathrm{Rs}.75$

#### Page No 20.9:

#### Question 20:

A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m × 1.5 m and three windows each of dimensions 1.5 m × 1 m. Find the cost of white washing the walls at Rs 3.80 per square metre.

#### Answer:

We have,

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Area of 4 walls = 2(l + b)h

= 2(9 m + 8 m) x 6.5 m = 2 x 17 m x 6.5 m = 221 m^{2}

Length of a door = 2 m

Breadth of a door = 1.5 m

Area of a door = Length x Breadth = 2 m x 1.5 m = 3 m^{2}

Length of a window = 1.5 m

Breadth of a window = 1 m

Since, area of one window = Length x Breadth = 1.5 m x 1 m = 1.5 m^{2}

Thus,

Area of 3 such windows = 3 x 1.5 m^{2} = 4.5 m^{2}

Area to be white-washed = Area of 4 walls − (Area of one door + Area of 3 windows)

Area to be white-washed = [221 − (3 + 4.5^{ })] m^{2}

= (221 − 7.5 ) m^{2} = 213.5 m^{2}

Cost of white-washing for 1 m^{2} area = Rs. 3.80

∴ Cost of white-washing for 213.5 m^{2} area = Rs. (213.5 x 3.80) = Rs. 811.30

#### Page No 20.9:

#### Question 21:

A hall 36 m long and 24 m broad allowing 80 m^{2} for doors and windows, the cost of papering the walls at Rs 8.40 per m^{2} is Rs 9408. Find the height of the hall.

#### Answer:

We have,

Length of the hall = 36 m

Breadth of the hall = 24 m

Let h be the height of the hall.

Now, in papering the wall, we need to paper the four walls excluding the floor and roof of the hall.

So, the area of the wall which is to be papered = Area of 4 walls

= 2h(l + b)

= 2h (36 + 24) = 120h m^{2}

Now, area left for the door and the windows = 80 m^{2}

So, the area which is actually papered = (120h − 80) m^{2}

Again,

The cost of papering the walls at Rs 8.40 per m^{2} = Rs. 9408.

⇒ (120h − 80) m^{2} x Rs. 8.40 per m^{2}^{ }= Rs. 9408

⇒ (120h* *− 80) m^{2} = $\frac{\mathrm{Rs}.9408}{\mathrm{Rs}.8.40}$

⇒ (120h − 80) m^{2} = 1120 m^{2}

⇒ 120h* *m^{2} = (1120 + 80) m^{2}

⇒ 120h* *m^{2}^{ }= 1200 m^{2}

∴ h =$\frac{1200{\mathrm{m}}^{2}}{120\mathrm{m}}=10\mathrm{m}$

Hence, the height of the wall would be 10 m.

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