The document RD Sharma Solutions (Part - 2)- Ex-20.2, Mensuration - I, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.

All you need of Class 7 at this link: Class 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.**

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}

**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.**

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* = −11/2

The width cannot be a negative value.

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

**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.**

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}

**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}.**

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

**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.**

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}

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

- RD Sharma Solutions (Part - 3)- Ex-20.2, Mensuration - I, Class 7, Math
- RD Sharma Solutions (Part - 1) - Ex-20.3, Mensuration - I, Class 7, Math
- RD Sharma Solutions (Part - 2) - Ex-20.3, Mensuration - I, Class 7, Math
- RD Sharma Solutions (Part - 1) - Ex-20.4, Mensuration - I, Class 7, Math
- RD Sharma Solutions (Part - 2) - Ex-20.4, Mensuration - I, Class 7, Math