Class 7 maths Perimeter Area

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

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

• 23m²

• 2.7 m²

• 48m²

• 55m²

Answer explanation

Given Length = 5.5 m, Breadth = 2.4 m

We know that area of rectangle = Length x Breadth

= 5.5 m x 2.4 m = 13.2 m²

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

We know that area of rectangle = Length x Breadth

= 1.8 m x 1.5 m = 2.7 m²

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

(i) 1.2 dm

• 312cm²

• 108cm²

• 144 cm²

• 54cm²

Answer explanation

Given side of the square = 1.2 dm

= 1.2 x 10 cm = 12 cm [Since 1 dm = 10 cm]

We know that area of the square = (Side)²

= (12 cm)²

= 144 cm²

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

(i) 75 m 5 dm and 120 m [ 100 m² = 1 acre]

• 58.12 acres

• 90.6 acres

• 40.06 acres

• 51.6 acres

Answer explanation

Given 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

We know that area of the rectangular field = Length x Breadth

= 75.5 m x 120 m

= 9060 m²

= 90.6 acres [Since 100 m² = 1 acre]

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

• RS 674

• RS 235

• RS 127

• RS 408

Answer explanation

Given length of the door = 3 m

Also given that breadth of the door = 2 m

Side of the wall = 10 m

We know that Area of the square = (Side)²

Area of the wall = Side x Side = 10 m x 10 m

= 100 m²

We know that the area of the rectangle = length x breadth

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

Thus, required area of the wall for painting = Area of the wall – Area of the door

= (100 – 6) m²

= 94 m²

Rate of painting per square metre = Rs. 2.50

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

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?

• 564 cm²

• 197 cm²

• 961cm²

• 335 cm²

Answer explanation

Given length of rectangular shaped wire = 40cm

Breadth of rectangular shaped wire = 22cm

Perimeter of the 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

We know that perimeter of square = 4 x side

Therefore 4 x side = 124 cm

Side = 124/4 = 31 cm

We know that area of rectangle = length x breadth

Now, Area of the rectangle

= 40 cm x 22 cm

= 880 cm²

We know that area of the square = (Side)²

= (31 cm)²

= 961 cm².

Therefore, the square-shaped wire encloses more area.

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?

• 147

• 221

• 125

• 654

Answer explanation

Given length of the sheet of paper = 125 cm

Also given that breadth of the sheet of paper = 85 cm

Area of a sheet of paper = Length x Breadth

= 125 cm x 85 cm

= 10,625 cm²

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 c m²

Thus, required number of envelopes = 10,625/85

= 125

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 leveling the path at the rate of Rs 8.25 per square meter.

• RS 6623

• RS 1901

• RS 1755

• RS 2277

Answer explanation

Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC

Given Length PQ = 40 m and breadth QR = 25 m

Area of lawn PQRS = 40 m x 25 m = 1000 m²

Length AB = (40 + 2 + 2) m = 44 m

Breadth BC = (25 + 2 + 2) m = 29 m

Area of ABCD = 44 m x 29 m = 1276 m²

Now, Area of the path = Area of ABCD – Area of the lawn PQRS

= 1276 m² – 1000 m²

= 276 m²

Rate of levelling the path = Rs. 8.25 per m²

Cost of levelling the path = (8.25 x 276)

= Rs. 2277

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 manures. Find the cost of applying the manures at the rate of Rs 300 per area.

• RS 10745

• RS 14244

• RS 13252

• RS 7845

Answer explanation

Given that length of a rectangular field = 80 m

Breadth of a rectangular field = 60 m

Area of rectangular field = length x breadth

= (80 x 60) m

= 4800 m²

Again, Area of the garden = 10 m x 4 m = 40 m²

Area of one flower bed = 4 m x 1.5 m = 6 m²

Thus, Area of two flower beds = 2 x 6 m² = 12 m²

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 m² – (40 + 12) m²

= (4800 – 52) m²

= 4748 m²

Since 100 m² = 1 acre

Therefore by using the above

4748 m² = 47.48 acres

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

= Rs. 14244

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

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

• 0.96m²

• 0.45 m²

• 0.78 m²

Answer explanation

Given base =15 dm, altitude = 6.4 dm

By converting these to standard form we get,

Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m

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

We know that area of the parallelogram = Base x Altitude

= 1.5 m x 0.64 m

Area of parallelogram = 0.96 m²

 Find the altitude of a parallelogram whose area is 54 d m² and base is 12 dm.

• 8.8 dm

• 6.5 dm

• 4.5 dm

• 3.8 dm

Answer explanation

Given area of the given parallelogram = 54 d m²

Base of the given parallelogram = 12 dm

We know that area of the parallelogram = Base x Altitude

Therefore altitude of the given parallelogram = Area/Base

= 54/12 dm

= 4.5 dm