# Free Perimeter and Area 01 Practice Test - 7th grade

### Question 1

Puja is the owner of a farm as shown in the figure. She goes to the market to buy the fence wire for the field. Calculate the minimum length of the wire to be bought by Puja so that the entire farm is covered. (All dimensions are in m)

55 m

70 m

60 m

45 m

#### SOLUTION

Solution :B

To find out the minimum length of the wire to be bought by Puja so that the entire farm is covered, we have to find out the perimeter of the entire farm.

From the above figure, the length of the side AF = (Length of side BC) - (Length of side ED) = (20-5) m = 15m.

Length of side AB = (Length of side DC + Length of side FE) = (5+10) m = 15mAs we know, the perimeter of a plane figure is the length of its boundary.

Perimeter of the farm = (15+20+5+5+10+15) m = 70 m.

So, the minimum length of the wire to be bought by Puja is 70 m.

### Question 2

Puja has two ropes measuring 5 m and 6 m. She ties a goat at point A with the 5 m rope and ties another goat at point B with the 6 m rope. Find out the ratio of areas grazed at point A and point B. The goats can only move inside the field.

25:36

26:37

24:36

25:37

#### SOLUTION

Solution :A

The area grazed by both the goats will be circular in shape as they are moving at a constant distance from a fixed point.

Hence, the area grazed when a goat is tied at a point with a rope would be a circle with radius equal to the length of the rope.

Area of a circle = πr2.

Hence, the required ratio = Area at AArea at B

= (πrAπrB)2

=(rArB)2

= (56)2 = 25:36 .

### Question 3

Raj is living in a big room. The dimension of the room is 15.6 m x 19.5 m. He wants to cover the floor with the tiles, each of them measuring 120 cm x 150 cm. Find the exact number of tiles required by Raj to cover the floor.

169

#### SOLUTION

Solution :A

First of all, we have to convert all the dimensions into the same unit. ( 1 m = 100 cm )

The number of tiles × Area of each tile = Area of the entire room.

⇒ Number of tiles = 15.6×19.51.2×1.5 = 169 tiles.

### Question 4

A hall is in the shape of a parallelogram. One of the parallel walls measures 8 m. The perpendicular distance corresponding to that wall is 1.75 times the length of this wall. If the cost of carpeting the hall is **₹** 2 per m2 . Find the total cost of carpeting the hall.

**₹** 224

**₹**264

**₹**642

**₹**292

#### SOLUTION

Solution :A

Area of parallelogram = Base × corresponding height

Area of parallelogram = 8 × (1.75 × 8)

Area of parallelogram = 112 m2 .

Total cost of carpeting the hall = Cost per m2 × Area

=₹2/m2 × 112 m2 =₹224.

### Question 5

Find the base length of a parallelogram if the area is 24 m2 ^{ }and the corresponding height is 8 m.

3 m

#### SOLUTION

Solution :A

Area of a parallelogram = Base × corresponding height .

Therefore, Base = Area of a parallelogramCorresponding height

⇒ Base = 248 = 3 m.

### Question 6

A car has to travel 0.121 km to reach from A to B. If the radius of the tyre is 7 cm. How many times will the tire rotate while the car travels from A to B ? (Use π = 227)

#### SOLUTION

Solution :Circumference of tyre = 2 πr = Distance travelled in each rotation = 2×227×7 = 44 cm.

Total distance covered = Number of rotations × Circumference of tyre

Number of rotations = 0.121 km44 cm=0.121×1000 m44×0.01 m = 275

### Question 7

There is a square park in the society. The management wants to put a circular fountain at the centre of the park with diameter 14 m. The remaining area is to be covered with grass. If the cost of laying grass is **₹** 20 per m2 and the cost of constructing the fountain is 10 per m2 . Find the length of the field if the total cost incurred is **₹** 6460 .( Use π = 22/7)

#### SOLUTION

Solution :Diameter of a circular fountain= 14 m, so radius = 7 m.

Let the length of the square be 'a'.

Total cost = cost of fountain + cost of grass.

Hence,

₹6460 = (area of fountain x cost per m2) + (area of square park -area of circular fountain) x (cost for grass per m2)6460 = (π(7)2×10)+((a2−π(7)2×20)

6460 = (154×10)+(a2−154)×20

6460 =1540+(a2−154)×20

a = 20 m.

### Question 8

Raj runs and covers a distance of 0.616 km in the ground which is in the shape of semi-circle mounted on the rectangle as shown below. Find the number of rounds taken by Raj to cover the distance

#### SOLUTION

Solution :Perimeter of half circle = π r + d .

But according to the figure given here ,

the perimeter of the part of the semi circle = π×142=7π=7× 227 =22 m. (Radius is half the diameter)

Hence, total perimeter = Perimeter of the half circle + side of 14 m + side of 10 m + side of 10 m

= (22 + 14 + 10 +10) m = 56 m.

Number of rounds he will have to make to complete the punishment = 0.616 × 1000 m56 m = 11 rounds.

### Question 9

A rectangular park is having a grass field of 18cm long and 15 cm wide such that there is a path of 1.5 cm along each of its sides extending outside on each side of the park. The area of the path is 108 cm2.

True

#### SOLUTION

Solution :A

Area of the path = Area of the outer rectangle - Area of the inner rectangle

The path so formed is effectively adding 1.5m + 1.5m = 3m to the length and breadth of the park, as shown in the figure.

So, area of the path = 21 × 18 - 18 × 15 = 18 × (21 - 15) = 18 × 6 = 108 cm2

### Question 10

One hectare = ________m2. (Enter a numeric value)

#### SOLUTION

Solution :The

hectare(symbolha) is an SI accepted metric system unit of area equal a square with 100-metre sides (10,000 m2) and primarily used in the measurement of land.

Hence, 1 hectare = 10000m2.