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

### Question 1

Find the perimeter of a square of side 10m. Express your answer in cm. [1 MARK]

#### SOLUTION

Solution :Result : 1 Mark

Let the square given be of side A cm.Perimeter of the square whose side is A cm is 4 × A

Perimeter of the given square

= 4 × 10

= 40 m

= 4000 cm

### Question 2

The perimeter of a rectangular field is 50 m. If the length of the field is 13 m. Find the area of the field. [2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

Given that,

The perimeter of a rectangular field is 50 m.

The length of the field is 13 m.

Perimeter of rectangular field = 2( length + breadth)

Or, Breadth=perimeter2−length

Or, Breadth = 25 - 13 = 12 m

Area of rectangle = length × breadth =13×12 = 156 sq. m

The area of the rectangular field is 156 m2.

### Question 3

Find the area of the parallelogram whose side is 3m with an altitude of 38m. Express your answer up to 3 decimals.

[2 MARKS]#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

Given that,

The side of a parallelogram is 3 m and the corresponding altitude = 38m

Let the side and altitude of the parallelogram be A and BArea of the parallelogram = A×B

Given side and altitude are 3 m and 38m

Area of the given parallelogram

=(3)×38=1.125m2

The area of the parallelogram is 1.125m2.

### Question 4

Find the height ‘x’ if the area of the parallelogram is 24 cm2 and the base is 4cm. [2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

Given that,

Area of a parallelogram is 24 cm2

Area of parallelogram = Base × height

Or, Height = AreaBase

On substituting the values, we get

Height = 244 = 6 m

So, the height of the parallelogram is 6 m.

### Question 5

The radius of a circle is 10 meters. Find the area of the circle in m2 , Take π = 3.14 [2 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Answer: 1 Mark

Area of the circle of radius R is π×R2Radius of the given circle = 10 m

Area of the given circle = π×102

= 314 m2

The area of the circle is 314 m2.

### Question 6

Area of a pizza is one-fourth of that of a cake. What is the ratio of their radii? [3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Answer: 1 Mark

Let the radius of the pizza be r1 and radius of the cake be r2.

Given: area of the pizzaarea of the cake=14

Now, area of a circle = πr2

⇒πr12πr22=14

⇒r21r22=12

⇒r1r2=12

Therefore, the ratio of their radii is 12

### Question 7

Before a cricket match, the officials wanted to check the distance of the boundary from the centre of the pitch, given that the ground is circular in shape. The entire boundary length is 628 m. Also find its area in m2 . Take π=3.14. [3 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Radius: 1 Mark

Area: 1 Mark

Given that,

The entire length of the boundary is 628 m.

The perimeter of a cricket of radius r meters is 2×π×r metersGiven perimeter of the ground = 628 m

r = 628÷(2×π) =100 m

The distance of the boundary from the centre of the pitch is 100 m.

The area of a circle is π×(r)2Area of the ground = π×(100)2

= 31400 m2

The area of the ground is 31400 m2.

### Question 8

From the given information calculate the area of the triangle. Given the height of the triangle is 5 cm and it divides the base into two equal parts of 3 cm each. Express your answer in m2. [3 MARKS]

#### SOLUTION

Solution :Formula : 1 Mark

Process : 1 Mark

Result : 1 Mark

Height = 5cmGiven base is divided into two equal parts, so base length = 3+3 = 6cm

Area of the triangle = 12×base×height

= 12×5100×6100

= 1510000

= 0.0015

= 0.0015 m2

### Question 9

A man is walking in a circular park. In one round he completes 314 m. If there is a man at the opposite end of the diameter which is drawn from the point where the man is standing. Find the minimum distance he has to walk to meet his friend who is sitting at the described point? Also, find the area of the circular park. Take π=3.14 [3 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Shortest Distance: 1 Mark

Area: 1 Mark

If the man has to cover the minimum distance, he has to walk along the diameter.Circumference of the circle = 314 m

Radius of the given circle = 3142π

= 50 m

Diameter = (2)×(50) = 100 m

The radius of the circular park = 1002 = 50 m

The area of the circular park = π×r2

On substituting the values, we get

Area = π×502 = 7850m2

### Question 10

If a given stick of length 628 m is bent in the form a circle and later bent into a square. Which of them, circle or square will have greater area? [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Result: 1 Mark

Length of the stick is equal to the perimeter of circle and square individually.

Circle of given radius has a perimeter of 628 m.

Side of square = 157 m

Radius of the given circle:

= 6282×π

= 100mArea of the circle:

= π × r2

= π × 1002

= 3.14× 1002

= 31400m2^{ }Area of the square

= length×length

= 157×157

= 24649m2The area of circle is greater than area of square.

### Question 11

Find the area of the shaded portion. A circle is inscribed in a square such that both their centres coincide. BDE is an isosceles triangle. The side of a square is 10m.

[4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Application: 1 Mark

Answer: 1 Marks

Given that,

The side of the square ABCD = 10 m.

Now the diameter of the circle will be equal to the side of the square ABCD = 10 m.

Area of the square = a2

Area of the square is (10) × (10) = 100 m2Given diameter of circle = 10 m

Radius of the circle = 102 = 5 m

Area of the circle

= π × (r2)

= π × (52)= 78.5m2

Area of the triangle BDE

= 12 × (BD) × (DE)

= 0.5 × (10) × (10)

= 50 m2

Area of the shaded region

= Area of square ABCD - Area of circle + Area of triangle BDE

= 71.5 m2

So, the area of the shaded region is 71.5 m2.

### Question 12

A man wants to paint two of his walls. The cost of painting is Rs10 per cm2. How much money is required by him to paint a rectangular wall of length 400 cm and breadth 900 cm and a circular wall of diameter 126 cm? [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

Given that

The cost of painting is Rs10 per cm2.

The length of the rectangular wall=400 cm

The breadth of the rectangular wall=900 cm

Area of the rectangle = length × breadth

On substituting the values we getArea of the rectangle = (400) × (900)

= 360000 cm2

Money required to paint the rectangular wall = (10) × (360000)

= Rs.3600000

The diameter of the circular wall = 126 cm.

Radius of the wall, r = 1262 = 63 cm

The area a circle is given by, Area = π×r2

On substituting the values we get,

Area of the circular wall

= 227×63×63 = 12474 cm2

Money required to paint the circular wall

= 12474×10 = Rs 124740

The total cost of painting

= 3600000 + 124740 = Rs 3724740

Hence, the total cost of painting the walls is Rs 37,24,740.

### Question 13

A rectangular park is 40m wide and 35m long. A path 2.5m wide is constructed outside the path. Find the area of the path. [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Application: 1 Mark

Answer: 1 Mark

The given area will be inscribed between two rectangles of length and breadth.

We can imagine as a rectangle whose length has been increased on both the sides.

So, the sides of the outer triangle are,

Length = 35 + 5 = 40 m

And breadth = 40 + 5 = 45 m40m and 45m, 40m and 35m respectively.

The area of the footpath = Area of outer rectangle - Area of inner rectangle

Area of the rectangle = length×breadth

∴ On substituting the values we get;Area required = (45) × (40) - (40) × (35)

= 1800 - 1400

= 400m2

The area of the footpath is 400m2.

### Question 14

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL. [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 1 Mark

Each Answer: 1 Mark

Given: area of parallelogram = 147 cm

Base (AB) = 35 cm and base (AD) = 49 cm

Since area of parallelogram = base×height

On substituting the values we get,

⇒ 1470=35×DL

⇒ DL=147035

⇒ DL = 42 cm

Again, Area of parallelogram = base×height

⇒ 1470=49×BM

⇒ BM=147049

⇒ BM = 30 cm

Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

### Question 15

A man wants to cover his floor with carpet which costs Rs100 per square meters. His floor is in the form of a square which has a perimeter of 24 meters. What will be the cost of carpeting the floor? [4 MARKS]

#### SOLUTION

Solution :Formula: 1 Mark

Steps: 2 Marks

Result: 1 Mark

Given that :

A man wants to cover the floor with carpet which costs Rs 100.

His floor is in the form of a square.

Perimeter of the floor = 24 meters

Let the length of each side of the square be 'a'.

Perimeter of a square = 4a = 24 meters

Length of one side, a = 244 = 6 meters

Area of a sqaure = a × a = 6 × 6 = 36 square meters

Total Cost of carpeting the floor = 36 × 100 = Rs 3600.

The total cost of carpeting the floor is Rs 3600.