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

Result : 1 Mark

Let the square given be of side A cm.

Perimeter of the square whose side is A cm is 4 $\times$ A

Perimeter of the given square

= 4 $\times$ 10

= 40 m

= 4000 cm

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, $Breadth=\frac{perimeter}{2}-length$
Or, Breadth = 25 - 13 = 12 m
Area of rectangle = length $\times$ breadth $=13\times 12$ = 156 sq. m
The area of the rectangular field is 156 $m^2$.

Formula: 1 Mark
Answer: 1 Mark

Given that,
The side of a parallelogram is 3 m and the corresponding altitude = $\frac{3}{8}m$
Let the side and altitude of the parallelogram be A and B

Area of the parallelogram = A$\times$B

Given side and altitude are 3 m and $\frac{3}{8}m$

Area of the given parallelogram
$=\left(3\right)\times \frac{3}{8}=1.125m^2$
The area of the parallelogram is $1.125m^2$.

Formula: 1 Mark
Answer: 1 Mark

Area of the circle of radius R is $\pi \times R^2$

Radius of the given circle = 10 m

Area of the given circle  = $\pi \times {10}^2$
                                       = 314 $m^2$
The area of the circle is 314 $m^2$.

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 $2\times \pi \times r$ meters

Given perimeter of the ground = 628 m

r = $628÷(2\times \pi )$ =100 m
The distance of the boundary from the centre of the pitch is 100 m.
The area of a circle is $\pi \times (r{)}^2$

Area of the ground = $\pi \times (100{)}^2$

= 31400 $m^2$ 
The area of the ground is 31400 $m^2$.

Formula : 1 Mark
Process : 1 Mark
Result : 1 Mark

Height = 5cm

 Given base is divided into two equal parts, so base length = 3+3 = 6cm

Area of the triangle = $\frac{1}{2}\times base\times height$

                                     = $\frac{1}{2}\times \frac{5}{100}\times \frac{6}{100}$

                                      = $\frac{15}{10000}$

                                       = 0.0015

                                      = 0.0015 $m^2$

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 = $\frac{314}{2\pi }$

= 50 m

Diameter = (2)$\times$(50) = 100 m
The radius of the circular park = $\frac{100}{2}$ = 50 m

The area of the circular park = $\pi \times r^2$

On substituting the values, we get
Area = $\pi \times {50}^2$ = $7850m^2$

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:
= $\frac{628}{2\times \pi }$
= 100m

 Area of the circle:                
= $\pi$ $\times$ $r^2$
= $\pi$ $\times$ ${100}^2$
= $3.14\times$ ${100}^2$
= $31400m^2$ 

Area of the square
= $\text{length}\times \text{length}$
= $157\times 157$
= $24649m^2$

The area of circle is greater than area of square.

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 = $a^2$
Area of the square is (10) $\times$ (10) = 100 $m^2$

Given diameter of circle = 10 m

Radius of the circle = $\frac{10}{2}$ = 5 m

Area of the circle 

= $\pi$ $\times$ ($r^2$)
= $\pi$ $\times$ ($5^2$)

= 78.5$m^2$   

Area of the triangle BDE

= $12$ $\times$ (BD) $\times$ (DE)

= 0.5 $\times$ (10) $\times$ (10)

= 50 $m^2$

Area of the shaded region

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

= 71.5 $m^2$

So, the area of the shaded region is 71.5 $m^2$.

Formula: 1 Mark
Steps: 2 Marks
Answer: 1 Mark

Given that 
The cost of painting is Rs10 per $c{m}^2$.
The length of the rectangular wall=400 cm
The breadth of the rectangular wall=900 cm

Area of the rectangle =  length $\times$ breadth
On substituting the values we get

Area of the rectangle = (400) $\times$ (900)

  = 360000 $c{m}^2$

Money required to paint the rectangular wall = (10) $\times$ (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 = $\pi \times r^2$
On substituting the values we get,
Area of the circular wall
= $\frac{22}{7}\times 63\times 63$ = 12474 $c{m}^2$

Money required to paint the circular wall
 = 12474$\times$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.

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 m

40m and 45m, 40m and 35m respectively.
The area of the footpath = Area of outer rectangle - Area of inner rectangle
Area of the rectangle = length$\times$breadth
$\therefore$ On substituting the values we get;

Area required = (45) $\times$ (40) - (40) $\times$ (35)

= 1800 - 1400

= 400$m^2$
The area of the footpath is 400$m^2$.

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 = $\frac{24}{4}$ = 6 meters

Area of a sqaure = a $\times$ a = 6 $\times$ 6 = 36 square meters
Total Cost of carpeting the floor = 36 $\times$ 100 = Rs 3600. 
The total cost of carpeting the floor is  Rs 3600.