Perimeter Area Volume Questions and Answers

Let us assume the length as 2x and breadth as 3x.

Now, 1/6 hectares = 1/6 * 10000 = 5000/3 m²

Then, 2x * 3x = 5000/3

6x² = 5000/3

x² = 5000/18 or 2500/9

x = 50/3

Thus, length = 100/3 m

Breadth = 150/3 or 50 m

The Diameter of the circle is equals to the smallest side of the rectangle.

Therefore, diameter = 22 cm.

Then, radius = 22/2 = 11 cm

Area = πr² = 22/7*11*11 = 380.28 cm²

Given, length = 29.5

Given breadth = 21.8

Area of rectangular = l * b = 29.5 * 21.8= 643.1m²

Perimeter of rectangle = 2 * (29.5 + 21.8) = 2 * 51.3

= 102.6 m

Perimeter = Area covered in 8 minutes = 10000/60 * 12

= 2400 m

Let us assume the length as 5x and breadth as 4x.

Then, 2(5x + 4x) = 2400

2(9x) = 2400

18x = 2400

x = 133.33

Then, length = 666.65

Breadth = 533.32

Therefore, area = length * breadth

= 666.65 * 533.32 = 355537.78 m²

Let us assume the length as x, breadth as 3x/2, and height as h.

Then the area of the floor = Cost of flooring/Rate meter per square

= 540/10 = 54 m².

Therefore, x * 3x/2 = 54

x² = 54 * 2/3 = 36

x = 6

Thus, length = 6

Breadth = 3 * 6/2 = 9

Now, area of four walls = Cost of painting/Rate = 3440/20 = 172 m²

Given, area of two windows and one door = 8 m²

Therefore, total area of walls = 172 + 8 = 180 m²

Thus, 180 = 2(l+b) * h

2(6+9)* h = 180

30 * h = 180

30h = 180

h = 6 m

As we know that diagonals of a rhombus bisect at a right angle. Therefore, one side and half of each diagonal form a right angle triangle.



Given, one side = 5 cms

Another side = 8/2 = 4 cms

Third side = (5)² – (4)² = = 3

Therefore, half of the other diagonal is 3

Thus, other diagonal = 3*2 = 6 cms

Hence, area = ½ * 8*6 = 24 cms²

 

Given, AB = 80 m

CD = 50 m

BC = 70 m

AD = 48 m

CE is parallel to AD, therefore, CE = 48 m

Now, AE = CD = 50 m

BE = (AB – AE) = (80-50) = 30 m

CE = AD = 48 m

Then, s = 30 + 70 + 48/2 = 74 m

Area of BCE = ⎷74 * 30 * 70 * 48 = 2731.15 m²

So, ½ * 30 * height = 2731.15

Height 182.07 m

Therefore, area of trapezium = ½ * (80+50) * 182.07 = 11834.55 m²

Therefore, the cost of cultivating = 11834.55 * 95/100 = Rs. 11242.82

Let us assume the original length as x and breadth as y.

Then area = xy

If the side is increased by 42%, then new length = 142/100x = 71x/50

New breadth = 71x/50

Then, new area = 71x/50* 71x/50

= 5041/2500 xy

The difference between new and original area = 5041/2500 xy – xy

= 2541xy/2500

Therefore, percent increase = 2541xy/2500 * 1/xy * 100 = 101.64%

Let us assume PQR as an isosceles triangle and PS as altitude.

Let PQ = PR = x

Then, QR = (40-x)

As we know that in an isosceles triangle the altitude bisects the base,

Then, QS = SR = (20 – x)

In triangle PSR, PR² = PS² + SR²

Then, x² = (10)² + (20 – x)²

x² = 100 + 400 – 40x + x²

40x = 500

x = 25/2

Thus, QR = (40 – 2x) = 40 – 25 = 15

Hence, area = ½ * QR * PS

= ½ * 25/2 * 15 = 93.75 cm²

Area = (total cost)/ (rate/hectare)

Hectares = 1880/117.5 = 16

16 hectares = 160000 m²

Side of the square = 160000 = 400m

Perimeter = 4 * 400 = 1600 m

Cost of fencing = 1600 * 80/100 = Rs. 1280.

The area of the plot = 150 * 75 = 11250 m²

Area of plot without the path = [(150-6) * (75-6)] = 9936 m²

Therefore, area of the path = 11250 – 9936 = 1314 m = 13140000cm²

Cost of 1000 cm² = 78 paisa

Cost of 13140000 cm = 78/1000 * 13140000/100 = Rs. 10249.2

Area of Rhombus = Area of 2 triangles (with one Diagonal as the base and half of other

Diagonal a height)

95 =  * 10 *  +  * 10 *

95 = 2 *  * 10 *

5L = 95 cm

L = 19 cm

Hence, the length of other Diagonal is 19 cm

Let us assume the base of both the triangle as p and q, respectively.

Then,  = 7/5

p/q = 7/5 * 7/5 = 49/25

Therefore, ratio = 49: 25

The distance covered in a single revolution: 176 * 2000/1000

= 352 m

Area of the circle = 2πr

= 352 = 2 * 22/7 * r

2464 = 44r

r = 2464/44

r = 56

Let us assume the other diagonal as 2x cm

As we know that the diagonals bisect one another at the right angle, then

(24)² = (25)² + (x)²

(x)² = (25)² – (24)²

(x)² = 49

x = 9 cm

Therefore, another diagonal = 18 cm

Area = ½ (18 * 50) = 450 cm²

Area of floor = 15 * 10 = 150 m²

Area of carpet = 150 m²

Breadth of carpet = 50cm = ½ m

Length of carpet = 150*1/2 = 75 m

Cost of carpet = 75 * 3.50 = 262.5

Area of walls = 2*(15+10) * 12 = 600 m²

Area to be painted = 600 – 50 = 550 m²

Cost of painting = 550 * 5.65 = Rs. 3107.5

No. of artificial grass block = Rectangular field/ Area of each artificial grass block

= (36* 6 m^2)/(160*60 cm^2)

= (36* 6*100*100 cm^2)/(160*60 cm^2)

= 2160000/9600

=225 blocks

Square perimeter = 4 * triangle Area
As it is given that the area of triangle and perimeter of circle is equal,
Then, Square perimeter = 4 * circle perimeter
= 4 * 2πr
= 4 * 2 * π * 21
= 528 cm
Side of the square = 528/4 = 132 cm

Let us assume the width of the boundary as x.

Then, (20 – 2x) (10 – 2x) = 96

4x² + 60x -104 = 0

x² + 15x – 26 = 0

x² - 13x - 2x – 26 = 0

(x-13) (x-2)

x = 13, 2

As, x cannot be 13, we will take the value of x as 2

Radius of plot = 140/2 = 70

Radius of plot including path = 70 + 7 = 77

Area of plot excluding path = 22/7 * 70 * 70 = 15400 m²

Area of plot including path = 22/7 * 77 *77 = 18634 m²

Area of path = 18634 – 15400 = 3234 cm² = 32340000 m²

Area of one tile = 50*22 = 1100

Number of stones = 32340000 /1100= 29400

Side of the square = 52/4 = 13

Given, Diameter of the circle = 13

Then, radius = 13/2

Area = πr²

=22/7 *13/2  *13/2 = 3718/28

= 132.78 cm²

Area of road PQRS = 160 * 24 = 3840 m²

Area of road UVWX = 100 * 24 = 2400 m²

It is clearly depicted that area JKLM is common to both PQRS and UVWX road.

Area of ABCD = 24*24 = 576 m²

Therefore, area of path to be gravelled = 3840 + 2400 – 576 = 5664

Cost of levelling = 5664 * 25/100 = Rs. 1416.

Let us assume the breadth as x meters

Then, length = x + 110

Perimeter = 18500/65.5 = 282.44

Therefore, 2[(x+110) + x] = 282.44

= 2x + 220 + 2x = 282.44

= 4x + 220 = 282.44

= 4x = 282.44 – 220

4x = 62.44

x = 15.61 meters

Thus, length = 15.61 + 110 = 125.61 meters

Area to be coloured: [2(l+b) * h] + (l*b)

= [2(14+8)*4] + (14 * 8)

= [2(22)*4] + 112

= 176 + 112

= 288 m²

Cost of colouring = 288 * 99/100

= Rs. 285.12

2P + 5P + 9P = 1022

16P = 1024

P = 64

Side are, 2P = 2*64 = 128

5P = 5 * 64 = 320

Area =   * b * h

=   * 128 * 320

= 20480 cm²

As we know that diagonals of a rhombus bisect at a right angle. Therefore, one side and half of each diagonal form a right angle triangle.



Given, one side = 32 cm

Given, diagonal = 18 cm

Then, half of the diagonal = 18/2 = 9 cm = another side of the right angle triangle

Third side = (32)² – (9)² = = 30.70 cm

Therefore, half of the other diagonal is 30.70

Thus, other diagonal = 30.70 * 2 = 61.4 cm

Hence, area = ½ * 18 * 61.4 = 552.6 cm²

Area of PQRS = 2 * (area of POR)

PQ = 42, QR = 56, PR = 60

S = ½ PQ + QR + PR = ½ * 42 + 56 + 60 = 79 Cm

Therefore, (S-42) = 37

(S – 56) = 23

(S – 60) = 19

Area of PQR = ⎷S (37) (23) (19) = ⎷79*37*23*19 = ⎷1277351 = 1130.19

Area of parallelogram PQRS = 1130.19 * 2 = 2260.38 cm²

Side of the square = 20/4 = 5

Given, Diameter of the circle = 5

Area = πr²

= 22/7 * 5/2 * 5/2 = 19.64 cm^2

Square Diagonal =  * 25 = 25 * 2

= 50 cm

Therefore, diameter = 50 cm

Area of the circle = 22/7 *  (50)^2/ 4

= 22/7 * 625

= 1964.28 cm²

Area of paper = 22 * 22 = 484 m²

No. of triangles = Paper Area / triangular print area

= 484 m2/ 16 cm2

=(484 * 100 * 100 cm2)/ 16 cm2

= 4840000/ 16

= 302500 cm.