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Mensuration

Mensuration is one of the branches of mathematics. This means measurement. It is being done in our life in many situations. For example, length of cloth we need for stitching, the area of a wall which is being painted, perimeter of a rectangular plot to be fenced, quantity of water needed to fill the tank etc. For these kinds of activities, we are doing measurements for further needs. F Here, we cover three areas
 

1. Perimeter        
2. Area        
3. Volume
To find these things we have to remember the formulae of plane figures and solid figures. Try to understand the concept behind reaching the formula.


Mensuration - 2D (Plane Figures)

1. The length of a rectangular field is four times its width. If the perimeter of the field is 30 m, find the area of the field?
a) 108 m2                b) 27 m2              c) 9 m2             d) 36 m2

 

2. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio 6 : 5. Find the smaller side of the rectangle?
a) 72 cm            b) 120 cm             c) 60 cm            d) 144 cm

 

3. The radius of a circle is 14 cm. Find the area of the sector with central angle 36°?
a) 61.6 cm2            b) 30.8 cm2             c) 15.4 cm2            d) 308 cm2

 

4. In the given figure, if AB = 4 cm and BD = 4√3 cm, then the area of the shaded region will be?

a) 48 π cm2            b) 24  π cm2              c) 16 π cm2               d) 12 π cm2 

 

5. A rectangular garden has an area of 2000 m2 and its length and breadth are in the respective ratio 5 : 4. A road of uniform width runs inside the garden around the perimeter and has an area of 344 m2. The width of the road is......
a) 3 m               b) 3.5 m              c) 4 m                 d) 2 m

 

6. If the perimeters of a rectangle and a square are equal and the ratio of two adjacent sides of the rectangle is 1 : 2 then the ratio of area of the rectangle and that of the square is?
a) 1 : 1            b) 1 : 2             c) 2 : 3            d) 8 : 9

 

7. If a wire is bent into the shape of a square, then the area of the square so formed is 81 cm2. When the wire is rebent into a semicircular shape, then the area (in cm2) of the semicircle will be (Take π = 22/7)
a) 22            b) 44             c) 77             d) 154

 

8. At each corner of a triangular field of sides 26 m, 28 m and 30 m, a cow is tethered by a rope of length 7 m. The ungrazed area (in m2) by the cow is.......
a) 336            b) 259           c) 154           d) 77

 

9. How many rotations will a wheel of a car make in a journey of 88 km, if the radius of the wheel is 56 cm?
a) 60000            b) 50000           c) 30000             d) 25000

 

10. A cow is tied by a rope to one of the vertices of a square of side 14 m. The length of the rope is 7 m. What percentage of the field is grazed by the cow?
a) 15%           b) 25%          c) 18%          d) 19.6%

 

11. A horse is tethered by a rope. What is the length of the rope so that the horse can graze over an area of 154 m2?
a) 2 m           b) 7 m           c) 14 m           d) 49 m

 

12. A circular grass lawn of 28 m radius, has a path of 14 m wide running around it on the outside. Find the area of the path?
a) 3920 m2             b) 3080 m2               c) 3980 m2            d) 8030 m2             

 

13. The area of a circular plot is 154 m2. Find the cost of fencing it at the rate of Rs. 2.75 per metre?
a) Rs.56             b) Rs.100             c) Rs.121              d) Rs.423.50

 

14. A wire bent in the form of a square, encloses an area of 484 cm2. If the same wire is bent to form a circle, then the area enclosed will be
a) 1232 cm2             b) 616 cm2              c) 2464 cm2             d) 4312 cm2           

 

15. A copper wire is bent in the form of an equilateral triangle and has area 121√3 cm2. If the same wire is bent into the form of a circle, the area (in cm2) enclosed by the wire is....... (Take π = 22/7)
a) 364.5               b) 693.5              c) 346.5               d) 639.5

 

16. The area of a rhombus is 150 cm2. The length of one of its diagonals is 10 cm. The length of the other diagonal is........
a) 25 cm           b) 30 cm          c) 35 cm             d) 40 cm

 

17. The area of the ring between two concentric circles, with circumference 88 cm and 132 cm respectively is......
a) 780 cm2                 b) 770 cm2             c) 715 cm2               d) 660 cm2

 

Key With Explanations

1-d; 2(l + b) = 30
⇒ 2(4b + b) = 30
⇒ 10b = 30 ⇒ b = 3 m and l = 12 m
So, area of the field = l × b = 3 × 12 = 36 m2.


2-c; Perimeter of the rectangle = Circumference of the circular wire 
= 2 × 22/7 42 cm = 264 cm. 
Let the dimensions of the rectangle be 6x and 5x respectively.
∴ 2 × (6x + 5x) = 264 ⇒ x = 12
∴ Smaller side = 5x = 60 cm

 

3-a; Radius of the circle = 14 cm
Angle of the sector = 36° 
Area of the sector = θ/360° × π r2 
∴ Area of the sector 
= 36/360°  ×  22/7 × 14 × 14 = 61.6 cm2

 

4-d; ∠ADC is a right angle (Angle in a semicircle)
So, BD2 = AB × BC ⇒ 16 × 3 = 4 × BC
⇒ BC = 12 cm
Now shaded area = Area of bigger semicircle − Areas of smaller semicircles 
= 1/2  π (8)2 - 1/2 π (2)2 - 1/2 π (6)2 
= 32 π − 2 π −18 π = 12 π cm2

 

5-d; Let 5x and 4x be the length and breadth of the garden.

Then, 5x × 4x = 2000
⇒ x2 = 100 ⇒ x = 10
∴ Length = 50 m and breadth = 40 m
Let ‘d’ be the width of the road. Then,
(50 − 2d) (40 − 2d) = 2000 − 344 ⇒ d = 2 m

 

6-d; Let side of rectangle be 2x and x units. and side of square = y units 
∴ 4y = 6x ⇒ x/y = 4/6 = 2/3
∴ 2x × x/y2 = 2x2/ y2 = 2 × 4/9 = 8 : 9 


7-c; Area of square = a2 = 81 cm⇒ a = 9 cm
Perimeter of square = 4 × a = 4 × 9 = 36 cm
Perimeter of semi circle = πr + 2r ⇒ 36 = r (π + 2)  
⇒ 36 = r (36/7)  ⇒ r = 36 × 7/36 = 7
Area of semicircle = 1/2 πr
= 1/2 × 22/7 × 7 × 7 = 77 cm2 

 


 


Hence, ungrazed area = 336 − 77 = 259 m2

 


 

11-b; Let the length of the rope be r. Horse can graze an area equal to area of the circle of radius r.
Then, πr2 = 154 ⇒ r = 7 m

 

12-b; Radius of the outer circle = 28 + 14 = 42 m
Area of the path = Area of outer circle − Area of inner circle
= π (42)2 − π (28)2 = 3080 m2

 

13-c; Let r be the radius of the circular plot.
Then, πr2 = 154 ⇒ r = 7 m 
Circumference of the plot = 2 × 22/7 × 7 = 44 m
Cost of fencing the plot = Rs.44 × 2.75 = Rs.121 

 

14-b; Side of the square =  = 22 cm
Perimeter of the square = 4 × 22 = 88 cm
Perimeter of the circle = 2π × radius = 88 cm
⇒ Radius = 14 cm 
Area of the circle = π × (14)2 
= 22/7 × 14 × 14 = 616 cm2 

 

15-c; Area of equilateral triangle = √3/4  a
[where a is side of triangle] 
∴ √3/4  a = 121√3 ⇒ a = 22 cm 
Perimeter of triangle = 22 × 3 = 66 cm
Perimeter of circle = 2πr i.e. 2πr = 66
 
∴ Area of outer circle = πr2 = 22/7 × 21 × 21 = 1386 cm2 
area of inner circle = πr12 = 22/7 × 14 × 14 = 616 cm2 
Hence, area of ring = 1386 − 616 = 770 cm2

 

PLANE FIGURES


 

SOLID FIGURES
 


 

EXERCISE

1. What will be the area (in square cm) of a triangle with base 10.2 cm and height 3.5 cm?
A) 15.65           B) 17.45            C) 18.25              D) 17.85

 

2. A square field has an area of 50625 sqm. Find the cost of fencing around it at Rs.15 per metre?
A) Rs.13,225           B) Rs.13,500           C) Rs.12,750            D) Rs.12,800

 

3. The breadth of a rectangular plot is 75% of its length. If the perimeter of the plot be 560 m, what is its area (in sqm)?
A) 17500              B) 22500                 C) 19200                 D) 16800

 

4. The perimeter of a square is 64 cm. Its area will be (in square cm) (SSC-CGL 2019)
A) 32         B) 256         C) 8         D) 128

 

5. The perimeter of an isosceles triangle is 32 cm and each of the equal sides is 5/6 times of the base. What is the area (in cm2) of the triangle? (SSC-CGL 2017)
A) 48           B) 39              C) 57                D) 64

 

6. The radius of a circular ground is 42 m. The distance (in meters) covered by running 8 rounds around it is … (SSC-CGL 2019)
A) 3248           B) 4262              C) 1124          D) 2112

 

7. A room is 7.5 m long, 5.5 m broad and 5 m high. What will be the expenditure in covering the walls by paper 40 cm broad at the rate of 75 paise per meter?
A) Rs.245.25           B) Rs.243.75          C) Rs.275.50            D) Rs.275.45

 

8. Find the length of the longest stick that can be placed in a room of 12 m long, 9 m broad and 8 m high?
A) 18 m             B) 19 m             C) 17 m            D) 32 m

 

9. Two cubes have their volumes in the ratio 8 : 27. Find the ratio of their surface areas?
A) 2 : 3                B) 4 : 9                 C) 16 : 81               D) 8 : 9

 

10. If the diameter of a sphere is 14 cm, then what is the surface area (in cm2) of the sphere? (SSC-CGL 2017)
A) 616            B) 308           C) 462           D) 636

 

11. How many tiles of 25 cm length and 15 cm width are required to pave the floor of a hall 9 m long and 6 m wide?
A) 1950           B) 1550         C) 1440          D) 2000

 

12. If the base radius of 2 cylinders are in the ratio 3 : 4 and their heights are in the ratio 4 : 9, then the ratio of their volumes is..... (SSC-CGL 2019)
A) 1 : 2           B) 2 : 1            C) 4 : 1              D) 1 : 4

 

13. A rectangular tank measuring 10 m × 8 m × 6 m with a circular hole of one-meter diameter, how much iron sheets (m2) will be needed to make such a tank? (Correct to one decimal place) (SSC-CGL 2018)
A) 370.4          B) 371.6           C) 375.2            D) 370.8

 

14. A rectangular field of 60 metres length and 40 meters wide is to be surrounded by a road 5 meter wide. If the cost of making one square meter road is Rs.500, what would be the cost of the entire road?
A) Rs.13,75,000            B) Rs.12,45,000             C) Rs.17,55,000             D) Rs.15,45,500

 

Key With Explanations

1-D; Area of triangle = 1/2 × base × height  2 
⇒ 1/2 × 10.2 × 3.5 = 17.85 cm2 


2-B; Side of the square field = √50625 = 225 m
Length of the wire required = 4 × 225 = 900 m
Total cost = Rs.15 × 900 = Rs.13,500

 

3-C; Let the length of the plot be 100, then breadth will be 75
∴ Perimeter = 2(100 + 75) = 350 ATQ, perimeter is 560
When the perimeter is 350, length is 100.
If the perimeter is 560, then the length = 560/350 × 100 = 160 
breadth = 3/4 × 160 = 120 
∴ Area = 160 × 120 = 19,200 sqmt

 

4-B; Let the length of the square be 'a'
∴ 4a = 64 ⇒ a =16  
∴ Area = 162 = 256

 

5-A; Let the base be 6 units then other 2 equal sides are 5 units each perimeter of the triangle = 16 units = 32 cm 
∴ base = 12 cm and other sides = 10 cm each
∴ Height = √102 - 62 = 8 (using pythagorous theorem) 
∴ Area of the triangle = 1/2 × base × height 
⇒ 1/2 × 12 × 8 = 48 cm2 

 

6-D; Circumference of the ground = 2 × π × 42
Required distance = 8 × (2 × π × 42) 
⇒ 8 × 2 × 22/7 × 42 = 2112 m 

 

7-B; Area of four walls = 2 × 5 (7.5 + 5.5) = 130 m2
Area of required paper = 130 m2
Breadth of the paper = 40 cm = 0.4 m 
∴ Length of the paper = 130/0.4 = 325 m 
∴ Cost of the paper = 325 × 0.75
= Rs.243.75

 

8-C; Length of the stick = length of the diagonal of the room 

 


∴ Surface areas ratio

 

10-A; Diameter = 14 cm, then radius will be 7 cm
Surface area = 4πr2 = 4 × 22/7  × 7 × 7 = 616 cm2 


11-C; Area of the floor of the hall = 900 × 600
Area of the tile = 25 × 15 cm2
∴ Number of tiles required = 900 × 600/25 × 15 = 1440 

 

12-D; Volume of a  cylinder = πr2h
∴ Required ratio is π × (3)2 × 4 : π × (4)2 × 9
⇒ 9 × 4 : 16 × 9 = 1 : 4

 

13-C; Required area of iron sheets = TSA of tank - area of circular hole
⇒ 2(10 × 8 + 8 × 6 + 10 × 6) - π (1/2)2
⇒ 2(188) - 22/7 × 1/4 ⇒ 376 - 0.785 = 375.2 

 

14-A; Area of the road = Area of outside rectangle - Area of inside rectangle
⇒ 70 × 50 - 60 × 40 
⇒ 3500 - 2400 = 1100 sqm
∴ Cost = 1100 × 1250 = Rs.13,75,000

Posted Date : 03-05-2021

 

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