## Mensuration

**Mensuration Notes**

**Area: A**rea of a bounded figure is the space covered by it.

**Perimeter: **The length of the boundary of a closed figure is called perimeter

**Triangle:**

**Area of a triangle =
**Where a, b, c are the lengths of the sides of triangle

- Area of Isosceles triangle =

And height of Isosceles triangle,

Perimeter of Isosceles triangle = 2b + a

**Square:**

where d is the diagonal of a square

Perimeter of a square = 4a

Diagonal of a square =

Area of a path which is outside of a square of uniform width,

Area of a path which is inside of a square of uniform width,

Perimeter of a path which is outside of a square of uniform width,

Perimeter of a path which is inside of a square of uniform width,

**Rectangle:**

Area of a rectangle = l * b, where l is the length of a rectangle and b is the breadth of a rectangle

Perimeter of a rectangle = 2(l + b)

Diagonal of a rectangle,

Area of a path which is outside of a rectangle of uniform width,

Area of a path which is inside of a rectangle of uniform width,

Perimeter of a path which is outside of a square of uniform width,

Perimeter of a path which is outside of a square of uniform width,

**Parallelogram:**

**Rhombus:**

- It is a parallelogram whose all sides are equal
- Its diagonal bisect each other at right angle.

Where b is the side of rhombus, h is the height and d1,d2are the diagonals of the rhombus

Perimeter of rhombus = 4b

**Trapezium:**

Area of trapezium = 12height (sum of parallel sides)

Median of trapezium = 12(sum of parallel sides)

Median is the line segment joining the midpoints of non parallel side

**Circle:**

Area of a circle= r2

Diameter (d) = 2r

Circumference = 2r

Area of a sector =360 r2

Length of arc = 360 2r

Area of the ring or circular path = (R+r) (R-r)

**Cube: **

Volume of a cube = a3

Diagonal of a cube = 3a

Surface area of a cube = 6a2

Area of four walls = 4a2

**Cuboid: **

Volume of a cuboid = lbh

Diagonal of a cuboid = l2+ b2+h2

Surface area of a cuboid = 2(lb + bh + hl)

Area of four walls = 2(lh + bh)

**Cylinder:**

Volume of a cylinder = r2h

Curved surface area = 2rh

Total surface area = 2r(r + h)

**Cone: **

Volume of a cone = 13r2h

Curved surface area = rl

Total surface area = r (l + r)

Slant height = h2+ r2

**Sphere: **

Volume of a sphere = 43r3

Surface area = 4r2

**Hemisphere:**

Volume of a hemisphere = 23r3

Curved surface area = 2r2

Total surface area = 3r2

**Frustum of a cone**

Volume = 13h(r2+R2+ r.R)

Surface area = (r + R)(R -r)2+h2

Total surface area = (r + R)(R -r)2+h2 + r2 + R2

Slant height, l = (R -r)2+h2

**Right pyramid**

Volume of a pyramid = 13(area of the base)height

Lateral surface area of a pyramid = 12(perimeter of base) slant height

Total surface area of a pyramid = lateral surface area + area of the base

Work out problems

- The area of a rectangular field is 460 square metres. If the length is 15 percent more than the breadth, what is breadth of the rectangular field ?

- 15
- 26
- 34.5
- Cannot be determined
- None of these

Answer: e) none of these

**Solution:** Let breadth of a rectangle be x

Then length = x + 15x/100

= x + 3x/20

= 23x/20

Area of a rectangle = 460

x 23x20 = 460

xx = 400

x = 20

- If the length of the diagonal AC of a square ABCD is 5.2 cm, then the area of the square is :

- 15.12 sq. cm
- 13..52 sq.cm
- 12.62 sq.cm
- 10.00 sq. cm

**Answer:** b) 13.52 sq.cm

**Solution:** since the square is a rhombus and the diagonals of a square are equal

Area of a rhombus = 12 product of diagonals

= 125.2 5.2

= 2.65.2

= 13.52

**3.**if the diagonal of two squares are in the ratio of 2:5, their area will be in the ratio of

- 2:5
- 2:5
- 4:25
- 4:5

**Answer:** c) 4:25

**Solution:** let the diagonals of the square be 2x and 5x respectively

Area of one square = 122x 2x

= 4x2/2

Area of other square = 125x 5x

= 25x2/2

Ratio of areas of both the square = 4x22/ 25x22

= 4/25

= 4:25

- 8 –
- 16 – 4
- 16 – 8
- 4 – 2

**Answer: **b) 16 – 4

**Solution:** area of a square sheet = 44 = 16 sq. cm

Since the angle between two sides of a square is 90 degree and the radius of arc is 2 cm

S, area of 4 arc will be = 49036022

= 4

Area of remaining portion = 16 – 4

- The cost of building a fence around a circular field is rupee 7700 at the rate of rupee 14 per foot . what is the area of the circular field ?

- 24062.5
- 23864.4
- 24644.5
- Cannot be determined
- None of these

**Answer:** a) 24062.5

**Solution:** length of a fence = 7700/14

= 550

Circumference of a circle field = 2r

2r = 550

2 227r = 550

r = 175/2

Area of circular field = r2

= 22717521752

= 24062.5

- The sides of a triangle are 6 cm, 8 cm and 10 cm. The area (in cm2) of the triangle formed by joining the mid_points of this triangle is:

- 12
- 24
- 6
- 18

**Answer:** c) 6

Solution: a = 10, b = 8, c = 6

s= (6 + 8 + 10)/2

S = 12

Area of a triangle ABC = 12(12-10) (12-6) (12-8)

= 12(2) (6)) (4)

= 24

All the four triangles ADE, BDF,CEF and DEF are congruent, so their area are equal

Hence, area of triangle DEF = 24/4

= 6 sq cm.

- If D and E are the mid_points of the side AB and AC respectively of the ABC in the figure given here, the shaded region of the triangle is what percent of the whole triangular region?

- 50%
- 25%
- 80%
- 75%

**Answer:** d) 75%

Solution: Since D and E are the mid_ points of AB and AC respectively

So, DE is parallel to BC and 2DE = BC

ADE ABC

area ADEarea ABC = (DE)2 (BC)2

= (DE)2 (2DE)2

= 1/4

area of shaded partarea ABC = ¾

% = 3/4100

= 75%

- If the length of a rectangle is increased by 20% and breadth is decreased by 20% . then its area

- Increases by 4%
- Decreases by 4%
- Decreases by 1%
- Remains unchanged

**Answer: **b) decreased by 4%

**Solution:** let the length of rectangle be 10

And its breadth be 10

Then, area = 100

When the length is increased by 20%, new length = 12

When its breadth is decreased by 20%, new breadth = 8

then , new area = 96

So, area decreased by 4%

- If the surface area of a cube is 294 sq. cm, then its volume(in sq. cm) is

- 216
- 125
- 343
- 512

**Answer **c) 343

**Solution** surface area of a cube = 294

6a2= 294

a2= 49

a = 7

Volume of a cube = a3

= 777

= 343

- A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side 3 cm to form a cone, the volume of the cone so formed is

- 16
- 12
- 15
- 20

**Answer:** a) 16

**Solution:**

when a triangle of side 3cm, 4cm and 5cm is rotated about the side 3 cm, then a cone of radius 4 cm and height 3 cm is formed

Volume of a cone = 13r2h

= 13443

= 16

- A wooden box measures 20 cm by 12 cm by 10 cm. Thickness of the wood is 1 cm. Volume of wood to make the box (in cubic cm) is

- 960
- 519
- 2400
- 1120

**Answer** a) 960

**Solution** outer dimension of wooden box = 20 cm , 12 cm and 10 cm

Outer volume of box = 201210

= 2400 cc ( cubic centimeter)

Thickness of wood = 1 cm

Inner dimension of wooden box = 20 – 2 = 18 cm, 12 – 2 = 10 cm and 10 – 2 = 8 cm

Inner volume of box = 18108

= 1440 cc

Volume of wood in the box = 2400 – 1440 = 960 cc

- Water flows into a tank which is 200m long and 150m wide, through a pipe of cross-section 0.3m0.2m at 20 km/hour. Then the time (in hours) for the water level in the tank to reach 8m is

- 50
- 120
- 150
- 200

**Answer** d) 200

**Solution: **speed of water = 20 km/hour

= 20000 m/hour

Area of base = 0.3m0.2m

= 0.06 sq. meter

Volume of water flows in 1 hour = area of base speed of water in 1 hour

= 0.06 20000

= 1200 cubic meter

Volume of tank upto 8 m height = 2001508

= 240000 cubic meter

Time to reach the water 8 m high = 240000/1200

= 200 hours

- The base of a right prism is a quadrilateral ABCD. given that AB = 9 cm, BC = 14 cm, CD = 13 cm, DA = 12 cm and angle DAB = 90 degree. If the volume of the prism be 2070 cm3, then the area of the lateral surface is

- 720
- 810
- 1260
- 2070

**Answer** a) 720

**Solution** Area of quadrilateral ABCD = area of ABD + area of BCD

Since ABD is a right angled triangle

BD = AB2 + AD2

= 92 + 122

= 225

= 15 cm

Area of ABD = 12912

= 54 sq.cm

For BCD, s = (13 + 15 + 14)/2

= 21

area of BCD = 21(21 – 13) (21 – 14) (21 – 15)

= 21876

= 84 sq. cm

Area of quadrilateral ABCD = 54 + 84

= 138 sq. cm

Volume of prism = area of baseheight

2070 = 138h

Height = 2070/138

= 15 cm

Perimeter of quadrilateral ABCD = 9 + 14 + 13 + 12

= 48 cm

Area of lateral surface = perimeter of baseheight

= 48 15

= 720 sq. cm

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