# Formula - Mensuration

**MENSURATION**

Mensuration is the science of measurement of the lenghts of lines, areas of surfaces and volumes of solids.

**Perimeter :** Perimeter is sum of all the sides. It is measured in cm,m, etc.

**Area :** The area of any figure is the amount of surface enclosed within its boundary lines. This is measured in square unit like cm^{2}, m^{2}, etc.

**Volume :** If an object is solid, then the space occupied by such an object is called its volume. This is measured in cubic unit like cm^{3}, m^{3}, etc.

**Basic Conversions :**

I. 1 km = 10 hm

1 hm = 10 dam

1 dam = 10 m

1 m = 10 dm

1 dm = 10 cm

1 cm = 10 mm

1 m = 100 cm = 1000 mm

1 km = 1000 m

II. 1 km = ^{5}⁄_{8} miles

1 mile = 1.6 km

1 inch = 2.54 cm

1 mile = 1760 yd = 5280 ft.

1 nautical mile (knot) = 6080 ft

III. 100 kg = 1 quintal

10 quintal = 1 tonne

1 kg = 2.2 pounds (approx.)

IV. l litre = 1000 cc

1 acre = 100 m^{2}

1 hectare = 10000 m^{2}(100 acre)

**Part I : Plane Figures**

**TRIANGLE**

**Right triangle**

**Example 1**

**1.Find the area of a triangle whose sides are 50 m, 78 m, 112 m respectively and also find the perpendicular from the opposite angle on the side 112 m.**

**Sol.** Here a = 50 m, b = 78 m, c = 112 m

s = ½(50 + 78 + 112) = 120 m

s – a = 120 – 50 = 70 m

s – b = 120 – 78 = 42 m

s – c = 120 – 112 = 8 m

∴ Area = \(\sqrt{120 \times 70 \times 42 \times 8} = 1680 \; sq.m.\)

∵ Area = ½ base × perpendicular

∴ perpendicular = \(\frac{2Area}{Base} = \frac{1680 \times 2}{112} = 30m.\)

**Example 2**

**2. The base of a triangular field is 880 m and its height 550 m. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of 24.25 per sq. hectometre.**

**Sol.** Area of the field = \(\frac{Base \times Height}{2}\)

= \(\frac{880 \times 550}{2} = 242000 \; sq.m. = 24.20 \; sq.hm.\)

Cost of supplying water to 1 sq. hm = 24.25

∴ Cost of supplying water to the whole field

= 24.20 × 24.25 = 586.85

**SOME IMPORTANT RESULT**

In a rectangle, ^{(Perimeter)2}⁄_{4} = (diagonal)^{2} + 2 × Area

In an isosceles right angled triangle,

Area = 23.3 × (perimeter)^{2}

In a parallelogram,

Area = Diagonal × length of perpendicular on it

If area of circle is decreased by x%, then the radius of circle is decreased by (100 - 10√(100 - x))%

**RECTANGLE**

Perimeter = 2 (l + b)

Area = l × b; where l → length

b → breadth

**SQUARE**

Perimeter = 4 × side = 4a

Area = (side)^{2} = a^{2}; where a → side

**PARALLELOGRAM**

Perimeter = 2 (a + b)

Area = b × h;

where a → breadth

b → base (or length)

h → altitude

**RHOMBUS**

Perimeter = 4a

Area = ½ d_{1} × d_{2}

where a → side and

d_{1} and d_{2} are diagonals.

**IRREGULAR QUADRILATERAL**

Perimeter = p + q + r + s

Area = ½ × d × (h_{1} + h_{2})

**TRAPEZIUM**

Perimeter = a + b + m + n

Area = ½ (a + b)h;

where a and b are two parallel sides;

m and n are two non-parallel sides;

h → perpendicular distance between two parallel sides.

Area of pathways running across the middle of a rectangle

A = a (l + b) – a^{2};

where l → length

b → breadth,

a → width of the pathway.

**PATHWAY SOUTSIDE**

A = (l + 2a) (b + 2a) – lb;

where l → length

b → breadth

a → width of the pathway

**PATHWAYS INSIDE**

A = lb – (l – 2a) (b – 2a);

where l → length

b → breadth

a → width of the pathway

**Example 3**

**3. A 5100 sq.cm trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40 m then find the length of the other parallel side.**

**Sol.** Since, A = ½ (a + b)h;

=> 5100 = ½(40 + x) × 60

=> 170 = 40 + x

∴ other parallel side = 170 – 40 = 130 m

**Example 4**

**4. A rectangular grassy plot is 112 m by 78 m. It has a gravel path 2.5 m wide all round it on the inside. Find the area of the path and the cost of constructing it at 2 per square metre?**

**Sol.** A = lb – (l – 2a) (b – 2a)

= 112 × 78 – (112 – 5) (78 – 5)

= 112 × 78 – 107 × 73 = 8736 – 7811

= 925 sq.m

∴ Cost of construction = rate × area

= 2 × 925 = Rs. 1850

**Example 5**

**5. The perimeter of a rhombus is 146 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus.**

**Sol.** Let ABCD be the rhombus in which AC = 55 cm.

and AB = ^{146}⁄_{4} = 36.5 cm.

Also, AO = ^{55}⁄_{2} = 27.5 cm.

∴ BO = \(\sqrt{\left ( 36.5 \right )^{2} ? \left ( 27.5 \right )^{2}} = 24 \; cm\)

Hence, the other diagonal BD = 48 cm

Now, Area of the rhombus = ½ AC × BD

= ½ × 55 × 48 = 1320 sq.cm.

**Example 6**

**6. Find the area of a quadrilateral piece of ground,one of whose diagonals is 60 m long and the perpendicular from the other two vertices are 38 and 22 m respectively.**

**Sol.** Area = ½ × d × (h_{1} + h_{2})

= ½ × 60 × (38 + 22) = 1800 sq.m.

**CIRCLE**

Perimeter (Circumference) = 2πr = πd

Area = πr^{2};

where r → radius

d → diameter

and π = ^{22}⁄_{7} or 3.14

**SEMICIRCLE**

Perimeter = πr + 2r

Area = ½ × πr^{2}

**SECTOR OF A CIRCLE**

Area of sector OAB = \(\frac{\theta}{360} \times \pi r^{2}\)

Length of an arc (l) = \(\frac{\theta}{360} \times 2\pi r\)

Area of segment = Area of sector – Area of triangle OAB

= \(\frac{\theta}{360^{\cdot}} \times \pi r^{2} - \frac{1}{2} r^{2} sin\theta\)

Perimeter of segment = length of the arc + length of segment

AB = \(\frac{\pi r\theta}{180} + 2r sin\frac{\theta}{2}\)

**RING**

Area of ring = π(R_{2}^{2} - R_{1}^{2})

**Example 7**

**7. A wire is looped in the form of a circle of radius 28 cm. It is re-bent into a square form. Determine the length of a side of the square.**

(a) 44 cm

(b) 45 cm

(c) 46 cm

(d) 48 cm

**Sol.(a)** Length of the wire = Perimeter of the circle

= 2π × 28

= 176 cm^{2}

Side of the square = ^{176}⁄_{4} = 44 cm

**Example 8**

**8.The radius of a circular wheel is 1 ^{3}⁄_{4} m. How many revolutions will it make in travelling 11 km ?**

**Sol.** Distance to be travelled = 11 km = 11000 m

Radius of the wheel = 1^{3}⁄_{4} = ^{7}⁄_{4}

∴ Circumference of the wheel = 2 × ^{22}⁄_{7} × ^{7}⁄_{4} = 11 m

∴ In travelling 11 m, wheel makes 1 revolution.

∴ In travelling 11000 m the wheel makes ^{1}⁄_{11} × 11000 revolutions, i.e., 1000 revolutions.

**Example 9**

**9. The circumference of a circular garden is 1012 m. Find the area of outsider road of 3.5 m width runs around it. Calculate the area of this road and find the cost of travelling the road at 32 per 100 sqm.**

**Solution :** A = πr^{2}, C = 2πr = 1012

=> r = 1012 × ^{1}⁄_{2} × ^{7}⁄_{22} = 161 m

Area of garden = ^{22}⁄_{7} × 161 × 161 = 81466 sqm

Area of the road = area of bigger circle – area of the garden

Now, radius of bigger circle = 161 + 3.5 = ^{329}⁄_{2} m

∴ Area of bigger circle = ^{22}⁄_{7} × ^{329}⁄_{2} × ^{329}⁄_{2} = 85046½ sq.m.

Thus, area of the road = 85046½ – 81466 = 3580½ sqm.

Hence, cost = ^{7161}⁄_{2} × ^{32}⁄_{100} = 1145.76

**Example 10**

**10. There is an equilateral triangle of which each side is 2m. With all the three corners as centres, circles each of radius 1 m are described.**

(i) Calculate the area common to all the circles and the triangle.

(ii) Find the area of the remaining portion of the triangle.

**Sol.**

Area of each sector = \(\frac{\theta}{360} \times \pi r^{2} = \frac{60}{360} \times \frac{22}{7} \times 1 \times 1\)

= \(\frac{1}{6} \times \frac{22}{7} = \frac{11}{21} m^{2}\)

Area of equilateral triangle = \(\frac{\sqrt{3}}{4}a^{2}\)

= \(\frac{\sqrt{3}}{4} \times 2 \times 2 = \sqrt{3} m^{2}\)

(i) Common area = 3 × Area of each sector

= 3 × ^{11}⁄_{21} = ^{11}⁄_{7} = 1.57 m_{2}

(ii) Area of the remaining portion of the triangle = Ar. of equilateral triangle – 3(Ar. of each sector)

√3 – 1.57 = 1.73 – 1.57 = 0.16 m_{2}

**PART-II SOLID FIGURE**

**CUBOID**

A cuboid is a three dimensional box.

Total surface area of a cuboid = 2 (lb + bh + lh)

Volume of the cuboid = lbh

Area of four walls = 2(l + b) × h

**CUBE**

A cube is a cuboid which has all its edges equal.

Total surface area of a cube = 6a^{2}

Volume of the cube = a^{3}

**RIGHT PRISM**

A prism is a solid which can have any polygon at both its ends.

Lateral or curved surface area = Perimeter of base × height

Total surface area = Lateral surface area + 2 (area of the end)

Volume = Area of base × height

**RIGHT CIRCULAR CYLINDER**

It is a solid which has both its ends in the form of a circle.

Lateral surface area = 2πrh

Total surface area = 2πr (r + h)

Volume = πr^{2}h; where r is radius of the base h is height

**PYRAMID**

A pyramid is a solid which can have any polygon at its base and its edges converge to single apex.

Lateral or curved surface area

= ½ (perimeter of base) × slant height

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

Volume = ⅓(area of the base) × height

**RIGHT CIRCULAR CONE**

It is a solid which has a circle as its base and a slanting lateral surface that converges at the apex.

Lateral surface area = πrl

Total surface area = πr (l + r)

Volume = ⅓ πr^{2}h;

where r : radius of the base

h : height

l : slant height

**SPHERE**

It is a solid in the form of a ball with radius r.

Lateral surface area = Total surface area = 4πr^{2}

Volume = ^{4}⁄_{3}πr^{3}

where r is radius.

**HEMISPHERE**

It is a solid half of the sphere.

Lateral surface area = 2πr^{2}

Total surface area = 3πr^{2}

Volume = ⅔πr^{3}

where r is radius

**FRUSTUM OF A CONE**

When a cone cut the left over part is called the frustum of the cone.

Curved surface area = πl(r_{1} + r_{2})

Total surface area = πl(r_{1} + r_{2}) + πr_{1}^{2} + πr_{2}^{2}

where \(l = \sqrt{h^{2} + \left ( r_{1} - r_{2} \right )^{2}}\)

Volume = \(\frac{1}{3}\pi h\left ( r_{1}^{2} + r_{1}r_{2} + r_{2}^{2} \right )\)

**Example 11**

**11. The sum of length, breadth and height of a room is 19 m. The length of the diagonal is 11 m.The cost of painting the total surface area of the room at the rate of 10 per m ^{2} is :**

(a)240

(b)2400

(c)420

(d)4200

**Sol.(b)** Let length, breadth and height of the room be l, b and h, respectively. Then,

l + b + h = 19

and \(\sqrt{l^{2} + b^{2} + h^{2}} = 11\)

=> \(l^{2} + b^{2} + h^{2} = 121\)

Area of the surface to be painted

= 2(lb + bh + hl)

(l + b + h)^{2} = l^{2} + b^{2} + h^{2} + 2 (lb + lh + hl)

=> 2(lb + bh + hl) = (19)^{2} – 121 = 341 – 121 = 240

Surface area of the room = 240 m^{2}.

Cost of painting the required area = 12 × 240 = 2400

**Example 12**

**12. ABCD is a parallelogram. P,Q, R and S are points on sides AB, BC, CD and DA, respectively such that AP = DR. If the area of the rectangle ABCD is 16 cm ^{2}, then the area of the quadrilateral PQRS is:**

(a) 6 cm^{2}

(b) 6.4 cm^{2}

(c) 4 cm^{2}

(d) 8 cm^{2}

**Sol.(d)** Area of the quadrilateral PQRS

= Area of ΔSPR + Area of ΔPQR

= ½ × PR × AP + ½ × PR × PB

= ½ × PR(AP + PB) = ½ × AD × AB

(PR = AD and AP + PB = AB)

= ½ × Area of rectangle ABCD = ½ × 16 = 8 cm^{2}

**Example 13**

**13. A road roller of diameter 1.75 m and length 1 m has to press a ground of area 1100 sq m. How many revolutions does it make ?**

**Sol.** Area covered in one revolution = curved surface area

∴ Number of revolutions = \(\frac{Total \; area \; to \; be \; pressed}{Curved \; surface \; area}\)

= \(\frac{1100}{2\pi rh} = \frac{1100}{2 \times \frac{22}{7} \times \frac{1.75}{2} \times 1}\)

= 200

**Example 14**

**14. The annual rainfall at a place is 43 cm. Find the weight in metric tonnes of the annual rain falling there on a hectare of land, taking the weight of water to be 1 metric tonne to the cubic metre.**

**Sol.** Area of land = 10000 sqm

Volume of rainfall = \(\frac{10000 \times 43}{100} = 4300 \, m^{3}\)

Weight of water = 4300 × 1 m tonnes = 4300 m tonnes

**Example 15**

**15. The height of a bucket is 45 cm. The radii of the two circular ends are 28 cm and 7 cm, respectively. The volume of the bucket is :**

(a) 38610 cm^{3}

(b) 48600 cm^{3}

(c) 48510 cm^{3}

(d) None of these

**Sol.(c)** Here r_{1} = 7 cm, r_{2} = 28 cm and h = 45 cm

Volume of the frustum of a cone

Volume of the bucket = \(\frac{1}{3} \pi h\left ( r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}\right )\)

Hence, the required volume

= \(\frac{1}{3} \times \frac{22}{7} \times 45\left ( 28^{2} + 7^{2} + 28 \times 7 \right ) = 48510 \, cm^{3}\)

**Example 16**

**16. A hollow cylindrical tube open at both ends is made of iron 2 cm thick. If the external diameter be 50 cm and the length of the tube be 140 cm,find the number of cubic cm of iron in it.**

**Sol.** Height = 140 cm

External diameter = 50 cm

∴ External radius = 25 cm

Also, internal radius OA = OB – AB = 25 – 2 = 23 cm

∴ Volume of iron = V_{external} – V_{internal}

= \(\frac{22}{7} \times 140\left ( 25^{2} ? 23^{2}\right) = 42240 \, cu.cm.\)

**Example 17**

**17. A cylindrical bath tub of radius 12 contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?**

(a) 80 cm^{2}

(b) 84 cm^{2}

(c) 104 cm^{2}

(d) 76 cm^{2}

**Sol.(b)** Volume of the spherical ball = volume of the water displaced.

=> ^{4}⁄_{3} πr^{3} = π(12)^{3} × 6.75

=> r^{3} = \(\frac{144 \times 6.75 \times 3}{4} = 729\)

or r = 9 cm.

**Example 18**

**18. A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy. (Use π = 3.14).**

**Sol.** The radius of the hemisphere = ½ × 6 = 3 cm

Now, slant height of cone = \(\sqrt{3^{2} + 4^{2}} = 5 \, cm\)

The surface area of the toy

= Curved surface of the conical portion + curved surface of the hemisphere

= (π × 3 × 5 + 2π × 3^{2}) cm^{2} = 3.14 × 3 (5 + 6) cm^{2} = 103.62 cm^{2}.

**Example 19**

**19. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Re. 1 per dm ^{2}.**

**Sol.**

Let the height of the cylinder be h cm.

Then h + 7 + 7 = 104

=> h = 90

Surface area of the solid

= 2 × curved surface area of hemisphere + curved surface area of the cylinder

= (2 × 2 × ^{22}⁄_{7} × 7 × 7 + 2 × ^{22}⁄_{7} × 7 × 90) cm^{2}

= 616 + 3960 cm^{2} = 4576 cm^{2}

Cost of polishing the surface of the solid

= \(\frac{4576 \times 1}{100} = Rs. 45.76\)

**Example 20**

**20. A regular hexagonal prism has perimeter of its base as 600 cm and height equal to 200 cm. How many litres of petrol can it hold ?Find the weight of petrol if density is 0.8 gm/cc.**

**Sol.** Side of hexagon = \(\frac{Perimeter}{Number \; of \; sides} = \frac{600}{6} = 100 \, cm\)

Area of regular hexagon = ^{3√3}⁄_{2} × 100 × 100 = 25950 sq.cm.

Volume = Base area × height

= 25950 × 200 = 5190000 cu.cm. = 5.19 cu.m.

Weight of petrol = Volume × Density

= 5190000 × 0.8 gm/cc

= 4152000 gm = 4152 kg.

**Example 21**

**21. A right pyramid, 12 cm high, has a square base each side of which is 10 cm. Find the volume of the pyramid.**

**Sol.** Area of the base = 10 × 10 = 100 sq.cm.

Height = 12 cm

∴ Volume of the pyramid = ^{1}⁄_{3} × 100 × 12 = 400 cu.cm.

**Example 22**

**22. Semi-circular lawns are attached to both the edges of a rectangular field measuring 42 m × 35 m. The area of the total field is :**

(a) 3818.5 m^{2}

(b) 8318 m^{2}

(c) 5813 m^{2}

(d) 1358 m^{2}

**Sol.(a)** Area of the field

= 42 × 35 + 2 × ^{1}⁄_{2} × ^{22}⁄_{7} × (21)^{2} + 2 × ^{1}⁄_{2} × ^{22}⁄_{7} × (17.5)^{2}

= 1470 + 1386 + 962.5 = 3818.5 m^{2}

**Example 23**

**23. A frustum of a right circular cone has a diameter of base 10 cm, of top 6 cm, and a height of 5 cm; find the area of its whole surface and volume.**

**Sol.** Here r_{1} = 5 cm, r_{2} = 3 cm and h = 5 cm.

∴ l = \(\sqrt{h^{2} + \left ( r_{1} ? r_{2} \right )^{2}}\)

= \(\sqrt{5^{2} + \left ( 5 ? 3 \right )^{2}} = \sqrt{29} = 5.385 \, cm\)

∴ Whole surface of the frustum

= πl(r_{1} + r_{2}) + πr_{1}^{2} + πr_{2}^{2}

= ^{22}⁄_{7} × 5.385(5 + 3) + ^{22}⁄_{7} × 5^{2} + ^{22}⁄_{7} × 3^{2} = 242. 25 sq.cm.

Volume = ^{πh}⁄_{3}(r_{1}^{2} + r_{1}r_{2} + r_{2}^{2})

= ^{22}⁄_{7} × ^{5}⁄_{3}[5^{2} + 5 × 3 + 3^{2}] = 256.67 cu.cm.

**Example 24**

**24. A cylinder is circumscribed about a hemisphere and a cone is inscribed in the cylinder so as to have its vertex at the centre of one end, and the other end as its base.The volume of the cylinder, hemisphere and the cone are,respectively in the ratio :**

(a) 2: 3 : 2

(b) 3 : 2 : 1

(c) 3 : 1 : 2

(d) 1 : 2 : 3

**Sol.(b)** We have,

radius of the hemisphere = radius of the cone

= height of the cone

= height of the cylinder = r (say)

Then, ratio of the volumes of cylinder, hemisphere and cone

= πr^{3} : ^{2}⁄_{3} πr^{3} : ^{1}⁄_{3} πr^{3} = 1 : ^{2}⁄_{3} : ^{1}⁄_{3} = 3 : 2 : 1