# Mensuration Problems

#### Chapter 35

5 Steps - 3 Clicks

# Mensuration Problems

### Introduction

Mensuration Problems deals with the area, perimeter, diagonal, height, circumference etc. of figures like square, rectangle, parallelogram, rhombus, trapezium, triangle, equilateral triangle, cube, cuboid, cylinder, cone, sphere, hemi-sphere, prism, circle, semi circle, sector.

Mensuration: It is a branch of mathematics which deals with the lengths of lines, areas of surfaces and volumes of solids.

Plane Mensuration: It deals with the sides, perimeters and areas of plane figures of different shapes.

Solid Mensuration: It deals with the areas and volumes of solid objects

### Methods

Square: It is a regular quadrilateral which consists four equal sides and four equal angles. Every angle is 90 degrees.

Here AB = BC = CD = AD = 5 m = a (Let)

1. Perimeter of square = 4 x sides = 4a
= 4 x 5 = 20m

2. Area of a square = $$(sides)^{2}$$
= $$a^{2}$$ = $$(5)^{2}$$ = 25 sq. m

3. Side of a square = $$\sqrt{area}$$ = $$\sqrt{25}$$ = 5m or,
$$\frac{Perimeter}{4}$$ = $$\frac{20}{4}$$ = 5m

4. Diagonal of a square = $$\sqrt{2}$$ x side = $$\sqrt{2}$$ a
= $$\sqrt{2}$$ x 5 = 5$$\sqrt{2}$$ m

5. Side of a square = $$\frac{diagonal}{\sqrt{2}}$$ = $$\frac{5\sqrt{2}}{\sqrt{2}}$$ = 5m

Rectangle: A rectangle is a plane, whose opposite sides are equal and diagonals are equal. Each angle is equal to 90˚.

Here AB = CD; length l = 4m

AD = BC; breadth b = 3m

1. Perimeter of a rectangle = 2(length + breadth)

= 2(l + b)

= 2(4 + 3) = 14 m

2. Area of rectangle = length x breadth = l x b = 4 x 3

= 12 $$m^{2}$$

3. Length of a rectangle : $$\frac{area}{breadth}$$ = $$\frac{A}{b}$$ = $$\frac{12}{3}$$ = 4 m

Or, [$$\frac{perimeter}{2}$$ – breadth] = ($$\frac{14}{2}$$ – 3) = 4m

Breath of a rectangle : $$\frac{area}{length}$$ = $$\frac{A}{l}$$ = $$\frac{12}{4}$$ = 3 m

Or, [$$\frac{perimeter}{2}$$ – length] = ($$\frac{14}{2}$$ – 4) = 3 m

Parallelogram: It is a rectilinear figure with opposite sides parallel.

(i) Area of parallelogram = base x height

= b x h

= 8 x 5 = 40 sq.cm.

(ii) Perimeter of a parallelogram = 2(AB + BC)

= 2( 8 + 5 ) = 26 cm

Rhombus: It is a simple quadrilateral in which all sides have same length. A rhombus is often called as ‘Diamond’.

i) Area of rhombus = $$\frac{1}{2}$$ x (product of diagonals)

= $$\frac{1}{2} (d_{1} . d_{2})$$ = $$\frac{1}{2}$$ x 8 x 6 = 24$$cm^{2}$$

ii) Perimeter of rhombus = 4 x side = 4a

here AB = BC = CD = AD = 4a

AC = $$d_{1}$$, BD = $$d_{2}$$

Trapezium: It is a quadrilateral with one of pair of sides parallel.

Area if a trapezium = $$\frac{1}{2}$$ x ( sum of parallel sides ) x height

= $$\frac{1}{2}$$ x ( a + b ) x h

= $$\frac{1}{2}$$ x (15 + 17) x 10

= $$\frac{1}{2}$$ x 32 x 10 = 160 $$cm^{2}$$

Triangle: It is a polygon with three edges and three vertices.

1. Area of triangle = $$\frac{1}{2}$$ x base x height

= $$\frac{1}{2}$$ x 15 x 12 = 90 sq. cm

here AD = 12 cm = height, BC = 15 cm = base

2. Semi perimeter of a triangle

S = $$\frac{a + b + c}{2}$$ = $$\frac{10 + 8 + 6}{2}$$ = 12 cm

here BC = a, AC = b, AB = c

3. Area of triangle = $$\sqrt{s(s – a)(s – b)(s – c)}$$

Where a = 10cm, b = 8cm, c = 6cm, s = 12cm

= $$\sqrt{12(12 – 10)(12 – 8)(12 – 6)}$$

= $$\sqrt{12 \times 2 \times 4 \times 6}$$

= 24 $$cm^{2}$$

4. Perimeter of a triangle = 2s = (a + b + c)

= 10 + 8 + 6 = 24 cm

Equilateral triangle: It is defined as a polygon with three equal sides. Each and every angle in it is 60 degrees.

1. Area of an equilateral triangle = $$\frac{\sqrt{3}}{4}$$ x $$(side)^{2}$$

= $$\frac{\sqrt{3}}{4}$$ x $$(4\sqrt{3})^{2}$$

= $$\frac{\sqrt{3}}{4}$$ x 48 = $$12\sqrt{3}$$ $$cm^{2}$$

2. Height of an equilateral triangle = $$\frac{\sqrt{3}}{2}$$ x $$(side)^{2}$$

= $$\frac{\sqrt{3}}{2}$$ x $$4\sqrt{3}$$

= 6cm

3. Perimeter of an equilateral triangle = 3 x (side)

= 3 x $$4\sqrt{3}$$ = $$12\sqrt{3}$$ cm

Cube: A cube is a three dimensional figure and has six square faces which meet each other at right angles. It has eight vertices and twelve edges.

1. Volume of cube = $$(side)^{3}$$

= $$12^{3}$$

= 1728 cubic cm

Cube: All sides are equal = 12 cm

2. Side of a cube = $$\sqrt[3]{Volume}$$

= $$\sqrt[3]{1728}$$

= 12 cm

3. Diagonal of cube = $$\sqrt{3}$$ x (Side) = $$\sqrt{3}$$ x 12 = 12 $$\sqrt{3}$$ cm

4. Total surface area of a cube = 6 x $$(Side)^{2}$$ = 6 x $$(12)^{2}$$ = 864 sq.cm

Cuboid (Rectangular parallelopiped): A solid body having six rectangular faces, is called cuboid. (or) A parallelopiped whose faces are rectangles is called rectangular parallelopiped or cuboid.

1. Total surface area of cuboid = 2 (lb + bh + hl) sq. unit

Here l = length, b = breadth, h = height

= 2(12 x 8 + 8 x 6 + 6 x 12)

= 2(96 + 48 + 72) = 2 x 216 = 432 sq. cm.

2. Volume of a cuboid = (length × breadth × height) = lbh

= 12 × 8 × 6 = 576 cuboic cm

3. Diagonal of a cuboid = $$\sqrt{l^{2} + b^{2} + h^{2}}$$ = $$\sqrt{12^{2} + 8^{2} + 6^{2}}$$

= $$\sqrt{144 + 64 + 36}$$ = $$\sqrt{244}$$ = $$2\sqrt{61}$$ cm.

4. Length of cuboid = $$\frac{Volume}{Breadth \times Height}$$ = $$\frac{v}{b \times h}$$

5. Breadth of cuboid = $$\frac{Volume}{Length \times Height}$$ = $$\frac{v}{l \times h}$$

6. Height of cuboid = $$\frac{Volume}{length \times breadth}$$ = $$\frac{v}{l \times b}$$

Cylinder: A solid geometrical figure with straight parallel sides and a circular or oval cross section is called cylinder.

1. Area of curved surface = (perimeter of base) x height

= $$2 \pi r h$$ sq. unit

= 2 x $$\frac{22}{7}$$ x 7 x 15 = 660 sq. cm

2. Total surface area = area of circular ends + curved surface area

= $$2 \pi r^{2}$$ + $$2 \pi r(r + h)$$ sq. unit

= 2 x $$\frac{22}{7}$$ x 7(15 + 7)

= 2 x 22 x 22

= 968 sq. cm.

3. Volume = (area of base) x height

= $$(\pi r^{2})$$ x h = $$\pi r^{2}h$$

= $$\frac{22}{7}$$ x 7 x 7 x 15 = 2310 cubic cm.

4. Volume of a hollow cylinder = $$\pi R^{2}h – \pi r^{2}h$$

= ($$\pi h (R^{2} – r^{2})$$) = $$\pi h(R + r)(R – r)$$

= $$\pi \times height \times (sum of radii)(difference of radii)$$

Here R, r are outer and inner radii respectively and h is the height.

Cone: A solid (3-dimensional) object with a circular flat base joined to a curved side that ends in an apex point is called cone.

1. In right angled $$\bigtriangleup$$ OAC, we have

$$l^{2}$$ = $$h^{2}$$ + $$r^{2}$$

(Here r = 35 cm, l = 37 cm, h = 12 cm)

Or, l = $$\sqrt{h^{2} + r^{2}}$$

h = $$\sqrt{l^{2} – r^{2}}$$, r = $$\sqrt{l^{2} – h^{2}}$$

Where l = slant height, h = height, r = radius of base

2. Curved surface area = $$\frac{1}{2}$$ x (perimeter of base) x slant height

= $$\frac{1}{2}$$ x $$2 \pi r$$ x l = $$\pi r l$$ sq. unit

= $$\frac{22}{7}$$ x 35 x 37 = 4070 sq. cm

3. Total surface area S = area of circular base + curved surface area = $$(\pi r^{2} + \pi r l)$$ = $$\pi r (r + l)$$ sq. unit

= $$\frac{22}{7}$$ x 35 (37 + 35) = 7920 sq. cm

4. Volume of cone = $$\frac{1}{3}$$ (area of base) x height

= $$\frac{1}{3} (\pi r^{2})$$ x h = $$\frac{1}{3} \pi r^{2} h$$ cubic unit

= $$\frac{1}{3}$$ x $$\frac{22}{7}$$ x 35 x 35 x 12

= 15400 cubic cm

Frustum of Cone:

5. Volume of frustum = $$\frac{1}{3} \pi h (R^{2} + r^{2} + Rr)$$ cubic unit

6. Lateral surface = $$\pi l (R + r)$$

Where $$l^{2}$$ = $$h^{2}$$ + $$(R – r)^{2}$$

7. Total surface area = $$\pi [R^{2} + r^{2} + l(R + r)]$$

R, r be the radius of base and top the frustum

ABB ‘A’ h and l be the vertical height and slant

Height respectively.

Sphere: A 3-dimensional object shaped like a ball is called sphere and every point on the surface is the same distance from the centre.

1. Surface area = $$4 \pi r^{2}$$

= 4 x $$\frac{22}{7}$$ x $$(10.5)^{2}$$ = 1386 sq. cm

Here, d = 21 cm

Therefore, r = 10.5 cm

2. Radius of sphere = $$\sqrt{ \frac{surface area}{4 \pi}}$$ = $$\sqrt{ \frac{1386 \times 7}{4 \times 22}}$$ = 10.5 cm

3. Diameter of sphere = $$\sqrt{ \frac{surface}{4 \pi}}$$ = $$\sqrt{ \frac{surface}{4 \pi}}$$ = $$\sqrt{ \frac{1386 /times 7}{22}}$$ = 21 cm

4. Volume of sphere V = $$\frac{4}{3} \pi r^{3}$$ = $$\frac{4}{3} \pi (\frac{d}{2})^{3}$$ = $$\frac{1}{6} \pi d^{3}$$

= $$\frac{1}{6}$$ x $$\frac{22}{7}$$ x 21 x 21 x 21 = 4831 cubic cm

5. Radius of sphere = $$\sqrt{ \frac{3}{4 \pi} \times volume of shpere}$$

6. Diameter = $$\sqrt[3]{ \frac{6 \times v}{\pi}}$$

7. Volume of spherical ring = $$\frac{4}{3} \pi (R^{3} – r^{3}$$

Hemisphere: In geometry, hemisphere is an exact half of sphere.

8. Curved surface of hemisphere = 2 $$\pi r^{2}$$

9. Volume of hemisphere = $$\frac{2}{3} \pi r^{3}$$

10. Total surface area of hemisphere = $$3 \pi r^{2}$$

Circle: It is defined as, if a straight line is bent until its ends join. It is a simple closed curve and the distance between any of the points and centre of circle is called radius.

1. Circumference of a circle = $$\pi$$ x diameter

= $$\pi$$ x $$2r$$ = $$2 \pi r$$

= 2 x $$\frac{22}{7}$$ x 42 = 264 cm

2. Radius of a circle = $$\frac{circumference}{2 \pi}$$ = $$\frac{264 \times 7}{2 \times 22}$$ = 42 cm

3. Area of a circle = $$\pi \times r^{2}$$ = $$\frac{22}{7}$$ x $$42^{2}$$ = $$\frac{22}{7}$$ x 42 x 42 = 5544 $$cm^{2}$$

4. Radius of a circle = $$\sqrt{\frac{area}{\pi}}$$

= $$\sqrt{\frac{5544}{22} \times 7}$$ = $$\sqrt{1764}$$ = 42cm

Semi-circle: It is a two-dimensional geometric shape that also includes the diameter segment from one end of the arc to the other as well as all the interior points.

5. Area of a semi circle = $$\frac{1}{2} \pi r^{2}$$ = $$\frac{1}{8} \pi d^{2}$$

= $$\frac{1}{2}$$ x $$\frac{22}{7}$$ x $$42^{2}$$ = $$2772 cm^{2}$$

6. Circumference of semi circle = $$\frac{22}{7}$$ x 42 = 132 cm

7. Perimeter of semi circle = $$(\pi r + 2r)$$ = $$(\pi + 2)r$$ = $$(\pi + 2)\frac{d}{2}$$

8. Area of sector OAB = $$\frac{x}{360}$$ x $$\pi r^{2}$$

(x being the central angle)

= $$\frac{30^{0}}{360^{0}}$$ x $$\frac{22}{7}$$ x 3.5 x 3.5 = 3.21 sq. m.

9. Central angle by arc AB = $$360^{0}$$ x $$\frac{area of OAB}{are of circle}$$

= $$360^{0}$$ x $$\frac{3.21}{\frac{22}{7} \times 3.5 x 3.5}$$ = $$\frac{360 \times 321}{22 \times 35 \times 5}$$ = $$30^{0}$$(approx)

10. Radius of circle = $$\sqrt{\frac{360^{0}}{central \ angle \ by \ arc} \times \frac{area \ of \ OAB}{\pi}}$$

= $$\sqrt{\frac{360^{0}}{30^{0}} \times \frac{3.21}{\frac{22}{7}}}$$ = $$\sqrt{\frac{134.82}{11}}$$ = $$\sqrt{12.23}$$ = 3.5 m.

11. Area of ring = difference of the area of two circle

= $$\pi R^{2}$$ – $$\pi r^{2}$$ = ($$R^{2} – r^{2}$$)

= $$\pi$$(R + r)(R – r)

= (sum of radius)(diff. of radius)

= $$\frac{22}{7}$$ x (4 + 3)(4 – 3) = $$\frac{22}{7}$$ x 7 x 1

= 22 sq. cm.

### Formulae

1. Square:
Area = $$(side)^2$$
Perimeter = 4 x side
Diagonal = $$\sqrt{2(side)^2}$$

2. Rectangle:
Area = Length x Breadth
Perimeter = 2(Length x Breadth)
Diagonal = $$\sqrt{l^2 + b^2}$$

3. Parallelogram:
Area = Base x Height
Perimeter = 2(Length x Breadth)
Base = $$\frac{Area}{Height}$$

4. Rhombus:
Area = $$\frac{1}{2}* d_{1} d_{2}$$
Where, $$d_{1} \ and \ d_{2}$$ are diagonals.
Perimeter = 4 x side
Diagonal = $$\frac{2 * area}{other diagonal}$$

5. Trapezium:
Area = $$\frac{1}{2}(a + b) * h$$
Perimeter = Sum of the sides

6. Triangle:
Area = $$\frac{1}{2} * Base * Height$$ (or) $$\sqrt{s(s – a)(s – b)(s – c)}$$
Where a, b, c are sides of triangle and s = $$\frac{a + b + c}{2}$$
Perimeter = Sum of the sides

7. Equilateral Triangle:
Area = $$\frac{\sqrt{3}}{2}(side)^2$$
Perimeter = 3(Side)

8. Cube:
Let each edge of a cube be of length $$a$$. Then,
Volume = $$a^3$$ cubic units
Surface area = 6$$a^2$$ sq. units
Diagonal = $$\sqrt{3}a$$units

9. Cuboid:
Let l – length, b – breadth, h – height. Then,
Volume = (l x b x h) cubic units
Surface area = 2(lb + bh + lh) sq. units
Diagonal = $$\sqrt{l^2 + b^2 + h^2}$$ units

10. Cylinder:
Let radius of base = r and height( or length) = h. Then,
Volume = $$\pi r^2 h$$ cubic units
Curved Surface area = 2$$\pi r h$$ sq. units
Total surface area = 2($$\pi rh + 2\pi r^2$$) = 2$$\pi r(h + r)$$sq. units

11. Cone:
Let radius of base = r and height = h. Then,
Slant height, l = $$\sqrt{h^2 + r^2}$$ units
Volume = $$\frac{1}{3}\pi r^2 h$$ cubic units
Curved surface area = $$\pi r l$$ sq. units
Total surface area = $$\pi r l + \pi r^2$$ sq. units

12. Sphere:
Let radius of the sphere be r. Then,
Volume = ($$\frac{4}{3} \pi r^3$$) cubic units
Surface area = $$\pi r^2$$ sq. units

13. Hemisphere:
Let the radius of a hemisphere be $$r$$.
Volume = $$\frac{2}{3} \pi r^3$$ cubic units.
Curved surface area = 2$$\pi r^2$$ sq. units.
Total surface area = 3$$\pi r^2$$ sq. units.

14. Prism:
Volume of right prism = (Area of the base * height)cu. units
Lateral surface area of a right prism = (perimeter of the base * height)sq. units
Total surface area of a right prism = lateral area + 2(area of one base)sq. units

15. Circle:
Area = $$\pi r^2$$
where, r = radius
Circumference = $$2 \pi r$$

16. Semi-circle:
Area = $$\frac{1}{2} \pi r^2$$
Circumference = $$\pi r + 2r$$

17. Sector:
Area = $$\frac{\theta}{360} * \pi r^2$$
Circumference = l + 2r[l = $$\frac{\theta}{360} * 2\pi r$$]
where l = length of arc.

### Samples

1. A wheel makes 1000 revolutions in covering a distance of 88 km. Find the radius of the wheel?

Solution:

Given that,

Number of revolutions made by wheel = 1000

Distance = 88 km

So, distance covered in 1 revolution = $$\frac{88 * 1000}{1000}$$ m = 88 m

Therefore, $$2 \pi r$$ = 88

⇒ 2 x $$\frac{22}{7}$$ x r = 88

⇒ r = 88 x $$\frac{7}{44}$$

⇒ r = 14 m

Therefore, radius of the wheel = 14m.

2. Expenditure incurred in cultivating a square field at the rate of Rs. 170 per hectare is Rs. 680. What would be the cost of fencing the field at the rate of Rs. 3 per metre.

Solution:

Given that,

Expenditure per hectare = Rs. 680

Area = $$\frac{680}{170}$$ = 4 hectares = 4 x 10000 sq.m.

So, Sides of the field = $$\sqrt{40000}$$ = 200 m.

Perimeter of the field = 4 x side = 4 x 200 = 800 m.

Therefore, Cost of putting the fence around it is

= 800 x 3 = Rs. 2400.

3. A man walking at the rate of 6 km per hour crosses a square field diagonally in 9 seconds. Find the area of the field ?

Solution:

Given that,

Time = 9 seconds

Rate = 6 km

Now, Distance covered in $$\frac{6 * 1000}{3600} * 9$$ = 15 m

Diagonal of the square field = 15 m.

Hence, Area of the square field = $$\frac{(15)^2}{2}$$ = $$\frac{225}{2}$$ = 112.5 sq.m.

4. Find the diagonal of a cuboid whose dimensions are 22 cm, 12 cm, and 7.5 cm?

Solution:

Given that,

Length = 22 cm

Breadth = 12 cm

Height = 7.5 cm

Now,

Diagonal of the cuboid = $$\sqrt{l^2 + b^2 + h^2}$$ = $$\sqrt{(22)^2 + (12)^2 + (7.5)^2}$$ = $$\sqrt{684.25}$$ = 26.15 cm.

Therefore, Diagonal of the cuboid = 26.15 cm.

5. The ratio between the curved surface area and the total surface area of a right circular cylinder is 1 : 2. If the total surface area is 616$${cm}^2$$, then find the volume of the cylinder?

Solution:

GGiven that,

Ratio between the curved surface area and total surface area of a right circular cylinder is 1 : 2

Total surface area = 616$${cm}^2$$

Now, $$\frac{2 \pi r h}{2 \pi r (h + r)}$$ = $$\frac{1}{2}$$

⇒ $$\frac{h}{h + r}$$ = $$\frac{1}{2}$$

⇒ h = r

Hence, total surface area = $$2 \pi r (h + r)$$ = 4 $$\pi r^2$$ (since, h = r)

Therefore, 4 $$\pi r^2$$ = 616

⇒ $$r^2$$ = 616 x $$\frac{1}{4}$$ x $$\frac{7}{22}$$ = 49

⇒ $$r$$ = 7

Thus, h = 7 cm and r = 7 cm

Therefore, Volume = $$\pi r^2 h$$

= $$\frac{22}{7}$$ x 7 x 7 x 7 = 1078 $${cm}^3$$.