 # Mensuration Problems

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# 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
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.

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

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

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.

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.

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.

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 [latex ]\bigtriangleup[/latex] 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.

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.

### 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
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 ${cm}^3$.