 # 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.

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

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.

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.

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:
Diagonal = $$\sqrt{l^2 + b^2}$$

3. Parallelogram:
Area = Base x Height
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 = $$\pi r^2$$
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 $${cm}^3$$.