# MESURATION

Mensuration

Area of Triangle

Area of the Area = $\frac{1}{2}{\rm{\: }}$base*height

If a , b c be the                 the sides of the traingle , then perimeter of the triangle = a + b+ c

${\rm{\: }}$Semi – perimeter=$\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}$

Area of the triangle = $\sqrt {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}$

Area of the equilateral triangle = $\frac{{\sqrt 3 }}{4}{{\rm{a}}^2}$

Rectangle:

The perimeter of rectangle = 2(l + b).
Area of rectangle = l × b; (l and b are the length and breadth of rectangle)
Diagonal of rectangle = $\sqrt {{{\rm{l}}^2}{\rm{\: }} + {\rm{\: }}{{\rm{b}}^2}}$

Square:

Perimeter of square = 4 × S.
Area of square = S × S.
Diagonal of square = S√2; (S is the side of square)

Parallelogram:

Perimeter of parallelogram = 2 (sum of adjacent sides)
Area of parallelogram = base × height

Rhombus:

Area of rhombus = base × height
Area of rhombus = $\frac{1}{2}{\rm{\: }}$× length of one diagonal × length of other diagonal
Perimeter of rhombus = 4 × side

Trapezium:

Area of trapezium =$\frac{1}{2}$ (sum of parallel sides) × (perpendicular distance between them)
= $\frac{1}{2}$ (p1 + p2) × h (p1, pare two parallel sides)

Circle:

Circumference of circle = 2πr
= πd
Where, π = 3.14 or π = 22/7
r is the radius of circle
d is the diameter of circle
Area of circle = πr2
Area of ring = Area of outer circle - Area of inner circle.

Area of the pathways

Area of the path running outside a rectangle

Let PQRS is a rectangle and a path of uniform thickness of d’ running outside the rectangle.

Area of the path running outside a rectangle = 2d(l + b +2d)

In the case of the square field l = b

Area of the path running outside a square = 4d(l + d)

Area of the path running inside a rectangle

Let ABCD is a rectangle and path of thickness dis running inside a rectangle

Area of the path running inside a rectangle = 2d (l + b-2d)

In the case of the square field l = b

Area of the path running inside a square = 4d(l- d)

Area of paths’s crossing each other perpendicularly

Area of                 crossing paths =d(l+ b-d)

In the case of the square field l = b

Area of                 crossing paths = d(2l –d)

Area of path running outside the circular field

Let r be the radius of the circle.

Area of the path = area of the PQR - Area of the ABC

Area of the path of thickness  d ‘running outside the circular field = π d (2r +d)

Area of path running inside a circular field

Let r be the radius of the circle.

Area of the path = area of the ABC - Area of the PQR

Area of the path of thickness ` d ‘running outside the circular field = π d (2r - d)

Area, cost and quantities

C = A * R

N = $\frac{{\rm{A}}}{{\rm{a}}}{\rm{\: }}$

C = N* R

A = area of paths

R = Rate

C = Total cost

N = number of bricks required to pave the path

A= surface area of each brick

Area of 4 walls floor and ceiling =2h(l+b)+2lb

Area of the 4 walls excluding a window and a door =2h (l +b)- l1h1- l2h2

Where h= height of wall

L = length of wall

b = breadth of wall

l1= length of window

l2= lengthof door

h1=heightof window

h2=heightof door

Area and volume of a solid

Solids have definite shape and size and occupy the space

Area of cuboid= 2(lb+bh+lh)

Volume of cubiod = l *b* h

Area of the volume of cube = l3

Area of total surface of the cube = 6l2

Volume of a prism = area of cross section *height

Lateral surface area of prism = perimeter of cross –section * length

Total surface area of a prism = lateral surface area + 2*area of the cross –section

Estimation of number of bricks and cost required for building.

Number of bricks= $\frac{{{\rm{volume\: of\: wall}}}}{{{\rm{volume\: of\: brick\: }}}}$ = $\frac{{\rm{V}}}{{\rm{v}}}$

Cost= N*R

R=$\frac{{\rm{N}}}{{\rm{C}}}$

Examples

1.

The length and breadth of the rectangular room are 12cm and 10cm respectively How many pieces of carpet Is Rs900 .Find the cost of the carpeting the floor / If the cost of each piece of carpet is Rs 900 .Find the cost of carpeting the floor .

Soln

Length of room (l) = 12m

Breadth of room (b) = 10m

Length of the carpet (l1) = 4m

Breadth of the carpet (b1) = 2.5m

Area of room = l * b = 12 * 10 = 120m

Now, Area of a piece of carpet (A) = l1 * b1 = 4 * 2.5 = 10m2.

Now, Pieces of carpet = $\frac{{120}}{{10}}$ = 12 pieces.

So, 1 piece costs Rs. 900

So, 12 piece costs Rs. 900 * 12 = Rs. 10,800.

2.

The cost of constructing a path 5mboard inside the boundary of a square lawn Rs 36.25per sq meter isRs90625. What is the cost of covering the empty space with turfs at the rate of Rs 20 per sqmeter?

Soln

Width (d) = 5m.

Now, Area of the broad = $\frac{{{\rm{Rs}}.90625}}{{{\rm{Rs}}.36.25}}$= 2,500 m2.

Now, 4d(l – d) = 2500

Or, 4d(l – 5) = 2500

or, 4 * 5(l – 5) = 2500

So, l = 130m

Now, Area of square = l2 = 16,900

Now, area of empty space = 16,900 – 2,500 = 14,400

Now, cost of covering the empty space = 14,400 * Rs. 20 = Rs. 2, 88,000.

3.

A room is 10m long and 8 m wide .It has two windows each of 2m × 2mand a door of size 1.5m × 4m .If the cost of plastering its wall s at Rs 20pper sqmeteris Rs 3320. Find the height of room

Here, l = 10m, b = 8m

l1 = 2m, b1 = 2m

l2 = 1.5m, b2 = 4m.

h=?

Now, Area of the room = $\frac{{{\rm{Total\: cost}}}}{{{\rm{Rate}}}}$ = $\frac{{3320}}{{20}}$

= 166m2

Now, Area of 4 walls = 166m2

Or, 2h (l + b) – 2l1b1 – l2b2 = 166

Or, 2h (10 + 8) – (2 * 2 * 2) – (1.5 * 4) = 166

Or, 2h * 18 – 8 – 6 = 166

Or, 36h – 14 = 166

Or, 36h = 180

So, h = 5m

So, the height of the room is 5m.

4.

Find the area of the cross –section , volume ,lateral surface area , total surface areaof the given figure .

Soln

Area of cross section = 8 * 8 + 8 * 4 = 96 cm2

Lateral surface area = Perimeter of base * height = (16 + 8 + 8 + 4) * 6 = 264 cm2

Volume = Area of cross-section * height = 96 * 6 = 576 cm2

Total surface area = Later surface area + 2 * area of cross-section = 264 + 2 * 96 = 456 cm2

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