Question

A court yard yard, 80 m long and 45 m wide is to be paved with bricks . If each brick is of lenght 50 cm and width 20 cm . Find the number of bricks required​

Answer 1

Answer:

Given:-

A court yard, 80 m long and 45 m wide is to be paved with bricks . If each brick is of length 50 cm and width 20 cm .

To find:-

The number of bricks required

Solution:-

First we need to convert all units to same measure(i.e.the measures of court yard)

Court yard dimensions measures:-

Length=80m

1m=100cm

80m=80×100cm

=8000cm

Breadth(wide) =45m

1m=100cm

45m=45×100cm

=4500cm

Second after covering we need to find area for court yard and area of each brick

I) Area of court yard=length×breadth

on putting values we get,

=8000cm×4500cm

$= 36,000,000 {cm}^{2}$

ii) Area of each brick=length×breadth(width)

on putting values we get,

=50cm×20cm

$= 1000 {cm}^{2}$

So now,

Number of bricks required=

$\frac{area \: of \: court \: yard}{area \: of \: each \: brick}$

on substituting values we get,

=

$\frac{36000000 {cm}^{2} }{1000 {cm}^{2} }$

By cancelling we get,

=36000

Therefore 36000 bricks required

Additional information:-

$\huge\mathfrak\blue{More \: formulas \: of \: areas:-}$

Area of square=side×side

Area of rectangle=length×breadth

Area of Parallelogram=Base×Height

Area of triangle=

$\frac{1}{2} \times base \times height$

Area of trapezium=

$\frac{a(base) + b(base)}{2} \times height$

Area of rhombus=

$\frac{1}{2} \times d_{1} \times d_{2}$

Area of ∆scalene=

$\sqrt{s(s - a)(s - b)(s - c)}$

Area of circle=πr^2

Area of semicircle=

$\frac{\pi \: {r}^{2} }{2}$

r=radius

Step-by-step explanation:

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