Question

The diameter of a wheel of a car is 42cm. Find the distance covered by the car during
the time in which the wheel makes 1000 revolutions.
HELP FAST PLEASE

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Diameter of wheel of a car = 42 cm

So,

Radius of wheel of a car, r = 21 cm

Now, we know

Distance covered by wheel in 1 revolution is equals to circumference of the wheel

So,

$\rm \: Distance \: covered \: in \: 1 \: revolution \: = \: 2\pi \: r$

$\rm \: = \: 2 \times \dfrac{22}{7} \times 21$

$\rm \: = \: 2 \times 22 \times 3$

$\rm \: = \: 132 \: cm$

Now,

$\rm \: Distance \: covered \: in \: 1 \: revolution \: = \: 132 \: cm$

So,

$\rm \: Distance \: covered \: in \: 1000 \: revolution \: = \: 132 \times 1000$

$\rm \: Distance \: covered \: in \: 1000 \: revolution \: = \: 132000 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

Information provided with us:

➪ The diameter of a wheel of a car .

$\implies \sf \:42\: cm$

What we have to calculate༄

➪ The required distance covered by the car during the time in which the wheel makes 1000 revolutions.

✰Note :

➪ Calculate first radius of wheel of car

✰Formula used ༄

$\clubsuit \: \rm \: Radius \: of \:wheel \: of \: car$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\orange{ \bigg(\dfrac{Diameter \: of \: wheel \: of \: car}{2}\bigg)}}}\: \: \: \bigstar$

✰Now ༄

➡ Substitute the given values in above formula and solve

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\green{ \bigg(\dfrac{42}{2}\bigg)}}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\red{21 \: cm}}}\: \: \: \bigstar$

$\rm \therefore \: Thus \: radius \: of \: wheel\:of \: car$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{21 \:cm}}}\: \: \: \bigstar$

✰Now ༄

➪ Find circumference of wheel

✰Formula used ༄

$\clubsuit \: \rm \: Circumference \: of \:wheel \: of \: car$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\orange{2\pi \: r}}}\: \: \: \bigstar$

✰Where ༄

➡ Value of π is taken as 22/7

➡ r stand for radius of wheel of car

✰Now ༄

➡ Substitute the given values in above formula and solve

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\green{ \bigg(2 \times \dfrac{22}{7} \times 21\bigg) \: cm}}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\pink{2 \times 22 \times 3 }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\red{132\: cm}}}\: \: \: \bigstar$

$\rm \therefore \: Thus \: circumference \: of \: wheel\:of \: car$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{132 \:cm}}}\: \: \: \bigstar$

✰Here ༄

• ➪ Covert 132 cm to m ,by dividing 132 by 100 we get

$\implies \sf \: \dfrac{132}{100} \: = 1.32 \: m$

✰Note ༄

• ➪ In 1 revolution the wheel of car covers a distance equal to its Circumference

✰Therefore ༄

$\rm \therefore \: The \: distance \: covered \:by\: wheel\:of \: car\: in\:one \: revolution$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{1.32 \:m}}}\: \: \: \bigstar$

$\rm \: Now \: the \: distance \: covered \: by\: wheel\:of \: car\: in\:1000\: revolution$

$\rm \implies \: (1.32 \times 1000)m$

$\rm \implies \:1,320 \: m$

✰We know that ༄

$\rm \implies \: 1 \: km = 1000 \:m$

✰So ༄

• ➪ By dividing 1,320 m by 1000 we get

$\rm \implies \:1.32 \: km$

✰Thus ༄

$\rm \therefore \: The \: distance \: covered \: by\: wheel\:of \: car\: in\:1000\: revolution \: is$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\orange{1.32 \: km}}}\: \: \: \bigstar$