Math, asked by imsanjanadhakad, 2 months ago

a rectangular field has its length and breadth in the ratio 6:5 respectively. a man riding bicycle complete one lap of the field along its perimeter at the speed of 12 kmph in 1.5m. what is the area of the field?​

Answers

Answered by george0096
3

Answer:

  • Area of the rectangular field is 5467.5 m².

Step-by-step explanation:

Given that:

  • A rectangular field has its length and breadth in the ratio 6 : 5.
  • A man riding bicycle completes one lap of the field along its perimeter at the speed of 12 km/h in 1.5 minutes.

To Find:

  • Area of the field.

Solution:

Converting 12 km/h to m/s:

To convert km/h to m/s we have to divide it by 18/5.

\sf{\longmapsto12\;km/h=\left(12\div\dfrac{18}{5}\right)m/s}

\sf{=\left(12\times\dfrac{5}{18}\right)m/s}}

Reducing the numbers,

\sf{=\left(2\times\dfrac{5}{3}\right)m/s}}

Multiplying the numbers,

\sf{=\dfrac{10}{3}\;m/s}}

Dividing the numbers,

\sf{=3.3\;m/s}}

Converting 1.5 minutes to seconds:

To convert minutes into seconds we have to multiply it by 60.

\sf{\longmapsto1.5\;minutes=(1.5\times60)\;seconds}}

\sf{=90\;seconds}}

Finding distance covered by the man:

As we know that,

  • Distance = Time × Speed

Substituting the values,

\sf{\longmapsto(3.3\times90)m}

\sf{\longmapsto297\;m}

Hence,

  • Distance covered by the man = 297 m

Therefore,

  • Perimeter of the field = 297 m

Let us assume:

  • Length and breadth of the field be 6x and 5x respectively.

As we know that:

  • Perimeter = 2(length + breadth) units

Substituting the values,

\sf{\longmapsto297=2(6x+5x)}

\sf{\longmapsto297=2\times11x}

\sf{\longmapsto297=22x}

\sf{\longmapsto\dfrac{297}{22}=x}

\sf{\longmapsto13.5=x}

Hence,

  • x = 13.5

Therefore,

  • Length of the field = 6x = 6 × 13.5 = 81 m
  • Breadth of the field = 5x = 5 × 13.5 = 67.5 m

Finding area of the field:

As we know that,

  • Area of rectangle = (length × breadth) sq. units

Substituting the values,

\sf{\longmapsto(81\times67.5)m^2}

\sf{\longmapsto5467.5\;m^2}

Hence,

  • Area of the field is 5467.5 m².
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