Question

The areas of a rectangular park and equilateral triangular park are same and the value is 40cm². The length of the rectangular park is 10cm and the height of the equilateral triangular park is 8cm . Which park will cost more for fencing?​

Answer 1

Answer:

side=40

perimeter=40×3=120

total cost=24000

cost per m =24000÷120

=₹200 per metre

Answer 2

${\underline{\underline{\bf{Given:}}}}$

• The area of a rectangular park and an equilateral triangular park are of same measure, that is 40 m².
• The length of rectangular park is 10 m.
• The height of the equilateral triangular park is 8 m.

${\underline{\underline{\bf{Need\:to\:find:}}}}$

• Which park costs more in fencing.

${\underline{\underline{\bf{Formulae\:to\:be\: used:}}}}$

$\underline{\boxed{\sf{{Area}_{[Rectangle]}=lb\:sq.units}}}$

$\underline{\boxed{\sf{{Perimeter}_{[Rectangle]}=2(l+b)\:units}}}$

$\:\:\:\:\:\:\:\:\:\bullet{\sf{\:l=length}}$

$\:\:\:\:\:\:\:\:\:\bullet{\sf{\:b=breadth}}$

$\underline{\boxed{\sf{{Area}_{[Equilateral\: triangle]}=\frac{1}{2}ah\:sq.units}}}$

$\underline{\boxed{\sf{{Perimeter}_{[Equilateral\: triangle]}=3a\:units}}}$

$\:\:\:\:\:\:\:\:\:\bullet{\sf{\:a=side}}$

$\:\:\:\:\:\:\:\:\:\bullet{\sf{\:h=height}}$

${\underline{\underline{\bf{Solution:}}}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎The area of both the parks are equal. We're given with the measures of length of rectangular park and height of equilateral triangular park. We're asked to find which park costs more in fencing. To find that, we need to find the perimeter of both the parks as fencing is done in the boundaries of the parks.

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ We know the formulae to find the perimeter of both the parks. By relating that, first we have to find the breadth of rectangular park and side of equilateral triangular park. They both can be found by putting the given measures in their respective formulae used to find area.

* The park with greater perimeter costs more in fencing.

Length of rectangular park:

$\longrightarrow{\sf{Area=lb\:sq.units}}$

$\:\:\:\:\:\:\implies{\sf{40=10\times b}}$

$\:\:\:\:\:\:\implies{\sf{\frac{4\cancel{0}}{1\cancel{0}}=b}}$

$\:\:\:\:\:\:\implies{\underline{\underline{\sf{4\:m=b}}}}$

Side of equilateral triangular park:

$\longrightarrow{\sf{Area=\frac{1}{2}ah\:sq.units}}$

$\:\:\:\:\:\:\implies{\sf{40=\frac{1}{\cancel{2}}\times a\times \cancel{8}}}$

$\:\:\:\:\:\:\implies{\sf{40=a\times 4}}$

$\:\:\:\:\:\:\implies{\sf{\frac{\cancel{40}}{\cancel{4}}=a}}$

$\:\:\:\:\:\:\implies{\underline{\underline{\sf{10\:m=a}}}}$

Perimeter of rectangular park:

$\longrightarrow{\sf{2(l+b)\:units}}$

By substituting the measures,

$\:\:\:\:\:\:\implies{\sf{2\times(10+4)}}$

$\:\:\:\:\:\:\implies{\sf{2\times 14}}$

$\:\:\:\:\:\:\implies{\sf{\red{\underline{28\:m}}}}$

Perimeter of equilateral triangular park:

$\longrightarrow{\sf{3a\:units}}$

By substituting the obtained measure,

$\:\:\:\:\:\:\implies{\sf{3\times 10}}$

$\:\:\:\:\:\:\implies{\sf{\red{\underline{30\:m}}}}$

The perimeter of rectangular park and equilateral triangular park are 28 m and 30 m respectively.

$\:\:\:\:\:\:\:\:\:\:\:\:{\boxed{\sf{28\:m<30\:m}}}$

Since, equilateral triangular park has the larger perimeter, it costs the more in fencing.

${\underline{\underline{\bf{Required\:answer:}}}}$

• Equilateral triangular park costs more in fencing.

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