Question

a wire bent in the form of an equilateral triangle encloses an area of 36 root 3 cm sq...find the area enclosed by the same wire when bent to form 1.)a square and 2.)a rectangle whose length is 2 cm more than its width....

Answer 1

Concept And formula used :-

• As we know that, when something is bent to form another shape , length or perimeter of both shapes remains same .

• Area of Equaliteral ∆ = (√3/4) * (side)²

• Perimeter of Equaliteral ∆ = 3 * side

• Perimeter of Square = 4 * (side of Square)

• Area of Square = (side of square)²

• Perimeter of Rectangle = 2(Length + Breadth)

• Area of Rectangle = Length * Breadth .

Solution :-

we have given that, when wire was bent in the form of an equilateral triangle it encloses an area of 36√3 cm².

So,

→ Area of Equaliteral ∆ = 36√3 cm²

→ (√3/4) * (side of Equaliteral ∆)² = 36√3

→ (side of Equaliteral ∆)² = 36 * 4

→ (side of Equaliteral ∆)² = (9*4) * 4

→ (side of Equaliteral ∆)² = (3)² * (4)²

→ (side of Equaliteral ∆)² = (3*4)²

→ (side of Equaliteral ∆)² = (12)²

Square - root both sides,

→ side of Equaliteral ∆ = 12 cm.

Therefore,

→ Perimeter of Equaliteral ∆ = 3 * sides = 3 * 12 = 36cm.

now, we can conclude that, Perimeter of all shape will be 36cm.

_________

when bent in the form of Square :-

→ Perimeter of Square = 36cm.

→ 4 * side of Square = 36

→ side of Square = (36/4)

→ side of Square = 9cm.

Therefore,

→ Area enclosed by Square shape = (side of Square)² = (9)² = 81 cm² (Ans.)

_________

when bent in the form of Rectangle :-

→ Perimeter of Rectangle = 36cm.

We have given that, length of rectangle is 2 cm more than its width.

→ Let, width of rectangle = x cm.

→ Length = (x + 2)cm.

So,

→ Periemeter of rectangle = 2(Length + Breadth).

Putting value of length and breadth and perimeter as 36cm Now,

→ 2(x + 2 + x ) = 36

→ 2(2x + 2) = 36

→ 4x + 4 = 36

→ 4x = 36 - 4

→ 4x = 32

→ x = (32/4)

→ x = 8 cm.

Therefore,

→ width of rectangle = x cm = 8cm.

→ Length of rectangle = (x + 2)cm = 8 + 2 = 10cm.

Hence ,

→ Area enclosed by Rectangle shape = Length * Breadth = 10 * 8 = 80 cm². (Ans.)

Answer 2

$\bigstar$ It's always better to know Key Points before Solving the Question! Here are some Key Points which helps us solve this Question easily

• The Area of an equilateral triangle is given by $\sf{\sqrt{3}/4\times Side^2}$ and The Perimeter of an equilateral triangle is given by $\sf{3\times Side}$ . The Area of a Square is given by the expression $\sf{Side\times Side}$ and The Perimeter of a Square is given by the expression $\sf{4\times Side}$ . The Area of a Rectangle is given by the expression  $\sf{Length\times Breadth}$ and The Perimeter of a Rectangle is given by the expression $\sf{2(Lengh+Breadth)}$

$\rule{315}{2}$

First, We will find the Measure of Each Side in the Equilateral Triangle

$\longrightarrow\sf{Area\:of\:Equilateral\:Triangle = \sqrt{3}/4\times Side^2}$

It is Clearly mentioned Wire bent in the form of an equilateral triangle encloses an area of $\sf{36\sqrt{3}\:cm^2}$ in the Question

$\longrightarrow\sf{36\sqrt{3}\:cm^2 = \sqrt{3}/4\times Side^2}$

Divide both Sides by the Equation by $\sf{\sqrt{3}/4}$

$\longrightarrow\sf{36\sqrt{3}\:cm^2\div\sqrt{3}/4 = (\sqrt{3}/4\times Side^2)\div\sqrt{3}/4}$

$\longrightarrow\sf{Side^2 = 36\sqrt{3}\div\sqrt{3}/4\:cm^2 =36\sqrt{3}\times4/\sqrt{3} \times cm^2 }$

$\longrightarrow\sf{Side^2 = 144\sqrt{3}/\sqrt{3} \times cm^2 = 144\:cm^2 }$

Take Square Root on Both Sides of the Equation

$\longrightarrow\sf{\sqrt{Side^2} = \sqrt{144\:cm^2}}\longrightarrow\sf{Side = 12\:cm}$

Now the Total Length of Wire will be Equal to Perimeter of Wire. Perimeter of Wire = Perimeter of Equilateral Triangle

$\boxed{\sf{Total\: Length\: of \:Wire = 3\times Side = 3\times 12\:cm = 36\:cm}}$

1) The length of the side of the square is obtained by Dividing Total Length of Wire/Number of Sides in a Square or Total Length of Wire/4

$\longrightarrow\textsf{Length of the side of the Square = \sf{36/4\:cm = 9\:cm} }$

Thus the Area of Square is obtained as follows : $\sf{Side\times Side}$

$\boxed{\longrightarrow\sf{Area\:of\:Square=Side\times Side = 9\:cm\times 9\:cm= 81\:cm^2}}$

2) The Perimeter of the rectangle should be 36 cm. Thus the relation obtained is $\sf{2(Lengh+Breadth)}$ . Since the difference between length and breadth is 2 cm, thus the relation obtained is $\sf{Length-Bredth=2}$ . Thus the values of the length and breadth are calculated as follows : $\sf{Length-Breadth=2\:cm\longrightarrow Length = Breadth + 2\:cm}$

$\longrightarrow\sf{Perimeter \:of \:Rectangle = 36\:cm}\longrightarrow\sf{2(Length+Breadth) = 36\:cm}$

$\longrightarrow\sf{2(Breadth+2\:cm+Breadth) = 36\:cm\quad...\:Since\:Length = Breadth+2\:cm}$

$\longrightarrow\sf{2(2\times Breadth+2\:cm) = 36\:cm}\longrightarrow\sf{4\times Breadth+4\:cm = 36\:cm}$

Subtract 4 cm from Both Sides of the Equation

$\longrightarrow\sf{4\times Breadth+4\:cm-4\:cm = 36\:cm-4\:cm }$

$\longrightarrow\sf{4\times Breadth = 36\:cm-4\:cm = 32\:cm }$

Divide Both Sides of the Equation by 4

$\longrightarrow\sf{(4\times Breadth)/4 = 32/4\:cm }\longrightarrow\sf{ Breadth = 8\:cm }$

Also $\sf{Length = Breadth + 2\:cm}\longrightarrow\sf{Length = 8\:cm + 2\:cm=10\:cm}$

Thus the Area of Rectangle is obtained as follows : $\sf{Length\times Breadth}$

$\boxed{\longrightarrow\sf{Area\:of\:Rectangle=Length\times Breadth = 10\:cm\times 8\:cm = 80\:cm}}$