Question

21 A wire when bent
when bent in the form of an equilateral triangle encloses an area of 36 v3 cm.
the area enclosed by the same wire when bent to form:
Find
(i) a square, and
(ii) a rectangle whose length is 2 cm more than its width.​

Answer 1

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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 : $\sf{Length\times Breadth}$

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