Question

Hi guys please help me in this question
Book: M.L. Aggrawal
Class:9
Question is in the attachment
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Answer 1

Answer:

81 and 80

Step-by-step explanation:

First we need to calculate the length of the wire.

Since the wire bends to form an equilateral triangle, the perimeter of the triangle is nothing but the lenght of the wire.

Area of Equilateral Triangle = $\frac{\sqrt{3} }{4} (side)^{2}$

= $36\sqrt{3} =$ $\frac{\sqrt{3} }{4} (side)^{2}$

= $side= 12 cm$

Therefore perimeter of the triangle= 12 x 3 = 36 cm

(i) A square:

Perimeter = 36 cm = 4 x (side)

Side = 9 cm

Area = $(side)^{2}$ = 81 sq.cm

(ii) A rectangle:

Let width = W cm

Then length = (W+2) cm

Perimeter = 36 cm = 2 (Length + Width)

= 2 (W+2+W) = 36

= W = 8 cm

So L = 10 cm

Area = L x W = 80 sq. cm

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 : $\sf{Length\times Breadth}$

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