Question

a wire of length 86 cm is bent in the form of a rectangle such that its length is 7 cm more than its breadth. find the length and the breadth of the rectangle so formed.​

Answer 1

AnswEr :

Let the Breadth of the Rectangle be n cm, then Length of the Rectangle be (n + 7) cm.

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(7.3,2){\mathsf{\large{n cm}}}\put(7.7,1){\large{B}}\put(9.2,0.7){\matsf{\large{(n + 7) cm}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,3){\large{D}}\end{picture}$

• we need to understand the concept behind this question.
• As per Question, Wire is bent to form a Rectangle. So the Length of the wire will be the Perimeter of Rectangle.

• Let's Head to the Question Now :

↠ Perimeter = 2(Length + Breadth)

↠ Length of Wire = 2(Length + Breadth)

↠ 86cm = 2{(n + 7) + n}

• Dividing both term by 2

↠ 43 = {(n + 7) + n}

↠ 43 = 2n + 7

↠ 43 – 7 = 2n

↠ 36 = 2n

• Dividing both term by 2

↠ n = 18

$\rule{300}{1}$

• D I M E N S I O N S :

◗ Length = (n + 7) = (18 + 7) = 25 cm

◗ Breadth = n = 18 cm

⠀

∴ Length and Breadth of the Rectangle will be 25 cm and, 18 cm respectively.

$\rule{300}{2}$

$\star \:\underline \text{Some \:Information \:about \: Rectangle :}$

⋆ Opposite sides are equal and parallel.

⋆ All angles are equal to 90 degrees.

⋆ The diagonals are equal and bisect each other.

⋆ The intersection of the diagonals is the circumcentre. That is you can draw a circle with that as centre to pass through the four corners.

⋆ Any two adjacent angles add up to 180 degrees.

⋆ Lines joining the mid points of the sides of a rectangle in an order form a rhombus of half the area of the rectangle.

⋆ The sum of the four exterior angles is 4 right angles.

⋆ Area of Rectangle = Length * Breadth

⋆ Perimeter of Rectangle = 2*(Length + Breadth)

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