Question

the the length and breadth of a rectangular field in the ratio of 7:4. the cost of fencing the field at rupees 25 per metre is rupees 3300 find the dimensions of the field​

Answer 1

ANSWER

Let length of rectangle be x m

Raito of length and Breadth=7:4

Breadth

x

​  

=  

4

7

​  

 

⇒ Breadth=  

7

4x

​  

 

Cost of fencing 1m=Rs. 25

Cost of fencing the fluid =Rs. 3300

Perimeter of rectangle × cost of fencing 1cm = cost of fencing the field.

2(l+b)×25=3300

50(l+b)=3300

l+b=66

x+  

7

4x

​  

=66

11x=66×7

x=42

Length =42m

Breadth=24m.

Answer 2

Answer :-

• Length of Rectangular Field = 42 m
• Width of Rectangular Field = 24 m

Given :-

• Ratio of Length & Width = 7 : 4    
• Rate of Fencing = 25 Rs Per meter
• Total cost of Painting = 3300 Rs

To Find :-

• Length of Rectangular Field =
• Width of Rectangular Field =

Explanation :-  

Let the Length & Width of the Field to be ‘7x’ and ‘4x’ respectively.

As per above given, the fencing rate is given; it means that Outer Side of Filed (Perimeter) is Fenced.

$\blue { \boxed { \bf \bigstar \: Perimeter \: of \: Rectangle \times Rate = Total \: Cost \: \bigstar }}$

$\rm : \: \longrightarrow 2(L + B) \times 25 = 3300$

$\rm : \: \longrightarrow 2(7x + 4x) \times 25 = 3300$

$\rm : \: \longrightarrow (11x) \times 50 = 3300$

$\rm : \: \longrightarrow 11x= \dfrac{3300}{50}$

$\rm : \: \longrightarrow 11x= 66$

$\rm : \: \longrightarrow x= \dfrac{66}{11}$

$\green { \boxed { \bf : \: \longrightarrow x= 6 \: m }}$

Now substituting the value of ‘x’,

$\rm \star \: Length = 7x = 7(6) = \red{ \bold{42 \: m} }$

$\rm \star \: Width = 4x = 4(6) = \red{ \bold{24 \: m} }$

Hence, the Length & Width of the Rectangular Field is 42 m and 24 m Respectively.  

‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒‒

$\underbrace { \overbrace { \huge \text{More To Know} }}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large Length}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large Width}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

$\rm \odot \: \text{Area of Rectangle} = \bold{Length \times Width}$

$\rm \odot \: \text{Perimeter of Rectangle} = \bold{2 \: (Length + Width)}$

$\rm \odot \: \text{Diagonal of Rectangle} = \bold{\sqrt{(Length)^2 + (Width)^2}}$