Question

fencing the compound of a house costs 5452 rupees.if the rate is 94 rupees per metre,find the perimeter of the compound .if the breadth is 10m ,find it's lenght​

Answer 1

Answer:

Step-by-step explanation:

perimeter =5452 / 94=58m. Ans

2 (I + b) = 58

2 (I +10)=58

I+10=58/2=29

I=29-10=19m

I=19m. .. Ans

Answer 2

➤ Given :-

Total cost of fencing a house :- ₹5452

Cost of fencing per metre :- ₹94

Breadth of rectangle :- 10 metres

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➤ To Find :-

The perimetre and the length of the boundary.

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★ How to do :-

Here, we are given with the total cost of fencing a whole boundary of a house and the cost of fencing the boundary per metre. We are also given with the breadth of the rectangular boundary. We are asked to find the perimetre and the length of that boundary around the house. So, first we should find the perimetre of the rectangle by using the total cost and the cost per metre of fencing. Later, we can find the length by using the perimetre and the breadth. So, let's solve!!

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➤ Solution :-

Perimetre of the rectangle :-

${\tt \leadsto \underline{\boxed{\tt Perimeter \times Cost \: per \: metre = Total \: cost}}}$

Substitute the given values.

${\tt \leadsto P \times 94 = 5452}$

Shift the number 94 from LHS to RHS, changing it's sign.

${\tt \leadsto P = \dfrac{5452}{94}}$

Simplify the obtained fraction to get the answer.

${\tt \leadsto \pink{\underline{\boxed{\tt Perimeter = 58 \: \: metres}}}}$

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Now, find the length of the rectangle.

Length of the rectangle :-

${\tt \leadsto \underline{\boxed{\tt 2 \: (Length + Breadth)}}}$

Substitute the given values.

${\tt \leadsto 58 = 2 \: (L + 10)}$

Multiply the number 2 with both numbers in bracket.

${\tt \leadsto 58 = 2L + 20}$

Shift the number 58 from LHS to RHS, changing it's sign.

${\tt \leadsto 2L = 58 - 20}$

Subtract the values in RHS to get the value of 2L.

${\tt \leadsto 2L = 38}$

Shift the number 2 from LHS to RHS, changing it's sign.

${\tt \leadsto L = \dfrac{38}{2}}$

Simplify the fraction to get the value of length.

${\tt \leadsto \pink{\underline{\boxed{\tt Length = 19 \: \: metres}}}}$

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Hence solved !!

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$\dashrightarrow$ Some related formulas :-

$\begin{gathered}\small\boxed{\begin{array}{cc}\large\sf\dag \: {\underline{More \: Formulae}} \\ \\ \bigstar \: \sf{{Area}_{(Rectangle)} = Length \times Breadth} \\ \\ \bigstar \:\sf{{Length}_{(Rectangle)} = \dfrac{Perimetre}{2} - Breadth} \\ \\ \bigstar \: \sf{{Breadth}_{(Rectangle)} = \dfrac{Perimetre}{2} - Length} \\ \\ \bigstar \: \sf{{Length}_{(Rectangle)} = \dfrac{Area}{Breadth}} \\ \\ \bigstar \: \sf{{Breadth}_{(Rectangle)} = \dfrac{Area}{Length}}\end{array}}\end{gathered}$