Question

Length is 4 times the breadth
I am a rectangle
breadth x
My perimeter is 70 cm
perimeter of the rectangle
2([
x ) = 70
(using formula)
x +
2x
x = 70
x= 70
x=
breadth of rectangle =
cm and
=
cm
length of rectangle​

Answer 1

➤ Correct Question :-

The length of a rectangle is 70 cm. It's length is 4 times greater than it's breadth. Then find the length and breadth of a rectangle.

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★ How to do :-

Here, we are given with the perimeter of a rectangle and it's said that the length is 4 times greater than it's breadth. We are asked to find the length and breadth of that rectangle. So, here let's consider that the value of breadth be x and the value of length is 4 times of x and then find the value of x by substitution method i.e, transportation of numbers from one hand side to other. By this we can find the value of x. So, let's solve!!

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➤ Solution :-

Value of breadth :-

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

Substitute the given values.

${\tt \leadsto 70 = 2 \: (4x + x)}$

Add the values given in the bracket.

${\tt \leadsto 70 = 2 \times 5x}$

Multiply the values on RHS.

${\tt \leadsto 70 = 10x}$

Shift the number 10 from RHS to LHS, changing it's sign.

${\tt \leadsto x = \dfrac{70}{10}}$

Simplify the fraction to get the value of x.

${\tt \leadsto x = \cancel \dfrac{70}{10} = \pink{\underline{\boxed{\tt 7 cm}}}}$

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We know that the length is four times the breadth. So,

Value of length :-

${\tt \leadsto 4x = 4 \times 7}$

${\tt \leadsto \pink{\underline{\boxed{\tt 28 cm}}}}$

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Hence solved !!

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$\dashrightarrow$ Some related formulas :-

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