Question

The length of a rectangle is 5 cm more than its breadth its area is 300 . find it lengh and breadth in algebra form

Answer 1

Solution :

Length of rectangle is 20 and breadth is 15

Step by step explanation :

Given :

• Area of rectangle is 300 cm²
• Length is 5 cm more then it's breadth

To Find :

• Length
• Breadth of rectangle

Solution :

Let the breadth of rectangle be 'b ' and length 'l'

According to the question :

Length is 5 cm more then it's breadth , then

$\sf\:l=b+5$

We know that

$\rm\:Area\:of\:rectangle=length\:\times\:breadth$

$\sf\implies\:300=b(b+5)$

$\sf\implies\:300=b^2+5b$

$\sf\implies\:b^2+5b-300$

By Quadratic equation Formula:

$\sf\:b=\dfrac{-5\pm\sqrt{5^2-4(300)(1)}}{2(1)}$

$\sf\implies\:b=\dfrac{-5\pm\sqrt{25+1200}}{2(1)}$

$\sf\implies\:b=\dfrac{-5\pm\sqrt{1225}}{2}$

$\sf\implies\:b=\dfrac{-5\pm35}{2}$

$\sf\implies\:b=\dfrac{-5+35}{2}\:or\dfrac{-5-35}{2}$

$\sf\implies\:b=15\:or\:-20$

Since ,breath can't be negative

Thus , b = 15

Then , length = l+5= 20

Answer 2

Given that,

• The length of a rectangle is 5 cm more than its breadth
• The area of this rectangle is 300 .

❍ We need to find the length and breadth of the rectangle.

_____________

Assume that the breadth of rectangle is 'b' and length is 'l'.

According to the question,

length of a rectangle is 5 cm more than its breadth, therefore

$\implies{\sf{I=b+5}}$

We already know that,

$\displaystyle{\boxed{\bf{Area\:of\:rectangle=length \times breadth}}}$

Substituting the values, we get

:$\implies{\sf{300=b \times (b+5)}}$

:$\implies{\sf{300=b^{2}+5b}}$

:$\implies{\sf{b^{2}+5b-300}}$

• As we have to find the value of b, we can't simplify this equation anymore and find the value of b.

Therefore, we need to find the value of b by using quadratic equation.

:$\large\implies{\boxed{\bf{x= \dfrac{-b± \sqrt{b^{2}-4ac}}{2a}}}}$

When ax² + bx + c = 0

:$\implies{\sf{b= \dfrac{-5± \sqrt{5^{2}-4(300)(1)}}{2(1)}}}$

:$\implies{\sf{b= \dfrac{-5± \sqrt{25+1200}}{2}}}$

:$\implies{\sf{b= \dfrac{-5± \sqrt{1225}}{2}}}$

:$\implies{\sf{b= \dfrac{-5± 35}{2}}}$

:$\implies{\sf{b= \dfrac{-5+ 35}{2} or \dfrac{-5-35}{2}}}$

:$\implies{\sf{b= 15\:or\:-20}}$

And we know that breadth can't be negative,

therefore $\large{\boxed{\sf{\red{breadth= 15}}}}$

That's Y $\large{\boxed{\sf{\red{length= 15+5 = 20}}}}$

And we are done! :D

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