Question

The length and breadth of a rectangle are (a + 5b) units and (7a - b) units respectively. The
perimeter of this rectangle is equal to the perimeter of a square. Find how much is the area of the rectangle less than that of the square?​

Answer 1

Answer:

answer of your question

Answer 2

To Find :-

The Difference of the Area and the area of square.

Given :-

• Length of the Rectangle = (a + 5b) units

• Breadth of the Rectangle = (7a - b) units

We Know :-

Perimeter of a Rectangle :-

$\bf{\underline{P = 2(Length + Breadth)}}$

Perimeter of a Square :-

$\bf{\underline{P = 4 \times side}}$

Area of a Rectangle :-

$\bf{\underline{A = Length \times Breadth}}$

Area of a Square :-

$\bf{\underline{A = (side)^{2}}}$

Concept :-

According to the question , the perimeter of the Rectangle is equal to the perimeter of the Square , so by the given Length and breadth , we can find the side of the Square .

And after that , by finding the Area of the Rectangle and the Square and by subtracting them we will get the required answer.

Solution :-

Perimeter of the Rectangle :-

Given :-

• l = (a + 5b)

• b = (7a - b)

Using the formula for Perimeter of the Rectangle and Substituting the values in it , we get :-

$:\implies \bf{P = 2(Length + Breadth)} \\ \\ \\ :\implies \bf{P = 2(a + 5b + 7a - b)} \\ \\ \\ :\implies \bf{P = 2(8a + 4b)} \\ \\ \\ :\implies \bf{P = 16a + 8b}$

Hence, the perimeter of the Rectangle is 16a + 8b units.

Side of the Square :

A/c , the Perimeter of the Square is equal to the Perimeter of the Rectangle . i.e,

$\bf{P_{(Rectangle)} = P_{(SQuare)}}$

Hence, the perimeter of the Square is also

16a + 8b.

• Perimeter of the Square = 16a + 8b.

Let the equal side of the Square be a units.

Using the formula for Perimeter of a Square and by substituting the values in it,we get :-

$:\implies \bf{P = 4 \times side} \\ \\ \\ :\implies \bf{16a + 8b = 4 \times a} \\ \\ \\ :\implies \bf{4(4a + 2b) = 4a} \\ \\ \\ :\implies \bf{\dfrac{4(4a + 2b)}{4} = a} \\ \\ \\ :\implies \bf{\dfrac{\not{4}(4a + 2b)}{\not{4}} = a} \\ \\ \\ :\implies \bf{4a + 2b = a} \\ \\ \\ :\therefore \purple{\bf{4a + 2b = a}}$

Hence, the side of the Square is 4a + 2b units.

Area of the Rectangle :-

Given :-

• l = a + 5b

• b = 7a - b

Using the formula for area of a Rectangle and by substituting the values in it , we get :-

$:\implies \bf{A = Length \times Breadth} \\ \\ \\ :\implies \bf{A = (a + 5b) \times (7a - b)} \\ \\ \\ :\implies \bf{A = 8a^{2} - ab + 35ab - 5b^{2}+ 5b} \\ \\ \\ :\implies \bf{A = 8a^{2} + 34ab - 5b^{2}+ 5b} \\ \\ \\ :\therefore \purple{\bf{A = 8a^{2} + 34ab - 5b^{2}}}$

Hence, the Area of the Rectangle is 8a² + 34ab + 5b².

Area of the Square :-

• Side = 4a + 2b

By using the formula for area of a Square and by substituting the values in it ,we get :-

$:\implies \bf{A = (side)^{2}} \\ \\ \\ :\implies \bf{A = (4a + 2b)^{2}}$

By using the identity , we get :-

$\bf{(a + b)^{2} = a^{2} + b^{2} + 2ab}$

$:\implies \bf{A = (4a)^{2} + 2 \times 4a \times 2b + (2b)^{2}} \\ \\ \\ :\implies \bf{A = 16a^{2} + 16ab + 4b^{2}} \\ \\ \\ \therefore \purple{\bf{A = 16a^{2} + 16ab + 4b^{2}}}$

Hence, the area of the Square is 16a² + 16ab + 4b².

Now, to Find that by how much the Area of the Rectangle exceeds the Area of Square , we need to Find the difference between their areas.

==> Area of Rectangle - Area of Square

==> (8a² + 34ab + 5b²) - (16a² + 16ab + 4b²)

==> 8a² + 34ab + 5b² - 16a² - 16ab - 4b²

==> 18ab + b² - 8a² .

Hence, by 18ab + b² - 8a² , the area of Rectangle exceeds the area of the Square.