Question

the sides of a heptagon are enlarged 3times . Determine the ratio of the areas of the old and new heptagon​

Answer 1

Given:

The sides of a heptagon are enlarged 3 times.

To be found: Ratio of the areas of the old heptagon and new heptagon.

Heptagon: Heptagon is a closed figure having seven sides with equal dimensions. It is a polygon with 7 equal sides.

Formula to be used:

Area of a heptagon with side 'a' = $7/4 . a^{2} . cot (\pi / 7)$

Solution:

Let the side of the old heptagon be 'a₁'.

Let the area of the old heptagon be denoted as 'A₁'

Then, the area of the old heptagon is given by:

$A_1 = 7 / 4 . a_{1} ^{2} . cot(\pi / 7)$

Let the side of the new heptagon be 'a₂'.

Let the area of the new heptagon be denoted by 'A₂'.

Then, the area of the new heptagon will be given by:

$A_2 = 7 / 4 . a_2^{2} . cot (\pi / 7)$

But, as given in the question, the sides of new heptagon are enlarged three times.

⇒$a_2 = 3 (a_1)$

Due to this change, the area of the new heptagon also gets changed. That area is given by:

$A_2 = 7 / 4. (3a_1)^{2} . cot (\pi / 7)$

⇒$A_2 = (7 / 4). (9a_1^{2}) . cot (\pi / 7)$

⇒$A_{2} = (63/4) . a_{1} ^{2} . cot(\pi / 7)$

Now, the required ratio of the areas of old heptagon and new heptagon is given by $A_1 / A_2$.

$A_1 / A_2= ((7 / 4). a_1^{2} . cot (\pi /7)) / ((63/4). a_1^{2} . cot(\pi /7))$

⇒$A_1 / A_2 = (7/4)/(63/4)$

⇒$A_1/A_2 = (7/4) . (4/63)$

⇒$A_1 / A_2 = 7/63 = 1/9$

∴ The required ratio of the areas of old heptagon to the new heptagon is 1/9.

⇒Due to the enlargement of the sides of a heptagon by 3 times, the area increases by 9 times, which is given by:

$A_2 = 9 A_1$