Question

the number of sides of two regular polygon are (n-2) and (n+1) . Their exterior angles differ by 27° . Find the number of sides of each polygon.

ans n=7,number of sides are 5 and 8​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Case ❶ :-

→ No. of sides of Regular Polygon = (n - 2)

→ Each Exterior Angle = 360°/(No. of sides) = 360/(n - 2)

Case ❷:-

→ No. of sides of Regular Polygon = (n + 1)

→ Each Exterior Angle = 360°/(No. of sides) = 360/(n + 1)

A/q,

→ 360/(n - 2) - 360/(n + 1) = 27

→ 360[1/(n - 2) - 1/(n +1)] = 27

→ [ {n + 1 - n + 2 } /(n - 2)(n + 1) ] = 27/360

→ 3/(n - 2)(n + 1) = 27/360

→ 1/(n - 2)(n + 1) = 9/360

→ 1/(n - 2)(n + 1) = 1/40

→ (n - 2)(n + 1) = 40

→ n² + n - 2n - 2 = 40

→ n² - n - 42 = 0

→ n² - 7n + 6n - 42 = 0

→ n(n - 7) + 6(n - 7) = 0

→ (n - 7)(n + 6) = 0

→ n = 7 or (-6) .

since Negative value of sides is Not Possible..

Therefore,

=> n = 7.

Hence,

=> No. of sides of First Polygon = (n - 2) = 7 - 2 = 5 (Ans.)

=> No. of sides of second Polygon = (n + 1) = 7 + 1 = 8 (Ans.)

Answer 2

$\huge\sf\pink{Answer}$

☞ Your Answer is 5 and 8 sides

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Number of sides of 2 regular polygon are (n - 2) and (n + 1)

✭ Their exterior angles differ by 27°

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

➼ Number of sides of each polygon?

$\rule{110}1$

$\huge\sf\purple{Steps}$

In thePolygons Each Exterior Angle is given by,

$\sf\underline{\boxed{\green{\sf\dfrac{360\degree}{No \ Of \ Sides}}}}$

Substituting the given values in first polynomial,

$\sf\longmapsto\dfrac{360\degree}{(n-2)}$

Substituting the given values in second polynomial,

$\sf\longmapsto\dfrac{360\degree}{(n+1)}$

❍ Given that their exterior angles differ by 27°

➝ $\sf\dfrac{360\degree}{(n-2)} - \dfrac{360\degree}{(n+1)} = 27$

➝ $\sf 360\degree\bigg\lgroup\dfrac{1}{n-2} - \dfrac{1}{n+1} \bigg\rgroup$

➝ $\sf \bigg\lgroup\dfrac{n+1-n+2}{(n-2)(n+1)}\bigg\rgroup = \frac{27}{360}$

➝ $\sf \frac{3}{(n-2)(n+1)} = \frac{27}{360}$

➝ $\sf \frac{1}{(n-2)(n+1)} = \frac{1}{40}$

➝ $\sf \bigg\lgroup n-2\bigg\rgroup\bigg\lgroup n+1\bigg\rgroup = 40$

➝ $\sf n^2 + n -2n -2 = 40$

➝ $\sf n^2 -n -42 = 0$

➝ $\sf n^2-7n+6n-42 = 0$

➝ $\sf n\bigg\lgroup n-7 \bigg\rgroup + 6\bigg\lgroup n-7\bigg\rgroup$

➝ $\sv \bigg\lgroup n+6 \bigg\rgroup \bigg\lgroup n-7\bigg\rgroup$

➝ $\sf\color{red}{n = -6 \ or \ 7}$

➝ $\sf\orange{ n=7}\qquad \bigg \lgroup -6 \ is \ negative \ value \bigg\rgroup$

So,Now

$\dashrightarrow\sf{Polygon_1 = (n-2)}$

$\sf\dashrightarrow 7-2$

$\sf\color{aqua}{\dashrightarrow Polygon_1 = 5 }$

$\dashrightarrow\sf{Polygon_2 = (n+1)}$

$\sf\dashrightarrow 7+1$

$\sf\color{lime}{\dashrightarrow Polygon_2 = 8 }$

$\rule{170}3$