Question : The interior angle of a regular polygon exceeds its exterior angle by 90°. The number of sides of the polygon is:
Option 1: 8
Option 2: 6
Option 3: 10
Option 4: 12
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 8
Solution : Given: The interior angle of a regular polygon exceeds its exterior angle by 90°. Let $x$ be the polygon's exterior angle and $n$ be its number of sides. A regular polygon's sum of its exterior and interior angles = 180°. A regular polygon's exterior angle = $\frac{360°}{n}$, where $n$ is the number of sides. According to the question, $x+x+90°=180°$ ⇒ $2x=90°$ ⇒ $x=45°$ The number of sides of the polygon is given as, $45°=\frac{360°}{n}$ ⇒ $n=8$ Hence, the correct answer is 8.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : The interior angle of a regular polygon exceeds its exterior angle by 108$^\circ$. The number of the sides of the polygon is:
Question : $A_1$ and $A_2$ are two regular polygons. The sum of all the interior angles of $A_1$ is $1080^{\circ}$. Each interior angle of $A_2$ exceeds its exterior angle by $132^{\circ}$. The sum of the number of sides $A_1$ and $A_2$ is:
Question : If the sum of the measures of all the interior angles of a polygon is 1440$^\circ$, find the number of sides of the polygon.
Question : Two regular polygons are such that the ratio between their number of sides is 1 : 2 and the ratio of measures of their interior angles is 3 : 4. Then the number of sides of each polygon is:
Question : The sum of the interior angles of a regular polygon A is 1260 degrees and each interior angle of a regular polygon B is $128 \frac{4}{7}$ degrees. The sum of the number of sides of polygons A and B is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile