GCSE OCR Maths Polygons: An equilateral triangle, a regul

An equilateral triangle, a regular 10-sided polygon and another regular polygon meet at a point - OCR - GCSE Maths - Question 10 - 2021 - Paper 1

Question 10

An equilateral triangle, a regular 10-sided polygon and another regular polygon meet at a point. (a) Show that angle A is 156°. (b) Work out the number of sides of... show full transcript

Worked Solution & Example Answer:An equilateral triangle, a regular 10-sided polygon and another regular polygon meet at a point - OCR - GCSE Maths - Question 10 - 2021 - Paper 1

Step 1

Show that angle A is 156°.

96%

114 rated

Answer

To find angle A formed at point A, we need to consider the characteristics of the polygons meeting at this point.

Angle in Equilateral Triangle: Each angle in an equilateral triangle is 60°.
Regular 10-sided Polygon (Decagon): The internal angle of a regular polygon is given by the formula:

$ext{Internal Angle} = \frac{(n-2) \times 180°}{n}$ where n is the number of sides.

For a decagon (n=10): $\text{Internal Angle} = \frac{(10-2) \times 180°}{10} = \frac{8 \times 180°}{10} = 144°$
Sum of Angles Around Point A: The sum of angles around point A must equal 360°:

$ext{Angle A} + ext{Angle of Triangle} + ext{Angle of Decagon} = 360°$

Substituting the known values gives:

$\text{Angle A} + 60° + 144° = 360°$

Simplifying this leads to:

$\text{Angle A} = 360° - 204° = 156°$

Thus, angle A is shown to be 156°.

Step 2

Work out the number of sides of the other regular polygon.

99%

104 rated

Answer

To find the number of sides of the other regular polygon, we know angle A is 156°. We will use the internal angle formula again:

Using the internal angle formula:

$\text{Internal Angle} = \frac{(n-2) \times 180°}{n}$

Set the internal angle equal to 156°:

$156° = \frac{(n-2) \times 180°}{n}$
Cross-multiplying to eliminate the fraction:

$156n = (n-2) \times 180$
Expanding and rearranging gives:

$156n = 180n - 360$

Combining like terms results in: $360 = 180n - 156n$

So, $360 = 24n$
Solving for n:

$n = \frac{360}{24} = 15$

Therefore, the other regular polygon has 15 sides.

An equilateral triangle, a regular 10-sided polygon and another regular polygon meet at a point - OCR - GCSE Maths - Question 10 - 2021 - Paper 1

Worked Solution & Example Answer:An equilateral triangle, a regular 10-sided polygon and another regular polygon meet at a point - OCR - GCSE Maths - Question 10 - 2021 - Paper 1

Show that angle A is 156°.

Work out the number of sides of the other regular polygon.

Join the GCSE students using SimpleStudy...