4.2.1 Polygons, PT3 Practice - Mathematics

4.2.1 Polygons, PT3 Practice

Question 1:

Diagram below shows a pentagon PQRST. TPU and RSV are straight lines.

Find the value of x.

Solution:

\begin{array}{l} Sum of interior angles of a pentagon \\ = (5 - 2) \times 180^{o} \\ = 3 \times 180^{o} \\ = 540^{o} \\ ∠ T S R = 180^{o} - 70^{o} \\ = 110^{o} \\ ∠ T P Q = 180^{o} - 85^{o} \\ = 95^{o} \\ x^{o} = 540^{o} - (110^{o} + 105^{o} + 115^{o} + 95^{o}) \\ = 540^{o} - 425^{o} \\ = 115^{o} \\ x = 115 \end{array}

$\begin{array}{l} Sum of interior angles of a pentagon \\ = (5 - 2) \times 180^{o} \\ = 3 \times 180^{o} \\ = 540^{o} \\ ∠ T S R = 180^{o} - 70^{o} \\ = 110^{o} \\ ∠ T P Q = 180^{o} - 85^{o} \\ = 95^{o} \\ x^{o} = 540^{o} - (110^{o} + 105^{o} + 115^{o} + 95^{o}) \\ = 540^{o} - 425^{o} \\ = 115^{o} \\ x = 115 \end{array}$

Question 2:

In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.

Find the value of x + y.

Solution:

\begin{array}{l} ∠ Q P U = 180^{o} - 160^{o} = 20^{o} \\ Reflex ∠ P U T = 360^{o} - 80^{o} = 280^{o} \\ ∠ U T S = 180^{o} - 120^{o} = 60^{o} \\ ∠ T S R = 180^{o} - 35^{o} = 145^{o} \\ Sum of interior angles of a hexagon \\ = (6 - 2) \times 180^{o} \\ = 720^{o} \\ x^{o} + y^{o} + 145^{o} + 60^{o} + 280^{o} + 20^{o} = 720^{o} \\ x^{o} + y^{o} = 720^{o} - 505^{o} \\ = 215^{o} \\ x + y = 215 \end{array}

$\begin{array}{l} ∠ Q P U = 180^{o} - 160^{o} = 20^{o} \\ Reflex ∠ P U T = 360^{o} - 80^{o} = 280^{o} \\ ∠ U T S = 180^{o} - 120^{o} = 60^{o} \\ ∠ T S R = 180^{o} - 35^{o} = 145^{o} \\ Sum of interior angles of a hexagon \\ = (6 - 2) \times 180^{o} \\ = 720^{o} \\ x^{o} + y^{o} + 145^{o} + 60^{o} + 280^{o} + 20^{o} = 720^{o} \\ x^{o} + y^{o} = 720^{o} - 505^{o} \\ = 215^{o} \\ x + y = 215 \end{array}$

Question 3:

Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.

Find the value of x + y.

Solution:

\begin{array}{l} Size of each interior angle of a regular hexagon \\ = \frac{(6 - 2) \times 180^{o}}{6} \\ = 120^{o} \\ x^{o} = \frac{180^{o} - 120^{o}}{2} = 30^{o} \\ y^{o} = 180^{o} - 120^{o} \\ = 60^{o} \\ x^{o} + y^{o} = 30^{o} + 60^{o} \\ = 90^{o} \\ x + y = 90 \end{array}

$\begin{array}{l} Size of each interior angle of a regular hexagon \\ = \frac{(6 - 2) \times 180^{o}}{6} \\ = 120^{o} \\ x^{o} = \frac{180^{o} - 120^{o}}{2} = 30^{o} \\ y^{o} = 180^{o} - 120^{o} \\ = 60^{o} \\ x^{o} + y^{o} = 30^{o} + 60^{o} \\ = 90^{o} \\ x + y = 90 \end{array}$

3 thoughts on “4.2.1 Polygons, PT3 Practice”

Xiying

April 30, 2019 at 8:54 am | Reply

answer in Question 1 is error. The answer is 115.
- myhometuition
  
  September 17, 2019 at 7:47 am | Reply
  
  Thanks for pointing out our mistake.
  We had done the correction accordingly.
chia zhong hern

September 26, 2020 at 7:16 am | Reply

I like these questions .

3 thoughts on “4.2.1 Polygons, PT3 Practice”

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