# Quadrilaterals

## Exercise 3.1 Part 2

Question 6: Find the angle measures x in the following figures.

**Solution:** We know that, angle sum of a quadrilateral = 360⁰

So, `50° + 130° + 120°+ x = 360°`

Or, `300°+ x = 360°`

Or, `x = 360°- 300°= 60°`

Question 6 (b)

**Solution:**

We know that, angle sum of a quadrilateral = 360⁰

So, `90° + 60°+ 70°+ x = 360°`

Or, `220° + x = 360°`

Or, `x = 360°- 220°= 140°`

Question 6 (c)

**Solution:**Solution

We know that angle sum of a pentagon = 540^{o}

`110° + 120° + 30° + x + x = 540°`

Or, `260° + 2x = 540°`

Or, `2x = 540° - 260° = 280°`

Or, `x = 280°÷2 = 140°`

Question 6 (d)

**Solution:**Solution: Angle sum of a pentagon `= (5 – 2) xx 180⁰ = 3 xx 180⁰ = 540⁰`

Since, it is a regular pentagon, thus, its angles are equal

So, `x + x + x + x + x = 540°`

Or, `5x = 540°`

Or, `x = 540°÷5 = 108°`

Question 7:

**Solution:**

We know that angle sum of a triangle = 180⁰

Thus, `30⁰ + 90⁰ + C = 180⁰`

Or, `120⁰ + C = 180⁰`

Or, `C = 180⁰ – 120⁰`

Or, `C = 60⁰`

Now, `y = 180° - C`

Or, `y = 180° - 60° = 120°`

Similarly, `z = 180°- 30° = 150°`

Similarly, `x = 180° - 90° = 90°`

Hence, `x + y + z = 90° + 120° + 150°= 360°`

**Alternate method:** We know that sum of external angles of a polygon = 360⁰

Hence, `x + y + z = 180°`

**Solution:**

We know that angle sum of a quadrilateral = 360⁰

`A + 60° + 80° + 120° = 360°`

Or, `A + 260° = 360°`

Or, `A = 360° - 260° = 100°`

Hence, `w = 180° - 100° = 80°`

Similarly, `x = 180° - 120° = 60°`

Similarly, `y = 180° - 80° = 100°`

Similarly, `z = 180° - 60° = 120°`

Hence, `x + y + z + w = 60° + 100° + 120° + 80° = 360°`

**Alternate method:** We know that sum of external angles of a polygon = 360⁰

Hence, `x + y + z + w = 360°`