Class 8 Maths


Quadrilaterals

Exercise 3.1 Part 2

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

Quadrilateral

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)

Quadrilateral

Solution:

Quadrilateral

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)

Pentagon

Solution:Solution

Pentagon

We know that angle sum of a pentagon = 540o
`110° + 120° + 30° + x + x = 540°`
Or, `260° + 2x = 540°`
Or, `2x = 540° - 260° = 280°`
Or, `x = 280°÷2 = 140°`

Question 6 (d)

Pentagon

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:

Triangle

Solution:

Triangle

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°`

Quadrilateral

Solution:

Quadrilateral

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°`