Let AB be the parabolic arch having O as the vertex and OY as the axis.
The parabola is of the form x2 = 4ay.
Now, CD = 5 m ⇒ OD = 2.5 m
BD = 10 m
⇒ Coordinates of point B are (2.5, 10)
Since the point B lies on the parabola x2 = 4ay

∴ (2.5)2 = 4a × 10 ⇒ a =

6.25

40

=

625

4000

=

5

32

∴ Equation of parabola is x2 = 4 ×

5

32

y
⇒ x2 =

5

8

y
Let PQ = d ⇒ NQ =

d

2

∴ Coordinates of point Q are (

d

2

,2)
Since point Q lies on the parabola x2 =

5

8

y
∴ (

d

2

)2 =

5

8

× 2 ⇒

d2

4

=

5

4

⇒ d2 = 5 ⇒ d = √5
Thus width of arch = √5 m = 2.23 m approx.