See the attached sketch. From their respective corners, each wiper sweeps out a portion of the total area of the windshield equal to the area of a quarter-circle with radius 1.5 m (shaded green), and they overlap in a half-lens-shaped region (shaded red). The area that the wipers together cover is equal to the sum of the green areas minus the area of the red overlap. It's easy to see that this red area is equal to the area of the highlighted circular segment (bordered in blue).
Not sure how exactly you can avoid finding the area of the segment without resorting to trigonometry, though... The radii AC and AD form an angle θ such that the segment has area
1/2 (1.5 m)² (θ - sin(θ))
but you can avoid mentioning the sine function by writing it as
1/2 (1.5 m)² (θ - √3/2)
which follows from
sin(θ) = 2 sin(θ/2) cos(θ/2) = 2 ((BD/2)/AB) ((AC/2)/AB)
and the fact that BD has length equal to the radii, since by the Pythagorean theorem
(AC/2)² + (BD/2)² = AB²
(whether this theorem falls under trigonometry is also up for debate)
You would need trig to find the measure of the angle θ as well, unless you just take for granted the fact that a triangle with side lengths 1.5√3/2, 1.5/2, and 1.5 (i.e. AC/2, BD/2, and AB), since they occur in a ratio of √3 : 1 : 2 is a 30°-60°-90° triangle, so that θ = 60° = π/3 rad.
At any rate, the percentage is about (2×(quarter-circle area) - (segment area))/(rectangle area) = 85.46%, since
• area of rectangle = (1.5 √3 m) • (1.5 m) ≈ 3.897 m²
• area of quarter-circle = π/4 (1.5 m)² ≈ 1.767 m²
• area of segment = 1/2 (1.5 m)² (π/3 - √3/2) ≈ 0.204 m²
