Geometric Probability Worksheet

Calculate the probability that a randomly thrown dart lands in a shaded region by forming the ratio of shaded area to total area

Transfer the same ratio idea to one dimension, comparing a shaded length such as a 13-unit hypotenuse or a 6π in arc against a total perimeter

Use the relationship shaded area + unshaded area = total area to reach an answer through whichever region is easier to compute

Write square roots in simplified radical form and recognise which of √40 to √49 are already simplified

Solve a right triangle for unknown sides, area and perimeter using cosine and sine of a given 28° angle, rounded to the nearest hundredth with units

The sheet opens with a blank "Appoint a reader" line, so decide before handing it out whether groups will name that reader or the line will simply be left unused

Problem 6 carries its own hint, telling students the radius of one of four circles across a 24 cm square is "NOT 12 cm"; it works well as the figure you model on the board first

Item 11 asks for the words "ALREADY SIMPLIFIED" on any radical that cannot be reduced, so remind students that leaving the entry blank is not the same response

Answers on the worked pages are given in exact form and then as a decimal, which lets you accept either 16 - 3π over 16 or 0.411 depending on what the class is practising

Problems 12 through 15 sit on the last page and can be assigned separately as a special-right-triangle and radicals review without touching the probability items

Geometric probability from area — students compute the shaded-to-total ratio for figures such as a 6 in circle inscribed in a square (36π/144) and four circles packed inside a 24 cm square (576 - 108π over 576).

Geometric probability from length — problems 7 to 9 switch the ratio to one dimension: a 13-unit hypotenuse against a 30-unit perimeter, a 6π in arc against a (12 + 6π) in semicircle perimeter, and 6 m of a 13 m segment.

Working with the complement — problem 6 directs students to decide whether the shaded or unshaded region is easier to calculate, then recover the other from the total 576 cm² square.

Simplifying radicals — item 11 has students reduce √40 to 2√10, √44 to 2√11 and √48 to 4√3, and mark the primes such as √41 and √47 as already simplified.

Right-triangle measurement — problems 14 and 15 apply the 30-60-90 and 45-45-90 side ratios and then cos 28° and sin 28° to find a = 6.18 ft, b = 3.29 ft, an area of 10.16 ft² and a perimeter of 16.47 ft.