Question

NO LINKS!!! Find the probability that a randomly chosen point in the figure lies in the shaded region area.​

Answer 1

#36

Area of rectangle

• 12(8)
• 96

Area of 6 circles.

• 6(π)(8/4)²
• 6π(4)
• 24π
• 75.4

Probability

• 75.4/96
• 0.78

#2

Area of whole

• (2+6)(6+10)
• 8(16)
• 144

Area of two rectangles

• 2(10)(2)
• 40

So

Probability

• 144-40/144
• 104/144
• 0.7

#3

Area of whole circle

• π(4+4+2)²
• 100π

Area of inner circle(not shaded)

• π(2)²
• 4π

Area of shaded circle

• π(4+2)²
• 36π

Probability

• 16π-4π/96π
• 12π/96π
• 3/8

Answer 2

Answer:

To find the probability that a randomly chosen point in the figure lies in the shaded region, we need to divide the area of shaded region by the total area of the figure.

Question 36

Total Area

Area of a rectangle = width × length = 8 × 12 = 96 units²

Shaded Region Area

This is made up of 6 congruent circles.

The radius of each circle is 1/6 of the length of the rectangle (or 1/4 of the width).

⇒ radius = 12/6 = 2 units

Area of a circle = πr² = π(2)² = 4π units²

⇒ Shaded region area = 6 circles = 6 × 4π = 24π units²

Probability

= Shaded region area ÷ total area

= 24π ÷ 96

= 0.7853981634...

= 78.5% (3 s.f.)

Question 37

Total Area

Area of a rectangle = width × length = 16 × 8 = 128 units²

Shaded Region Area

The easiest way to calculate this is to subtract the un-shaded areas from the total area:

⇒ Shaded region area = 128 - 2(2 · 10) = 88 units²

Probability

= Shaded region area ÷ total area

= 88 ÷ 128

= 0.6875

= 68.8% (3 s.f.)

Question 38

Total Area

The radius of the largest circle is the sum of the individual given radii.

⇒ Area of a circle = πr² = π(4 + 4 + 2)² = 100π units²

Shaded Region Area

= Area of a circle with radius (4 + 2) - area of circle with radius 2

= π(6)² - π(2)²

= 32π units²

Probability

= Shaded region area ÷ total area

= 32π ÷ 100π

= 0.32

= 32%