Geometric Probability Day 2 worksheet with problems and answer bank.

Let’s solve each problem one by one. We’re finding the probability that a randomly chosen point lies in the shaded region. That means:

> Probability = (Area of shaded region) ÷ (Total area of the figure)

We’ll calculate both areas for each figure, then divide.

---

Problem 1: Right triangle with smaller right triangle inside

- Big triangle: base = 12, height = 5 → Area = (1/2) × 12 × 5 = 30
- Small unshaded triangle: base = 4, height = 5 → Area = (1/2) × 4 × 5 = 10
- Shaded area = Total – Unshaded = 30 – 10 = 20
- Probability = 20 / 30 = 2/3 ≈ 66.67%

Wait — but looking at the diagram again: actually, the small triangle is *inside* and not overlapping? No — it looks like the big triangle has a vertical line splitting it into two parts: left part (base 4) is unshaded, right part (base 8) is shaded? Wait — no, let me re-read.

Actually, from the image description: “Find the probability that a randomly chosen point lies in the shaded region.”

In Problem 1: It's a large right triangle with legs 12 and 5. Inside, there’s a smaller right triangle on the left with base 4 and same height 5 — and that small triangle is UNSHADED. The rest (the trapezoid or larger triangle minus small one) is SHADED.

So yes:

Shaded area = total area – unshaded area = 30 – 10 = 20

Probability = 20/30 = 2/3 → which is about 66.67%, but that’s not in the answer bank. Hmm.

Wait — maybe I misread the diagram. Let me check again.

Alternative interpretation: Maybe the shaded region is the SMALL triangle? But the problem says “shaded region” — and typically in such problems, if not specified, we assume what’s colored is shaded. Since this is text-based, I have to rely on standard conventions.

But wait — looking at the answer bank: options are 5%, 9%, 25%, 30%, 2.26%, 42.86%, 35%, 43%

None match 66.67%. So perhaps I got it backwards.

Maybe the SMALL triangle is SHADED? Let’s try that.

If small triangle (area 10) is shaded, then probability = 10/30 = 1/3 ≈ 33.33% — still not in bank.

Wait — another possibility: maybe the figure is divided differently. Perhaps the 4 and 12 are along the base, and the height is shared.

Big triangle: base 12, height 5 → area 30

The unshaded part is a triangle with base 4, height 5 → area 10

So shaded is 20 → 20/30 = 2/3 — not matching.

Unless... is the shaded region the quadrilateral? Or maybe I need to look at problem 2 first.

---

Problem 2: Trapezoid with rectangle inside

Figure: A trapezoid with top base 20, bottom base 8, and height 8? Wait — labels: top side 20, bottom side 8, and vertical sides labeled 8? Actually, it looks like a trapezoid with parallel sides 20 (top) and 8 (bottom), and height 8? But then there’s a rectangle inside with width 8 and height 8? And the shaded regions are the two triangles on the sides.

Let me parse:

- The whole figure is a trapezoid: bases 20 and 8, height 8.
- Area of trapezoid = (1/2) × (sum of bases) × height = (1/2)(20 + 8) × 8 = (1/2)(28)(8) = 14 × 8 = 112

Inside, there’s a rectangle of width 8 and height 8 → area = 8 × 8 = 64

Then the shaded regions are the two triangles on the left and right.

Each triangle: base = (20 - 8)/2 = 6, height = 8 → area per triangle = (1/2)×6×8 = 24

Two triangles: 24 × 2 = 48

So shaded area = 48

Total area = 112

Probability = 48 / 112 = simplify: divide numerator and denominator by 16 → 3/7 ≈ 0.4286 → 42.86%

That’s in the answer bank! So Problem 2 is 42.86%

Okay, so now back to Problem 1 — maybe I misinterpreted.

Perhaps in Problem 1, the shaded region is the small triangle? But 10/30=33.33% not in bank.

Wait — another thought: maybe the big triangle is not 12x5, but the 12 is the hypotenuse? No, it’s drawn as legs.

Or perhaps the 4 and 12 are segments on the base, and the height is different.

Let me think differently. Maybe the figure is a right triangle with legs 5 and 12, so area 30. Then a line is drawn from the right angle vertex to a point on the hypotenuse? No, the diagram shows a vertical line from the top vertex down to the base, dividing the base into 4 and 8.

So the left part is a triangle with base 4, height 5 — area 10.

Right part is a triangle with base 8, height 5 — area 20.

If the right part is shaded, then shaded area = 20, total = 30, prob = 2/3 — not in bank.

But if the left part is shaded, prob = 10/30 = 1/3 — not in bank.

Unless... is the shaded region something else? Perhaps the entire figure is not the big triangle, but only the part shown? I'm stuck.

Let me skip to Problem 3.

---

Problem 3: Circle with sector shaded

Radius = 3.7 in. Angle of sector = 110 degrees.

Area of circle = πr² = π*(3.7)^2 = π*13.69 ≈ 3.1416*13.69 ≈ let's calculate:

3.1416 * 13 = 40.8408

3.1416 * 0.69 = approximately 2.1677

Total ≈ 43.0085 in²

Area of sector = (θ/360) * πr² = (110/360) * π*(3.7)^2

First, 110/360 = 11/36 ≈ 0.3056

So sector area ≈ 0.3056 * 43.0085 ≈ let's compute:

0.3 * 43.0085 = 12.90255

0.0056 * 43.0085 ≈ 0.2408

Total ≈ 13.14335

Probability = sector area / circle area = 110/360 = 11/36 ≈ 0.3056 → 30.56%, close to 30% or 35%? 30.56% is closest to 30%, but let's see exact fraction.

11/36 = ? As percentage: (11/36)*100 = 1100/36 = 275/9 ≈ 30.555...%

Answer bank has 30% and 35%. 30.56% is closer to 30%, but perhaps they want exact.

Wait — maybe I should keep it fractional.

Probability = 110/360 = 11/36 ≈ 30.56% — but 30% is an option. However, 42.86% was exact for problem 2, so perhaps here too.

But 11/36 is exactly 30.555...%, which rounds to 31%, not in bank. Options are 30% or 35%. Maybe I miscalculated.

Another thought: is the radius 3.7 or diameter? The label says "3.7 in." next to the radius arrow, so it's radius.

Perhaps the angle is not 110, but let's assume it is.

Maybe for probability, since it's uniform, it's just angle/360, so 110/360 = 11/36.

Now, 11/36 as decimal is 0.30555..., so 30.56%. Closest in bank is 30%, but let's see other problems.

---

Problem 4: Two concentric circles

Outer circle radius = 9, inner circle radius = 3.

Shaded region is the ring between them? Or the inner circle?

From typical diagrams, often the annulus (ring) is shaded, but here it might be the inner circle.

Label: "3" for inner radius, "9" for outer radius.

If shaded is the inner circle: area = π*3² = 9π

Total area = π*9² = 81π

Probability = 9π / 81π = 1/9 ≈ 0.1111 = 11.11% — not in bank.

If shaded is the ring: area = 81π - 9π = 72π

Probability = 72/81 = 8/9 ≈ 88.89% — not in bank.

But answer bank has 2.26%, 5%, etc. Too small.

Unless... is the shaded region a small part? The diagram might show a small sector or something.

Re-reading: "a circle with a smaller circle inside, and a dot in the center, and '3' and '9' labeled."

Perhaps the shaded region is a small circle of radius 3, but within a larger context? No.

Another idea: maybe the "3" is not radius, but diameter? But usually labeled as radius.

Or perhaps the figure is not full circles, but arcs.

I recall that in some problems, if there's a point selected in the large circle, and shaded is the small circle, prob = (r_small/r_large)^2 = (3/9)^2 = (1/3)^2 = 1/9 ≈ 11.11% — not in bank.

But 2.26% is there. How to get that?

(3/9)^2 = 1/9 ≈ 11.11%, not 2.26%.

2.26% is roughly 1/44.25, not nice.

Perhaps the shaded region is a segment or something else.

Let's look at Problem 5.

---

Problem 5: Regular pentagon with inscribed circle or something

It says "regular pentagon" with apothem or radius labeled 5, and a circle inside? Or the pentagon is shaded, and there's a circle?

From description: "pentagon with a circle inside, and '5' labeled from center to vertex or to side?"

Typically, for regular polygons, if a circle is inscribed, the radius given might be apothem.

Assume it's a regular pentagon with apothem 5, and the shaded region is the pentagon itself, and the circle is unshaded? Or vice versa.

The problem is to find probability that a random point in the figure is in the shaded region. What is the "figure"? Probably the pentagon, and shaded is part of it.

Commonly, in such diagrams, the pentagon is the outer shape, and there's an inscribed circle, and the shaded region might be the circle or the segments.

But let's assume the figure is the pentagon, and shaded is the inscribed circle.

For a regular pentagon with apothem a = 5.

First, area of pentagon = (1/2) * perimeter * apothem

Need side length. For regular pentagon, relationship between apothem and side.

The apothem a = (s / 2) / tan(π/5) , where s is side length.

tan(36°) since π/5 radians = 36 degrees.

tan(36°) ≈ 0.7265

So a = (s/2) / tan(36°) => s = 2 * a * tan(36°) = 2 * 5 * 0.7265 = 10 * 0.7265 = 7.265

Perimeter = 5 * s = 5 * 7.265 = 36.325

Area of pentagon = (1/2) * 36.325 * 5 = (1/2)*181.625 = 90.8125

Now, the inscribed circle has radius equal to apothem = 5, so area = π*5² = 25π ≈ 78.54

If shaded is the circle, probability = 78.54 / 90.8125 ≈ 0.8648 = 86.48% — not in bank.

If shaded is the pentagon minus the circle, i.e., the segments, then area = 90.8125 - 78.54 = 12.2725

Probability = 12.2725 / 90.8125 ≈ 0.1351 = 13.51% — not in bank.

But answer bank has 2.26%, which is very small.

Perhaps the "5" is the radius to vertex, not apothem.

Let me try that.

Suppose the distance from center to vertex (radius) R = 5.

For regular pentagon, area = (5/2) * R^2 * sin(72°) because each triangle from center has area (1/2)*R*R*sin(theta), theta=72° for pentagon.

Sin(72°) ≈ 0.9511

Area = (5/2) * 25 * 0.9511 = (2.5) * 25 * 0.9511 = 62.5 * 0.9511 ≈ 59.44375

Inscribed circle radius (apothem) a = R * cos(36°) because in the triangle, apothem = R * cos(36°)

Cos(36°) ≈ 0.8090

So a = 5 * 0.8090 = 4.045

Area of inscribed circle = π*a^2 = π*(4.045)^2 ≈ 3.1416*16.362025 ≈ 51.40

If shaded is the circle, prob = 51.40 / 59.44 ≈ 0.8648 again — same as before.

Not helping.

Perhaps the shaded region is a small part, like one triangle or something.

Another idea: in Problem 4, if the shaded region is a small circle of radius 3, but the total area is not the large circle, but something else.

Or perhaps in Problem 4, the "9" is diameter, not radius.

Let me try that.

Suppose in Problem 4, outer diameter is 9, so radius 4.5, inner radius 3.

Then area large circle = π*(4.5)^2 = 20.25π

Area small circle = π*9 = 9π

If shaded is small circle, prob = 9π / 20.25π = 9/20.25 = 900/2025 = 36/81 = 4/9 ≈ 44.44% — not in bank.

If shaded is ring, prob = (20.25 - 9)/20.25 = 11.25/20.25 = 1125/2025 = 45/81 = 5/9 ≈ 55.56% — not in bank.

Still not.

Let's go back to Problem 1. Perhaps the shaded region is the small triangle, and the total area is not the big triangle, but only the part from x=0 to x=12, but that doesn't make sense.

Another thought: in Problem 1, the big triangle has legs 5 and 12, area 30. The small triangle has base 4, but is it similar? Yes, because same height, so area ratio (4/12)^2 = (1/3)^2 = 1/9, but that's for similar figures, but here the small triangle is not similar to the big one if it's cut vertically.

In my initial calculation, the small triangle has base 4, height 5, area 10, big has base 12, height 5, area 30, so ratio 1/3.

But 1/3 is 33.33%, not in bank.

Unless the shaded region is the quadrilateral formed by the cut, but that would be the same as the right part, area 20, prob 2/3.

Perhaps for Problem 1, the answer is 30%, and they approximate.

But let's look at the answer bank: 5%, 9%, 25%, 30%, 2.26%, 42.86%, 35%, 43%

We have 42.86% for Problem 2.

For Problem 3, 110/360 = 11/36 ≈ 30.56%, so perhaps 30% or 35%. 30.56% is closer to 31%, but 30% is there.

For Problem 4, let's try if the shaded region is a small sector or something.

Perhaps in Problem 4, the "3" and "9" are radii, but the shaded region is the area between two radii or something.

Another idea: perhaps the figure is a circle of radius 9, and there is a smaller circle of radius 3 centered at the same point, and the shaded region is the smaller circle, but then prob = (3/9)^2 = 1/9 ≈ 11.11%, not in bank.

But 2.26% is there. 2.26% is approximately 1/44.25, or (1/6.65)^2, not nice.

2.26% = 0.0226, sqrt(0.0226) = 0.1503, so if r_small / r_large = 0.1503, then r_small = 0.1503*9 = 1.3527, not 3.

Perhaps the 3 is not radius, but diameter of the small circle.

Suppose small circle diameter 3, so radius 1.5, large circle radius 9.

Then area small = π*(1.5)^2 = 2.25π

Area large = π*81 = 81π

Prob = 2.25/81 = 225/8100 = 9/324 = 1/36 ≈ 0.02778 = 2.778% — close to 2.26%? Not really.

2.778% vs 2.26%, not close.

If large radius is 9, small radius is 3, but shaded is not the circle, but a point or something.

Perhaps in Problem 4, the shaded region is a small circle of radius 3, but the total area is the area of the large circle minus something, but that doesn't make sense.

Let's consider that in some problems, the "figure" might be the annulus, but then total area is 81π - 9π = 72π, shaded is 9π, prob = 9/72 = 1/8 = 12.5% — not in bank.

I recall that 2.26% might come from (3/9)^4 or something, but that's unlikely.

Another thought: perhaps for Problem 4, the "3" and "9" are not radii, but diameters, and the shaded region is the small circle, but then as above.

Let's calculate (3/9)^2 = 1/9 = 11.11%, not 2.26%.

2.26% = 0.0226, and 0.0226 * 81π / π = 0.0226*81 = 1.8306, so area shaded = 1.8306, while π*3^2 = 28.27, not match.

Perhaps the shaded region is a square or something inside.

I think I need to guess based on common problems.

Let me try Problem 5 again.

Suppose in Problem 5, the regular pentagon has side length or something, but labeled "5" from center to side, so apothem 5.

Then as before, area pentagon = (1/2)*perimeter*apothem.

Side s = 2 * a * tan(36°) = 2*5* tan(36°)

Tan(36°) = sqrt(5-2sqrt(5)) / something, but numerically tan(36°) = 0.726542528

s = 2*5*0.726542528 = 7.26542528

Perimeter = 5*7.26542528 = 36.3271264

Area = (1/2)*36.3271264*5 = 90.817816

Now, if there is an inscribed circle, radius = apothem = 5, area = 25π ≈ 78.53981634

If the shaded region is the pentagon minus the circle, area = 90.817816 - 78.53981634 = 12.278

Probability = 12.278 / 90.817816 ≈ 0.1352 = 13.52% — not in bank.

But 2.26% is there. 2.26% of 90.817816 = 0.0226*90.817816 ≈ 2.052, which is not matching any obvious area.

Perhaps the "5" is the side length.

Assume side length s = 5 for pentagon.

Then area of regular pentagon = (1/4) * sqrt(5(5+2sqrt(5)) ) * s^2

Formula: area = (5/4) * s^2 / tan(36°) or standard formula.

Area = (5 * s^2) / (4 * tan(36°))

Tan(36°) ≈ 0.7265

s=5, s^2=25

Area = (5*25)/(4*0.7265) = 125 / 2.906 = approximately 43.01

More accurately, tan(36°) = sqrt(5-2sqrt(5)) but use calculator value.

Actual formula: area = (1/4) * sqrt(5(5+2sqrt(5)) ) * s^2

sqrt(5)≈2.236, so 5+2*2.236 = 5+4.472 = 9.472

5*9.472 = 47.36

sqrt(47.36) ≈ 6.881

Then area = (1/4)*6.881*25 = (1/4)*172.025 = 43.00625

So area ≈ 43.006

Now, if there is an inscribed circle, radius r = (s/2) / tan(36°) = (2.5) / 0.7265 ≈ 3.441

Area circle = π*r^2 ≈ 3.1416*11.84 ≈ 37.2

If shaded is the circle, prob = 37.2 / 43.006 ≈ 0.865 = 86.5% — not in bank.

If shaded is the pentagon, and the circle is not part of the figure, but the figure is the pentagon, and shaded is the whole thing, prob=1, not in bank.

Perhaps the shaded region is one of the five triangles from center.

Each triangle has area (1/2)*r^2* sin(72°) , but r is distance to vertex.

If apothem is 5, then r = 5 / cos(36°) ≈ 5 / 0.8090 = 6.18

Then area of one triangle = (1/2)*r^2* sin(72°) = (1/2)*(6.18)^2*0.9511 ≈ (1/2)*38.1924*0.9511 ≈ (1/2)*36.32 = 18.16

Total area 5*18.16 = 90.8, same as before.

One triangle area 18.16, prob = 18.16/90.8 = 0.2 = 20% — not in bank.

2.26% is very small, so perhaps for Problem 4, if the shaded region is a small circle of radius 3, but the total area is a square or something, but the figure is a circle.

Let's look online or recall that sometimes in such problems, for concentric circles, if the shaded region is the small circle, and they ask for probability, it's (r/R)^2.

But here (3/9)^2 = 1/9 = 11.11%, not in bank.

Unless the 9 is not radius, but diameter.

Suppose large circle diameter 9, so radius 4.5, small circle radius 3.

Then prob = (3/4.5)^2 = (2/3)^2 = 4/9 ≈ 44.44% — not in bank.

If small circle diameter 3, radius 1.5, large radius 9, prob = (1.5/9)^2 = (1/6)^2 = 1/36 ≈ 2.778% — and 2.778% is close to 2.26%? No, 2.78 vs 2.26, difference of 0.52, not very close.

2.26% = 0.0226, 1/44.247, so (r/R)^2 = 0.0226, r/R = sqrt(0.0226) = 0.1503, so if R=9, r=1.3527, not 3.

Perhaps the "3" is the area or something.

Another idea: in Problem 4, the "3" and "9" are not radii, but the shaded region is a sector with angle corresponding to arc length or something.

Perhaps the figure is a circle, and there is a chord or something.

I recall that 2.26% might be for a different problem.

Let's try Problem 1 with different interpretation.

Suppose the big triangle has base 12, height 5, area 30.

The small triangle has base 4, but perhaps it's not with height 5, but with different height.

Or perhaps the 4 and 12 are not on the base, but on the height.

The diagram shows a right triangle with legs vertical and horizontal. Vertical leg 5, horizontal leg 12. Then a vertical line from the top vertex down to the base, at a distance 4 from the left, so it divides the base into 4 and 8.

So the left part is a triangle with base 4, height 5, area 10.

Right part is a triangle with base 8, height 5, area 20.

If the shaded region is the right part, prob = 20/30 = 2/3.

But perhaps in the diagram, the shaded region is the left part, and they want 10/30 = 1/3, and 33.33% is not in bank, but 30% is close.

Or perhaps for Problem 3, 110/360 = 11/36 = 30.555%, and they round to 31%, but 30% is there.

Let's list what we have:

Problem 2: 48/112 = 3/7 ≈ 42.857% -> 42.86% in bank.

Problem 3: 110/360 = 11/36 ≈ 30.555% -> perhaps 30% or 35%. 30.555% is closer to 31%, but 30% is available.

Problem 1: if shaded is small triangle, 10/30 = 33.33% -> not in bank, but 35% is there.

33.33% is closer to 33%, not 35%.

Perhaps for Problem 1, the total area is not 30, but something else.

Another thought: in Problem 1, the big triangle is 5-12-13, area 30, but the small triangle is similar, with base 4, so scale factor 4/12 = 1/3, so area ratio (1/3)^2 = 1/9, so area small = 30/9 = 10/3 ≈ 3.333, then if shaded is small, prob = (10/3)/30 = 10/90 = 1/9 ≈ 11.11% — not in bank.

If the small triangle is not similar, but in this case it is, because same angles.

In my initial setup, the small triangle has the same height, so it is not similar to the big triangle unless the cut is parallel, but here it's vertical, so for a right triangle with right angle at origin, cutting vertically at x=4, then the small triangle has vertices at (0,0), (4,0), (0,5)? No.

Let's define coordinates.

Put the right angle at (0,0), along x-axis to (12,0), along y-axis to (0,5). Then the hypotenuse from (0,5) to (12,0).

Then a vertical line at x=4, from (4,0) to the hypotenuse.

The hypotenuse equation: from (0,5) to (12,0), slope = (0-5)/(12-0) = -5/12

So y = 5 - (5/12)x

At x=4, y = 5 - (5/12)*4 = 5 - 20/12 = 5 - 5/3 = 10/3 ≈ 3.333

So the small triangle on the left is from (0,0) to (4,0) to (0,5)? No, the vertical line is at x=4, from (4,0) to (4,10/3) on the hypotenuse.

So the region to the left of x=4 is a triangle with vertices at (0,0), (4,0), and (0,5)? No, because the hypotenuse is not vertical.

Actually, the left region is a polygon: from (0,0) to (4,0) to (4,10/3) to (0,5) back to (0,0)? That's a quadrilateral.

I think I made a mistake earlier.

In a right triangle with legs on axes, from (0,0) to (12,0) to (0,5), then the hypotenuse from (12,0) to (0,5).

If we draw a vertical line at x=4, it intersects the hypotenuse at (4, y) where y = 5 - (5/12)*4 = 5 - 5/3 = 10/3, as above.

So the left region is a triangle with vertices at (0,0), (4,0), and (0,5)? No, because (0,5) is not connected directly; the boundary is from (0,0) to (0,5) to (4,10/3) to (4,0) back to (0,0)? That's not correct.

Actually, the left region bounded by x=0, x=4, y=0, and the hypotenuse.

So it is a trapezoid or triangle? From x=0 to x=4, the lower bound is y=0, upper bound is the line from (0,5) to (12,0), so y = 5 - (5/12)x.

So the left region is under the line from x=0 to x=4, so it is a triangle only if it goes to (0,5), but at x=0, y=5, at x=4, y=10/3, and at y=0, x=12, so for x from 0 to 4, the region is from y=0 to y=5-(5/12)x.

This is a trapezoid with parallel sides at x=0 and x=4.

At x=0, y from 0 to 5, so height 5.

At x=4, y from 0 to 10/3, so height 10/3.

Width in x is 4.

So area = average height times width = [(5 + 10/3)/2] * 4 = [(15/3 + 10/3)/2] * 4 = (25/3 / 2) * 4 = (25/6) * 4 = 100/6 = 50/3 ≈ 16.6667

Total area of big triangle = (1/2)*12*5 = 30

So if this left region is shaded, area = 50/3 ≈ 16.6667, prob = (50/3)/30 = 50/90 = 5/9 ≈ 55.56% — not in bank.

If the right region is shaded, from x=4 to x=12, at x=4, y=10/3, at x=12, y=0, so it is a triangle with base 8 (from x=4 to x=12), but the height varies.

The right region is a triangle with vertices at (4,0), (12,0), and (4,10/3)? No, because the hypotenuse is from (4,10/3) to (12,0), and the base from (4,0) to (12,0), so it is a triangle with base 8, and height 10/3? No, because the apex is at (4,10/3), but the base is from (4,0) to (12,0), so it's not a standard triangle.

Vertices: (4,0), (12,0), and (4,10/3)? But (4,10/3) is not on the base; the third vertex should be on the hypotenuse, which is (4,10/3) for the left, but for the right region, it is bounded by x=4, y=0, and the hypotenuse from (4,10/3) to (12,0).

So the right region is a triangle with vertices at (4,0), (12,0), and (4,10/3)? No, because (4,10/3) and (4,0) are on the same x, so it would be a line.

Actually, the right region is a triangle with vertices at (4,0), (12,0), and the point where? The hypotenuse from (4,10/3) to (12,0), so the region is from (4,0) to (12,0) to (12,0) wait.

Points: the right region is bounded by:
- from (4,0) to (12,0) along x-axis
- from (12,0) to (4,10/3) along the hypotenuse? No, the hypotenuse is from (0,5) to (12,0), so from (4,10/3) to (12,0) is part of it.
- from (4,10/3) to (4,0) along the vertical line.

So yes, it is a triangle with vertices at (4,0), (12,0), and (4,10/3)? But (4,0) and (4,10/3) have the same x, so the base is from (4,0) to (12,0), length 8, and the height is the y-distance, but the third vertex is at (4,10/3), which is not above the base; it's at the left end.

So this is a right triangle with legs: from (4,0) to (12,0) is 8 units, from (4,0) to (4,10/3) is 10/3 units, but then the hypotenuse would be from (4,10/3) to (12,0), which is correct, but this triangle has area (1/2)*base*height = (1/2)*8*(10/3) = 40/3 ≈ 13.333, but is that correct?

Let's calculate the area using shoelace formula.

Vertices of right region: A(4,0), B(12,0), C(4,10/3)

Shoelace: (4*0 + 12*(10/3) + 4*0) - (0*12 + 0*4 + (10/3)*4) = (0 + 40 + 0) - (0 + 0 + 40/3) = 40 - 40/3 = 80/3

Then area = |80/3| / 2 = 40/3 ≈ 13.333? Shoelace formula is half the absolute value of sum.

Standard shoelace: for points (x1,y1), (x2,y2), (x3,y3), area = |(x1(y2-y3) + x2(y3-y1) + x3(y1-y2))/2|

So A(4,0), B(12,0), C(4,10/3)

Area = |4*(0 - 10/3) + 12*(10/3 - 0) + 4*(0 - 0)| / 2 = |4*(-10/3) + 12*(10/3) + 0| / 2 = |-40/3 + 120/3| / 2 = |80/3| / 2 = 80/6 = 40/3 ≈ 13.333

But earlier total area is 30, left region should be 30 - 13.333 = 16.667, which matches my earlier calculation for left region as 50/3 = 16.6667.

So if the shaded region is the right part, area = 40/3 ≈ 13.333, prob = (40/3)/30 = 40/90 = 4/9 ≈ 44.44% — not in bank.

If shaded is left part, 50/3 / 30 = 50/90 = 5/9 ≈ 55.56% — not in bank.

This is frustrating.

Perhaps in the diagram, the shaded region is the small triangle formed by the cut, but in this case, the cut creates a small triangle on the left only if it's from the right angle, but here it's not.

Another possibility: perhaps the "4" is not on the base, but on the height.

Let's assume that the vertical leg is 5, and on it, at height 4 from the bottom, but the diagram shows "4" on the base.

Perhaps for Problem 1, the answer is 25%, and they have a different interpretation.

Let's look at Problem 4 again.

Suppose in Problem 4, the large circle has radius 9, small circle has radius 3, but the shaded region is the small circle, and they want the probability, but perhaps the "figure" is the large circle, so prob = (3/9)^2 = 1/9 = 11.11%, not in bank.

But 2.26% is there, and (1/6.65)^2 = 0.0226, and 6.65*1.35 = 9, not integer.

Perhaps the 3 and 9 are diameters, and shaded is small circle, so r_small = 1.5, r_large = 4.5, prob = (1.5/4.5)^2 = (1/3)^2 = 1/9 = 11.11% same.

Or if r_small = 3, r_large = 9, but shaded is a different region.

Another idea: perhaps in Problem 4, the "3" is the radius of the small circle, "9" is the radius of the large circle, but the shaded region is the area of the small circle divided