Question

A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section \(\sigma\) of a blueberry and the "target" density \(n_{\text {tar }}\) (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of \(\sigma\) and \(n_{\text {tar }}\) and then numerically)?

Short answer

The probability is approximately 0.0267 or 2.67%.

Step-by-step solution

Finding the Cross Section of a Blueberry

The cross section of one blueberry is the area of a circle with diameter 1 cm. To find the area, we use the formula for the area of a circle, which is \( \pi r^2 \), where \( r \) is the radius. Since the diameter is 1 cm, the radius \( r \) is 0.5 cm. Calculate the cross-sectional area: \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \).

Calculating the Area of the Pancake

The pancake is also circular, with a diameter of 15 cm, giving it a radius of 7.5 cm. Use the formula for the area of a circle: \( A = \pi r^2 \). Substitute the radius to find the area: \( A = \pi \times (7.5)^2 = 56.25\pi \text{ cm}^2 \).

Determining Target Density

Target density \( n_{\text{tar}} \) is defined as the number of blueberries per unit area of pancake. Since there are 6 blueberries, the target density is calculated as: \( n_{\text{tar}} = \frac{6}{56.25\pi} \text{ berries/cm}^2 \approx \frac{6}{176.715} \approx 0.03396 \text{ berries/cm}^2 \).

Computing the Probability of Hitting a Berry

To find the probability that a skewer hits a blueberry, we multiply the cross-sectional area \( \sigma \) of a berry by the target density \( n_{\text{tar}} \): \( P = \sigma \times n_{\text{tar}} = \frac{\pi}{4} \times 0.03396 \approx 0.0267 \).

Conclusion

The final probability that a skewer jabbed at random hits a berry within the pancake is approximately 0.0267, or 2.67%.

Key concepts

cross section

In geometry, a cross section is the area of an intersection made between a solid object and a plane. Imagine slicing through an object like a pancake. The resulting shape you see would be its cross section. In this exercise, we're interested in the cross section of a blueberry. A blueberry is roughly spherical, so its cross section—when viewed from above, as in this exercise—is a circle.

To find the cross section, we apply the formula for the area of a circle: \( A = \pi r^2 \). Here, \( r \) is the radius of the circle. Since the blueberry’s diameter is 1 cm, its radius is half of this, or 0.5 cm. Therefore, the cross-sectional area \( \sigma \) of a blueberry is calculated as:

• \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \)

This gives us the exact area of a typical blueberry's cross section when cut horizontally at its widest diameter.

target density

Target density is a useful measurement, especially in statistics and probability, for determining how densely packed objects are in a given area. It helps in understanding how likely it is for a certain outcome to occur—like hitting a blueberry with a skewer in a pancake.

In our context, target density \( n_{\text{tar}} \) is calculated by counting how many blueberries are spread across the pancake's total area. The pancake, having a diameter of 15 cm, forms a large circle with a calculated area of \( 56.25\pi \text{ cm}^2 \).

• The number of blueberries is given as 6.
• Therefore, the target density is \( n_{\text{tar}} = \frac{6}{56.25\pi} \approx 0.03396 \text{ berries/cm}^2 \).

This tells us that, on average, there are approximately 0.034 blueberries on every square centimeter of the pancake. This density plays a crucial role in determining the probability of a skewer hitting a berry.

circle area formula

The area of a circle is one of the most essential geometric formulas. Given a circle’s radius \( r \), the area \( A \) can be found using the formula: \( A = \pi r^2 \). This formula is applied both in finding the cross section of a blueberry and the full area of the pancake in the exercise.

For the blueberry, with a radius of 0.5 cm, the cross-sectional area becomes \( \sigma = \pi \times (0.5)^2 = \frac{\pi}{4} \text{ cm}^2 \). For the pancake, the radius is half of 15 cm, which is 7.5 cm. Thus, the pancake area is:

• \( A = \pi \times (7.5)^2 = 56.25\pi \text{ cm}^2 \).

Understanding and accurately applying the circle area formula is essential in many math problems involving circular shapes, whether those shapes appear alone or overlap with others in more complex arrangements.