Today’s word problem is deliciously seedy. Strawberries are strange. They are the only fruits with seeds on the outside. Can you estimate the number of seeds on our strawberry?
A strawberry has 1 seed per 5 square millimeters on its surface. Approximate the strawberry as a cone whose height is 36 mm and whose diameter at its widest is 54 mm. Round your answer to the nearest whole number.
It’s instructive to follow the video and create cones of your own. It’s easy. Get tactile and it will aid in visualizing and understanding the relationships that follow. Or come into Mathnasium and we’re happy to explain. We’re overflowing with scrap paper for scribbling, illustrating, slicing and dicing.
From the video above, we’ve learned that a cone is formed from a segment of a circle (A). When the circle-segment (A) is rolled up into a cone, the base of the cone forms a smaller circle (B). The area of the circle-segment (A) is proportional to its outside edge (E). It’s easiest to visualize this proportion when you fold a cone as shown in the video.
The base (B) is the widest part of the cone. The base (B) is also a smaller circle, and the circumference of B is also the outside edge (E) of the circle-segment (A).
We’re almost done solving for the area of a cone. We’ll need to know some elementary facts about a circle of radius r.
Let the radius of the large circle-segment (A) be big ‘R’ and the radius of the smaller base circle (B) be little ‘r’. The circumference of the complete circle (A) is C = 2πR. The outside edge of the circle-segment (A) is the circumference of the base (B), or E = 2πr. The area of the circle-segment (A) is the proportion (E/C )( πR2) = (2πr/2πR)( πR2) = πrR. The area of the base is πr2 and hence the cone’s base and top area combined = πr2 + πrR = πr(r + R).
But there is one last puzzle piece. We’re not given the radius (R) of the circle-segment (A). Instead we’re given the height of the cone (h) that is measured from the cone’s peak directly to the center of its base. Look sideways at the cone you’ve made. You’ll see that from the side, the cone is an isosceles triangle. The height is the triangle’s bisector from its peak to the middle of the base, and it forms a right-triangle of height (h), leg (r), and hypotenuse that is actually the radius (R) of the large circle-segment (A). Using Pythagoras theorem, R2 = h2 + r2, or R = sqrt(h2 + r2).
Hence the area of the cone of height (h) and base radius (r) = πr(r + sqrt(h2 + r2)).
Our strawberry has measurements r = half of 54 mm = 27 mm. h = 36 mm. Plugging in, we get the area of the cone is 1944π mm2. Divided into 5 mm2 segments, we get 388π or about 1,221 segments or seeds.
Fun fact, an average strawberry has about 200 seeds. That mean our strawberry is one huge deliciously seedy anomaly!
Contact:
Ruby Yao and Benedict Zoe, Mathnasium of Fort Lee
201-969-6284 (WOW-MATH), [email protected]
246 Main St. #A
Fort Lee, NJ 07024
Happily serving communities of Cliffside Park, Edgewater, Fort Lee, Leonia, Palisades Park, North Bergen, West New York, and Fairview.
(201) 461-6284
(201) 969-6284
