3440.31 – A Lovely Circular Lawn

Tags:Problem Set 10GeometryGeometry-Euclidean

A man has a circular lawn as in the figure below. Strange I know, but what can I do?

This lawn is twelve meters in diameter and is cut by a straight gravel path which is three meters wide (we wonder about this: why so wide, and what does the rest of his property look like?). One edge of the gravel path passes through the center of the lawn (curiouser and curiouser).

How many square meters of grass is in the lawn?

  1. 36π3436 \pi-34
  2. 30π1530 \pi-15
  3. 36π3336 \pi-33
  4. 35π9335 \pi-9\surd 3
  5. 30π9330 \pi-9 \surd 3


First, take the picture we’re given and add more information to make it more useful. We’ll label the two sections that actually contain grass Sections A and B:

Our task is to find the combined area of these two sections. Section A is simply half of the entire circle, so we can easily find its area:

Area of A = πr22=π622=18π\frac{\pi r^2}{2}=\frac{\pi 6^2}{2}=18 \pi

Finding the area of Section B will be a bit trickier. Let’s add some more information to our diagram:

We realize that Section B is a sector of a circle with a triangle cut out:

Area of B = sector AOB - △AOB

Triangle AOB can be further divided into right triangles AOC and AOB. These congruent triangles can be shown to be 30 degrees-60 degrees right triangles. We recall that for every such right triangle, the hypotenuse will be exactly twice the length of the shortest side. We know that line OC, the width of the path, is three meters, and lines OA and OB, the circle’s radius, are six each. Thus, angles AOC and BOC are each 60 degrees, meaning that angle AOB has a measurement of 120 degrees.

We can now calculate the area of the sector:

Area of sector AOB=120°360°πr2=120°360°π62=1336π=12π\frac{120°}{360°}\cdot\pi r^2=\frac{120°}{360°}\cdot \pi 6^2=\frac{1}{3}\cdot 36 \pi=12 \pi

We now need only subtract the area of triangle AOB, and we will have our answer. Because we know two sides of right triangle AOC, we can make use of the Pythagorean Theorem to find the length of side AC:

AC2=OA2OC2AC^2=OA^2- OC^2


AC=27=33AC=\surd 27=3\surd 3

Doubling this to find the length of line ABAB, we now know the height (OCOC) and base (ABAB) of triangle AOB: 3 meters and 63\surd 3 meters respectively, which we use to find the triangle’s area:

Area of AOB=(OC×AB)/2=(3×63)/2=93△AOB=(OC×AB)/2=(3×6\surd 3)/2=9\surd 3

Now that we have all the pieces to this puzzle, we can plug in the numbers we’ve found in order to find our answer:

Area of A + Area of B = 18π18 \pi +sector AOB-△AOB= 18π+12π93=30π9318\pi+12\pi-9\surd 3=30\pi-9\surd 3

Our final area of the grassy area in square meters is 30\pi-93\surd 3. The correct answer, therefore, is e.