Maralyn places her lawn sprinklers at the vertices of a tria

Maralyn places her lawn sprinklers at the vertices of a triangle that has sides 9m, 10m, and 11m. The sprinklers water in circular patterns with radii of 4, 5, and 6 meters. No area is watered by more than one sprinkler. The sprinklers do not water outside of the triangle. Draw a picture of this situation to scale. Label as much as possible. You can use the back of the paper if you wish. What is the total area that is watered? ___ What amount of area inside the triangle is not watered by any of the three sprinklers? ___ what percent of the whole triangle area is not watered? ___

Solution

Sides: a = 9, b = 10, c = 11
Angles A, B, C are across from sides a, b, c respectively.

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Total area watered
= (4)² + (5)² + (6)²
= 77 m^2

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Find angles A, B, C, using law of cosines:

cosA = (b²+c²a²)/(2bc) = (10²+11²9²)/(2×10×11) = 7/11
A = cos¹(7/11) 50.48°

cosB = (a²+c²b²)/(2ac) = (9²+11²10²)/(2×9×11) = 17/33
B = cos¹(17/33) 58.99°

cosC = (a²+b²c²)/(2ab) = (9²+10²11²)/(2×9×10) = 1/3
C = cos¹(1/3) 70.53°

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Area of triangle
= 1/2 * a * b * sinC
= 1/2 * 9 * 10 * 22/3
= 302

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Now we find areas of sectors inside triangle that are watered by each sprinkler.

Since a = 9, then angles B and C will have sprinklers with radii 4 and 5 in some order.
Since b = 10, then angles A and C will have sprinklers with radii 4 and 6 in some order.
Since c = 11, then angles A and B will have sprinklers with radii 5 and 6 in some order.

A - sprinkler with radius 6
B - sprinkler with radius 5
C - sprinkler with radius 4

Area of sector with central angle A
= A/360 * (6)² = cos¹(7/11)/360 * 36 = (/10) cos¹(7/11)

Area of sector with central angle B
= A/360 * (5)² = cos¹(17/33)/360 * 25 = (5/72) cos¹(17/33)

Area of sector with central angle C
= A/360 * (4)² = cos¹(1/3)/360 * 16 = (2/45) cos¹(1/3)

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Area inside triangle that is not watered by any of the three sprinklers
= 302 (/10) cos¹(7/11) (5/72) cos¹(17/33) (2/45) cos¹(1/3)
3.850198786 m^2