Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 months ago

# geometry problem? Relevance

Turn it 45 degrees and imagine that everything is centered at the origin of a coordinate plane

y = sqrt(9 - x^2)

y = (4 * sqrt(2)) / 2 = 2 * sqrt(2)

sqrt(9 - x^2) = 2 * sqrt(2)

9 - x^2 = 8

x^2 = 1

x = -1 , 1

The circle and square intersect at x = -1 and x = 1

What we have now is a circle with a radius of 3, and a triangle that measures 2 on one side and 3 on the other sides

h^2 = 3^2 - (2/2)^2 = 9 - 1 = 8

h = 2 * sqrt(2) (which we already knew)

A[triangle] = (1/2) * 2 * sqrt(2) * 2 = 2 * sqrt(2)

2^2 = 3^2 + 3^2 - 2 * 3 * 3 * cos(t)

4 = 18 - 18 * cos(t)

18 * cos(t) = 14

9 * cos(t) = 7

cos(t) = 7/9

t = cos-1(7/9)

A[sector] = pi * 3^2 * t / (2pi) = (9/2) * t

A[segment] = A[sector] - A[triangle]

A[segment] = (9/2) * cos-1(7/9) - 2 * sqrt(2)

We have 4 of these

4 * (9/2) * cos-1(7/9) - 8 * sqrt(2)

18 * cos-1(7/9) - 8 * sqrt(2)

‭0.92042024136362934505662068842592‬

a = 18 , b = 7 , c = 9 , d = 8

• Let θ be the central angle of each of the congruent segments.

apothem of square = 2√(2)

cos(θ/2) = 2√(2)/3

cos(θ) = 2cos²(θ) - 1 = 2[2√(2)/3]² = 7/9

sin(θ) = √[1 - (7/9)²] = 4√(2)/9

total area of segments

= 4[3²θ/2 - 3²sin(θ)/2]

= 18θ - 18sin(θ)

= 18cos⁻¹(7/9) - 18[4√(2)/9]

= 18cos⁻¹(7/9) - 8√(2)

Assuming b and c are positive, with no common divisors,

a = 18, b = 7, c = 9, d = 8

a + b + c + d = 42