Peter weighs 156 pounds, but he would like to wrestle in a lower weight class. He loses 4 pounds one week, gains back 2 pounds t
2 answers:
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Please see the figure. We'll first work out half the area of the rounded triangle, half the unshaded part, then double it, then subtract it from the big square.
Half the area is the circular sector PTQ (with center P, arc TQ) minus the right triangle PUT.
A/2 = area(sector PTQ) - area(triangle PUT)
The triangle is half of equilateral triangle PQT, so a 30/60/90 right triangle so we know the sides are in ratio 1:√3:2 so
TU = (7/2)√3
area(PUT) = (1/2) (7/2)(7/2)√3 = (49/8)√3
area(sector PTQ) = (angle TQP / 360°) πr^2
We know angle TQP is 60° because TQP is equilateral. r=7.
area(sector PTQ) = (60°/360°) π (7²) = 49π/6
Putting it together,
A/2 = area(sector PTQ) - area(triangle PUT)
A = 2(49π/6 - (49/8)√3)
A = 49(π/3 - √3/4) square cm
I hate ruining a nice exact answer with an approximation, but they seem to be asking.
A ≈ 30.095057615914535
Check:
I'm not sure how to check it. I'd estimate it's about 25% bigger than equilateral triangle PQT with area (√3/4)7² ≈ 21.2, so around 27. 30 seems reasonable.
Now the real area we seek is the big square PQRS minus A, so
area = 7² - 30.095057615914535 = 18.904942384086 sq cm
They want square meters for some reason; we scale by (1/100)²
Answer: 0.00189 square meters
