The area of the center section when the leaves are down is evaluated being of 932.84 in.²
What is Pythagoras Theorem?
If ABC is a triangle with AC as the hypotenuse and angle B with 90 degrees then we have:
[tex]|AC|^2 = |AB|^2 + |BC|^2[/tex]
where |AB| = length of line segment AB. (AB and BC are rest of the two sides of that triangle ABC, AC being the hypotenuse).
The simplified figure is attached below.
We need to find the area PQRS.
Area(PQRS) = 4(Area(DOQ) + Area(QOB)) (by symmetry.
The line OD and OT are perpendiculars on those 38 inch chords, and therefore D and T are bisecting them(by a theorem that perpendicular from center of a circle to its chords bisect them).
Using Pythagoras theorem, we get:
[tex]|DO| = \sqrt{|OQ|^2 -|DQ|^2 } = \sqrt{22^2 - 19^2} = \sqrt{123}[/tex]
Thus, we get:
[tex]\text{Area DOQ} = \dfrac{1}{2} \times 19 \times \sqrt{123} \approx 105.36 \: \rm in^2[/tex]
Since angle ∠DQO and ∠QOB are alternate interior angle, they are of same measure.
From triangle DOQ, from the perspective of ∠DQO, we get the sine ration as:
[tex]\sin(m\angle DQO) = \dfrac{|DO|}{|OQ|}\\m\angle DOQ = \sin^{-1}\left( \dfrac{\sqrt{123}}{22}\right) \approx 30.27^\circ \: \text{(chose acute angle)}[/tex]
Thus, we get:
[tex]m\angle QOB \approx 30.27^\circ[/tex]
360 degrees cover full area of circle,
1 degree will cover full area /360 of the circle
Thus, we get:
Aera of the sector QOB = Area covered by central angle ∠QOB[tex]\approx 30.27 \times \dfrac{\pi (22)^2}{360} \approx 127.85 \: \rm in^2[/tex]
Thus, as we have:
Area(PQRS) = 4(Area(DOQ) + Area(QOB))
we get:
Area(PQRS) ≈ 4(105.36 + 127.85) ≈ 932.84 in.²
Thus, the area of the center section when the leaves are down is evaluated being of 932.84 in.²
Learn more about Pythagoras theorem here:
https://brainly.com/question/12105522
