Na figura acima, qual é a área do quarto de círculo?
(In the figure above, what is the area of the quarter circle?)
PCFilho
(pescado no Twitter de Edwin Wallace)
quarta-feira, 14 de outubro de 2020
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First, we note that △JKL is a right triangle and ∠KJL is a right angle. This means that JK is orthogonal to JL. Then we draw segment AJ. Since AJ and AL are radii of the same circle, they are congruent and △AJL is an isosceles triangle. Then we draw the altitude of △AJL that is orthogonal to segment JL and label the its foot P and its point of intersection with KL Q. Since segment AP intersects the base of an isosceles triangle, it bisects both the base and the angle opposite the base. So, segments LP and PJ are congruent (meaning that P is the midpoint of segment JL) and ∠JAP and ∠PAL are congruent (this congruence will be relevant later).
ResponderExcluirSince segment AP is an altitude of △AJL, it is orthogonal to the side it intersects, which is segment JL. Since segments JK and AP are both orthogonal to segment AL, they are parallel. Since segment PQ is part of segment AP, segments PQ and JK are parallel. Since P is the midpoint of segment JL, this also means that Q is the midpoint of segment KL. Now we draw the altitude of △ALQ that is orthogonal to segment AL and label its foot R. Since Q is the midpoint of segment KL, which is also the hypotenuse of △AKL, this means that R is the midpoint of segment AL. So, segments AR and LR are congruent. Since segments RQ and AL are orthogonal, ∠ARQ and ∠LRQ are congruent (since both are right angles). Also, segment RQ is congruent to itself by the Reflexive Property of Congruence. This means that △ARQ and △LRQ are congruent by side-angle-side. Since segments AQ and LQ are corresponding parts of congruent triangles, they are congruent.
We established earlier that ∠JAP and ∠PAL are congruent. Since Q is on segment AP and R is on segment AL, we can also say that ∠JAP and ∠QAR are congruent. Since segments AP and PJ are orthogonal, and segments AR and RQ are orthogonal, that means that ∠APJ and ∠ARQ are both right angles and are therefore congruent. This means that △APJ ~ △ARQ by angle-angle. But this fact isn’t useful unless we know the lengths of some of these segments. We’ve established that P and Q are the midpoints of segments JL and KL, respectively. We also know that JK = 3 and JL = 4. Since (KL)² = (JK)² + (JL)², this means that KL = 5. So:
PJ = (1/2)(JL) = (1/2)(4) = 2
PQ = (1/2)(JK) = (1/2)(3) = 3/2
LQ = (1/2)(KL) = (1/2)(5) = 5/2
AQ = LQ = 5/2
AP = PQ + AQ = 3/2 + 5/2 = 4
AP/PJ = 4/2 = 2
So this means that AR/RQ = 2; that is, AR = 2(RQ). Since (AQ)² = (AR)² + (RQ)², that means that 5(RQ)² = 25/4. This means that RQ = (√5)/2. Since AR is twice RQ, AR = √5. AL is twice AR, so AL = 2√5. Since AL is the radius of the quarter-circle:
Area of quarter-circle = (π/4)(2√5)² = 5π
My answer = 5π square units
Brilliant, Jake! Well done!
ExcluirMy other post contains a small error that I didn't catch before it was published: It says "Since segments JK and AP are both orthogonal to segment AL" when it should say "Since segments JK and AP are both orthogonal to segment JL".
Excluir