Question

A circle with a radius 4 cm long is inscribed within a square (as shown below). Point A starts at the 3-o'clock position and moves CCW along the circle. Point B is fixed at the bottom-left comer of the square. Let o represent the given angle's measure (in radians) where Point A is the terminal point. In the applet below, you can move Point A to examine the situation a. Write an expression (in terms of 8) that represents Point A's distance to the right of the left side of the square in cm. Preview b. Write an expression in terms of that represents Point A's distance above the bottom side of the square in cm. that Point A has swept Preview c. Write a formula that expresses the distance between Points A and B (in cm), d, in terms of the number of radians out. (Hint: draw a diagram involving a right triangle, and use Pythagorean Theorem.) Preview

Answer 1

Solutions:-

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b] Point A's distance above the bottom side of the square We want only this pont CHA Given: Point A starts at the 3-o'clock position and moves Counter clockwise along the circle. where = angle's measure (in radians) Point A = terminal point Now, Distance from A to C is nothing but arc length (AC) where point touches bottom side of the square ... an expression (in terms ofo) ..1(AC) = Arc length (from AtoC) = 2 ( 2 ) = 2T (31/2) = 215r (34) = 2T (314) - 3 TUY = 3T (4) - 3T (2) = 6T = 18,84cm 2 2
continue ....

OR in clockwise direction $\Theta$ = -$\small \prod$/2.

then distance of point A above the bottom side of the square = Arc length(AC)

Arc length(AC)= 2$\small \prod$r$\times$((-$\small \prod$/2)/2$\small \prod$) = 2$\small \prod$r$\times$(-1/4) = -$\small \prod$r/2

here r = 4cm. so, Arc length(AC) = -$\small \prod$(4)/2 = -2$\small \prod$ = -(2$\times$3.14) = -(6.28)cm.

where, negative sign indicates that $\Theta$ rotates in clockwise direction

WE GET, by Counter Clockwise , Arc length(AC)= 18.84cm and by  Clockwise , Arc length(AC)= -6.28cm

Therefore, point A's distance above the bottom side of the square is the shortest distance above the bottom side of the square = 6.28cm ( we neglect negative sign because distance is always positive and also measure from bottom to top).

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c] T=4 cmA solution: A formala that expresses the distance between point A and B (in cmd, LADB = 90° By using Pythagorean Theorem . AB²= AD²+ BD2 ied² = AD²+ BD² id=VAD? + BD2 Here, length of seg BD = lengthot segA'A= 8 cm. length of segoc=lengthseg AD = 4cm Now, d-11834 +(42 (AD=4& BD =8) id=164+16 = 580 = 110x5 = 415 id=415cm = 8.94427 cm Distance between point A and B is 8,94427cm.

See it twice.