# Related rates calculus 1? A hexagon is formed by hinging six 13cm rods. ?

A hexagon is formed by hinging six 13cm rods. You start with a regular hexagon and then carefully push the opposite horizontal sides, keeping their alignment as shown in the picture. If you are pushing so that the distance between the opposite sides is decreasing at a rate of 3 cm/sec, at what rate is the area changing when the distance between your hands is 10cm?

### 3 Answers

- 1 month agoFavourite answer
So you really have 3 objects: 2 congruent triangles and a rectangle. The triangles have heights of h and bases of w. The rectangle has a length of 13 and a base of w

A[rectangle] + 2 * A[triangle]

13 * w + 2 * (1/2) * w * h

13w + w * h

w * (13 + h)

13^2 = (w/2)^2 + h^2

169 = w^2 / 4 + h^2

169 - w^2 / 4 = h^2

(4 * 169 - w^2) / 4 = h^2

(676 - w^2) / 4 = h^2

sqrt(676 - w^2) / 2 = h

w * (13 + (1/2) * sqrt(676 - w^2)) =>

w * (1/2) * (26 + sqrt(676 - w^2)) =>

(1/2) * (26 * w + w * (676 - w^2)^(1/2)) =>

(1/2) * (26 * w + (676 * w^2 - w^4)^(1/2))

A = (1/2) * (26w + (676w^2 - w^4)^(1/2))

2A = 26w + sqrt(676w^2 - w^4)

w = 10 cm

dw/dt = -3 cm/sec

2A = 26 * w + sqrt(676w^2 - w^4)

2 * dA/dt = 26 * dw/dt + (1/2) * (1352 * w - 4w^3) * dw/dt / sqrt(676 * w^2 - w^4)

2 * dA/dt = (26 + (676w - 2w^3) / sqrt(676 * w^2 - w^4)) * dw/dt

dA/dt = (13 + (338w - w^3) / sqrt(676w^2 - w^4)) * dw/dt

dA/dt = (13 + (338 - w^2) / sqrt(676 - w^2)) * dw/dt

dA/dt = (13 + (338 - 10^2) / sqrt(676 - 10^2)) * (-3)

dA/dt = (13 + (338 - 100) / sqrt(676 - 100)) * (-3)

dA/dt = -3 * (13 + 238 / sqrt(576))

dA/dt = -3 * (13 + 238 / 24)

dA/dt = -3 * (13 + 119 / 12)

dA/dt = -3 * (156 + 119) / 12

dA/dt = -(275 / 4)

dA/dt = -68.75

The area is decreasing at 68.75 square centimeters per second

- alexLv 71 month ago
x = the distance between the opposite sides , dx/dt = -3cm/s

Area of the Hexagon A = 13x + x√[13^2 -(x/2)^2]

find dA/dt

you can continue