Answered

3. Anton used strings to hang two small light balls on the ceiling
as shown in the figure on the right. The broken line represents
the distance from the point of tangency of the two light balls
to the ceiling

a. Suppose the diameter of each light ball is 10 cm and the
length of the string used to hang it is 40 cm. how far is the
point of tangency of the two light balls from the ceiling?

b. Suppose Anton hangs 40 pairs of light balls on the ceiling
of a hall in preparation for an event. How long is the string
that he needs to hang these light balls if each has a
diameter of 12 cm and the point of tangency of each pair
of balls is 30 cm from the ceiling?​

Sagot :

Answer:

A. [tex]\sf\large{ \sqrt{1575}}[/tex] cm

B. [tex]\sf\large{480 \sqrt{23}}[/tex] cm

[tex]{}[/tex]

Solution:

a) use Pythagorean theorem

lets r – balls radius, l – length of string,

a – length of dot line

[tex]\sf\large{r\:= \frac{d}{2} = \frac{10}{5}\:=\:5}[/tex] cm

l² = a² + r²

a² = l² - r² = 40² - 5² = 1600 - 25 = 1575

a = ± [tex]\sf{ \sqrt{1575}}[/tex]

a cant be negative, so a = [tex]\sf{ \sqrt{1575}}[/tex]

Answer: [tex]\sf{ \sqrt{1575}}[/tex] cm

[tex]{}[/tex]

b) Lets find length of one string

[tex]\sf\large{r\:= \frac{d}{2} = \frac{12}{2}\:=\:6}[/tex] cm

l² = r² + a² = 6² + 30² = 36 + 900 = 936

l = ± [tex]\sf{ \sqrt{936}}[/tex]

l cant be negative, so l = [tex]\sf{ \sqrt{936}}[/tex] cm

if he hangs 40 pairs, he hangs 40 × 2 = 80 balls

80 × [tex]\sf{ \sqrt{936}}[/tex] = 80[tex]\sf{ \sqrt{936}}[/tex] = 80[tex]\sf{ \sqrt{36 × 26}}[/tex] = 80 × 6[tex]\sf{ \sqrt{23}}[/tex]

= 480[tex]\sf{ \sqrt{23}}[/tex] cm

Answer: 480[tex]\sf{ \sqrt{23}}[/tex]cm

[tex]{}[/tex]

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[tex] \large\underline{\mathbb{ANSWER}:} [/tex]

[tex] \qquad \Large a) \: \rm{10\sqrt6 \: cm} [/tex]

[tex] \qquad \Large b) \: \rm{\approx 1,\!968 \: cm} [/tex]

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[tex] \large\underline{\mathbb{SOLUTIONS}:} [/tex]

Problem (a):

» Label the points:

  • A = Center of the ball
  • B and E = Point where the string is attached to the ball
  • C = Point of tangency or the point of contact of the balls
  • D = Point where the string is attached to the ceiling.

» Since the string used is 40cm, then each string used from a ball to the ceiling is 20cm.

  • [tex] BD = 20 [/tex]
  • [tex] ED = 20 [/tex]

» Draw two radii, from A to B, and from A to C. Since the diameter is 10icm, then each of them measures 5cm.

  • [tex] AB = 5 [/tex]
  • [tex] AC = 5 [/tex]

» Since AC is perpendicular to CD, then ∆ACD is a right triangle. Solve for CD using the Pythagorean Theorem.

  • [tex] (CD)^2 + (AC)^2 = (AD)^2 [/tex]

  • [tex] (CD)^2 + (AC)^2 = (AB + BD)^2 [/tex]

  • [tex] (CD)^2 + (5)^2 = (5 + 20)^2 [/tex]

  • [tex] (CD)^2 + (5)^2 = (25)^2 [/tex]

  • [tex] (CD)^2 + 25 = 625 [/tex]

  • [tex] (CD)^2 = 625 - 25 [/tex]

  • [tex] (CD)^2 = 600 [/tex]

  • [tex] \sqrt{(CD)^2} = \sqrt{600} [/tex]

  • [tex] CD = 10\sqrt6 [/tex]

[tex] \therefore [/tex] The distance between the point of tangency of the two balls from the ceiling is 10√6 cm.

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Problem (b):

» Label the points as used in Problem (a).

» Draw two radii from A to B, and from A to C. Since the diameter is 12cm, then each of them measures 6cm.

  • [tex] AB = 6 [/tex]
  • [tex] AC = 6 [/tex]

» The distance of the point of tangency of each pair of balls is 30 cm from the ceiling?

  • [tex] CD = 30 [/tex]

» Find the length of one string attached to one ball from the ceiling using the Pythagorean Theorem. Refer it as x.

  • [tex] (CD)^2 + (AC)^2 = (AD)^2 [/tex]

  • [tex] (CD)^2 + (AC)^2 = (AB + BD)^2 [/tex]

  • [tex] (30)^2 + (6)^2 = (6 + x)^2 [/tex]

  • [tex] 900 + 36 = 36 + 12x + x^2 [/tex]

  • [tex] 900 = 12x + x^2 [/tex]

  • [tex] x^2 + 12x - 900 = 0 [/tex]

» By solving the quadratic equation, we get the positive solution as:

  • [tex] x = 6\sqrt{26} - 6 [/tex]

  • [tex] BD = 6\sqrt{26} - 6 [/tex]

» Thus, the measure of string used to one ball is 6√26 - 6 cm. Therefore, each pair of balls will use a string of:

  • [tex] \rm Pair = 2(6\sqrt{26} - 6) [/tex]

  • [tex] \rm Pair = 12\sqrt{26} - 12 [/tex]

» And since there are 40 pairs to be used in the hall, then the total measure string to be used is:

  • [tex] \rm 40 \: Pairs = 40(12\sqrt{26} - 12) [/tex]

  • [tex] \rm 40 \: Pairs = 480\sqrt{26} - 480 [/tex]

» By getting the approximate value of √26.

  • [tex] \rm 40 \: Pairs \approx 480(5.1)- 480 [/tex]

  • [tex] \rm 40 \: Pairs \approx 2448 - 480 [/tex]

  • [tex] \rm 40 \: Pairs \approx 1968 [/tex]

[tex] \therefore [/tex] The measure of string used for 40 pairs of balls is about 1968 cm.

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