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}\:=\:5cm}[/tex]
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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