Question

This robotic arm is made up of two cylinders with equal volume and two triangular prism hands. The volume of each hand is

Answer 1

Answer:

r/(9.42H + r),

where H is each cylinder's height

Step-by-step explanation:

Volume of hands:

2 × ½t × ⅓r × 2r = ⅔r³

Let H be the height of each cylinder

Total volume:

2(3.14×r²×H) + ⅔r³

(6.28H + ⅔r)r²

Ratio:

[⅔r³] ÷ (6.28H + ⅔r)r²

r ÷ (9.42H + r)

r/(9.42H + r)

Answer 2

Answer:

$\frac{r}{3\pi h+r}$

Step-by-step explanation:

Since the height isn't given, we assume it to be "h" (of cylinders). And the answer will be in terms of "r" and "h".

The area of 1 arm is given, so the area of 2 arms would be:

$A_{arm}=2*(\frac{1}{2}r*\frac{1}{3}r*2r)=\frac{2r^3}{3}$

Now, area of 2 cylinders would be the formula:

$A_{cyl}=2*(\pi r^2 h)=2\pi r^2 h$

So, total area is A_arm PLUS A_cyl. The fractional area the arms are would be gotten by taking expression A_arm  divided  by A_total.

Shown below:

$\frac{A_{arm}}{A_{total}}=\frac{\frac{2r^3}{3}}{2\pi r^2 h + \frac{2r^3}{3}}$

We simplify further:

$\frac{\frac{2r^3}{3}}{2\pi r^2 h + \frac{2r^3}{3}}\\=\frac{\frac{2r^3}{3}}{2r^2(\pi h + \frac{r}{3})}\\=\frac{r}{3(\pi h + \frac{r}{3})}\\=\frac{r}{3\pi h+r}$

THis is the answer.