Question

1. a metal washer 1 inch. in diameter is pierced by a 1/2 - inch hole. what is the area of one of the washer?
2. the vertical end of a trough has the following dimensions width at top 4.4ft., width at bottom 3.2ft., depth 3.5ft. find the area of the end of the trough.
3. the surface of hemispherical dome whose diameter is 36ft. is to be covered with gold leaf which costa 15 cents per square inch. what must be paid for the golf leaf?
4. a cubic foot of ivory weighs 114Ib. find the weight of 1000 ivory billiard balls 21 in. in diameter?

*SHOW SOLUTION*

Answer 1

1.

A metal washer is made from two concentric circles and its a ring shape. 

Let $r_1=$ the radius of inner circle

      $r_2=$ the radius of outer circle

 Surface are

$$A=\pi r_2^2-\pi r_1^2$$

Given $2r_1=\dfrac{1}{2}in, 2r_2=1 in.$

Then

$$r_1=\dfrac{1}{4}in, r_2=\dfrac{1}{2}in$$

$$A=\pi(\dfrac{1}{2}in)^2-\pi(\dfrac{1}{4}in)^2=\dfrac{3\pi}{16}in^2$$

The area of one of the washer is $\dfrac{3\pi}{16}\ in^2.$

2. From the description of the given word problem, the vertical end of a trough is a trapezoid. 

Given $a=4.4ft, b=3.2ft, h=3.5ft.$

The area of a trapezoid is $A=\dfrac{a+b}{2}h.$

Substitute

$$A=\dfrac{4.4ft+3.2ft}{2}(3.5ft)=13.3ft^2$$

The area of the end of the trough is $13.3ft^2.$

3. The curved surface area of the dome is

$$A=\dfrac{1}{2}(4\pi r^2)=\dfrac{\pi d^2}{2}$$

Given $d=36ft=432in.$ Then

$$\dfrac{\pi (432in)^2}{2}=93 312\pi\ in^2$$

What must be paid for the golf leaf?

$$93 312\pi\ in^2(0.15\$/in^2)=\$43972.25$$

4. Volume of one billiard ball

$$V_1=\dfrac{4\pi}{3}r^3=\dfrac{\pi d^3}{6}$$

Given $d=21in=1.75ft.$ Then the volume of 1000 ivory billiard balls is

$$V=1000\cdot\dfrac{\pi(1.75ft)^3}{6}=\dfrac{5359.375\pi}{6}\ ft^3$$

The weight (mass) of 1000 ivory billiard balls is

$$M=\rho V$$

$$M=114lb/ft^3 \cdot(\dfrac{5359.375\pi}{6}\ ft^3)$$

$$=319902.5\ lb=145105.21\ kg$$

The weight (mass) of 1000 ivory billiard balls is $319902.5\ lb$ or $145105.21\ kg.$