Area and volume

Exercise 1

Area(I) $= \frac{1}{2} \cdot 6 \cdot 7 sin (20 °) \approx 7,18$ .

Area(II) $= \frac{1}{2} \cdot (7 + 3) \cdot 7 sin (50 °) \approx 26,81$ .

Figure III is the difference between an isosceles triangle with sides $13$ , $13$ and $2 \cdot 13 sin (15 °) \approx 6,73$ and an isosceles triangle with sides $5$ , $5$ and $\sqrt{5^{2} - {(13 sin (15 °))}^{2}} \approx 3,70$ .

Area(III) $= \frac{1}{2} \cdot 26 sin (15 °) \cdot 13 cos (15 °) - \frac{1}{2} \cdot 26 sin (15 °) \cdot \sqrt{5^{2} - {(13 sin (15 °))}^{2}} \approx 29,81$ .
Area(IV) $= 2 \cdot (\frac{1}{4} π \cdot 3^{2} - \frac{1}{2} \cdot 3 \cdot 3) \approx 5,14$ .
Area(V) $= \frac{1}{4} π \cdot 6^{2} - 2 \cdot \frac{1}{2} π \cdot 3^{2} + 2 \cdot (\frac{1}{4} π \cdot 3^{2} - \frac{1}{2} \cdot 3 \cdot 3) \approx 5,14$ .

Exercise 2

$π \cdot 5^{2} - 8 \cdot \frac{1}{2} \cdot 5 sin (22,5 °) \cdot 5 cos (22,5 °) \approx 43,18$

Exercise 3

The area is $3 \cdot \frac{1}{6} π \cdot 30^{2} + 3 \cdot 3 sin (30 °) \cdot 3 cos (30^{o}) \approx 1425,41$ cm².

The circumference is $3 \cdot 30 + \frac{1}{2} \cdot 2 π \cdot 30 = 90 + 30 π \approx 184,25$ cm.

Exercise 4

a

This is the case when the area of the trapezium that defines the front of the container is divided into two equal parts. The area of the trapezium is $\frac{1}{2} \cdot (40 + 50) \cdot 30 = 1350$ cm².
If the height of the lower half is $h$ , you can show by using similarity that the longest of the two parallel sides if this trapezium has a lenght of $40 + \frac{1}{3} h$ . The area of this lower trapezium is half the area of the total trapezium, so $\frac{1}{2} \cdot (40 + 40 + \frac{1}{3} h) \cdot h = 675$ .
This gives $80 h + \frac{1}{3} h^{2} = 1350$ or: $h^{2} + 240 h - 4050 = 0$ .
From this you find $h = \frac{-240 + \sqrt{73800}}{2} \approx 15,83$ cm.

b

The area of the water surface is approximately $(40 + \frac{1}{3} \cdot 15,83) \cdot 200 \approx 9055$ cm².

Solutions to the exercises