Area and volume > Area 2-D shapes
1234Area 2-D shapes

## Solutions to the exercises

Exercise 1

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

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

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

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

Exercise 2

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

Exercise 3

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

The circumference is $3\cdot 30+\frac{1}{2}\cdot 2\pi \cdot 30=90+30\pi \approx 184,25$ 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 $\frac{1}{2}\cdot \left(40+50\right)\cdot 30=1350$ 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 $\frac{1}{2}\cdot \left(40+40+\frac{1}{3}h\right)\cdot h=675$ .
This gives $80h+\frac{1}{3}{h}^{2}=1350$ or: ${h}^{2}+240h-4050=0$ .
From this you find $h=\frac{-240+\sqrt{73800}}{2}\approx 15,83$ cm.

b

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