Geometry exercise

Calculations from geometry:
The ratios of the sides of the quadrilateral are 3 : 6:4.5 : 3.5. Calculate their lengths if the circumference is 51 cm.

The sizes of the angles in the quadrilateral are equal to 29°30', 133°10', and 165°20'. What is the size of angle D?

Final Answer:

a =  9 cm
b =  18 cm
c =  13 ¹⁄₂ = 27/2 cm
d =  10 ¹⁄₂ = 21/2 cm
D =  32 °

Step-by-step explanation:

$$\begin{gathered} o=51\text{cm} \\ a:b:c:d=3:6:4.5:3.5 \\ o=a+b+c+d \\ a=\frac{3}{3+6+4.5+3.5}\cdot o=\frac{3}{3+6+4.5+3.5}\cdot 51=9\text{cm} \end{gathered}$$

$b=\frac{6}{3+6+4.5+3.5}\cdot o=\frac{6}{3+6+4.5+3.5}\cdot 51=18\text{cm}$

$c=\frac{4.5}{3+6+4.5+3.5}\cdot o=\frac{4.5}{3+6+4.5+3.5}\cdot 51=\frac{27}{2}\text{cm}=13.5\text{cm}$

$d=\frac{3.5}{3+6+4.5+3.5}\cdot o=\frac{3.5}{3+6+4.5+3.5}\cdot 51=\frac{21}{2}\text{cm}=10.5\text{cm}$

$$\begin{gathered} \alpha =29\frac{30}{60}=29+\frac{30}{60}=\frac{29\cdot 60+30}{60}=\frac{1740+30}{60}=\frac{1770}{60}=\frac{59}{2}=29.5{}^{\circ } \\ \beta =133\frac{10}{60}=133+\frac{10}{60}=\frac{133\cdot 60+10}{60}=\frac{7980+10}{60}=\frac{7990}{60}=\frac{799}{6}\doteq 133.1667{}^{\circ } \\ \gamma =165\frac{20}{60}=165+\frac{20}{60}=\frac{165\cdot 60+20}{60}=\frac{9900+20}{60}=\frac{9920}{60}=\frac{496}{3}\doteq 165.3333{}^{\circ } \\ D=180\cdot (4-2)-\alpha -\beta -\gamma =180\cdot (4-2)-\frac{59}{2}-\frac{799}{6}-\frac{496}{3}=32^{\circ } \end{gathered}$$

Try another example