Parallelogram ABCD

We have the parallelogram ABCD, where AB is 6.2 cm; BC is 5.4 cm and AC is 4.8 cm.
Calculate the height on the AB side and the angle DAB.

Final Answer:

h =  4.0307 cm
α =  36.7385 °

Step-by-step explanation:

$$\begin{gathered} a=6.2\text{cm} \\ b=5.4\text{cm} \\ u=4.8\text{cm} \\ s=\frac{a+b+u}{2}=\frac{6.2+5.4+4.8}{2}=\frac{41}{5}=8.2\text{cm} \\ {S}_1=\sqrt{s\cdot (s-a)\cdot (s-b)\cdot (s-u)}=\sqrt{8.2\cdot (8.2-6.2)\cdot (8.2-5.4)\cdot (8.2-4.8)}\doteq 12.4951{\text{cm}}^2 \\ S=2\cdot S_1=2\cdot 12.4951\doteq 24.9902{\text{cm}}^2 \\ S=a\cdot h \\ h=S/a=24.9902/6.2=4.0307\text{cm} \end{gathered}$$

$$\begin{gathered} \tan \alpha =h:b \\ A=\arctan (h/b)=\arctan (4.0307/5.4)\doteq 0.6412\text{rad} \\ \alpha =A\to \text{°}=A\cdot \frac{180}{\pi }\text{°}=0.6412\cdot \frac{180}{\pi }\text{°}=36.739\text{°}=36.7385^{\circ }=36°44^{\prime }19" \end{gathered}$$

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