Trapezoid - hard example

Base of the trapezoid are: 24, 16 cm.
Diagonal 22, 26 cm.
Calculate its area and perimeter.

Correct answer:

S =  264 cm²
o =  67.9663 cm

Step-by-step explanation:

$$\begin{gathered} a=24\text{ cm } \\ c=16\text{ cm } \\ u=22\text{ cm } \\ v=26\text{ cm } \\ w=a+c=24+16=40\text{ cm } \\ s={\displaystyle \frac{u+v+w}{2}}={\displaystyle \frac{22+26+40}{2}}=44\text{ cm } \\ S=\sqrt{s\cdot (s-w)\cdot (s-u)\cdot (s-v)}=\sqrt{44\cdot (44-40)\cdot (44-22)\cdot (44-26)}=264{\text{cm}}^2 \end{gathered}$$

Try calculation via our triangle calculator.

$$\begin{gathered} S={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{w}{2}}\cdot h \\ h=2\cdot S\mathrm{/}w=2\cdot 264\mathrm{/}40={\displaystyle \frac{66}{5}}=13.2\text{ cm } \\ {u}^2=h^2+a_1^2=13.2^2+17.6^2={\displaystyle \frac{4356}{25}}=174.24 \\ {a}_1=\sqrt{u^2-h^2}=\sqrt{22^2-13.2^2}={\displaystyle \frac{88}{5}}=17.6\text{ cm } \\ {a}_2=a-a_1=24-17.6={\displaystyle \frac{32}{5}}=6.4\text{ cm } \\ {b}^2=h^2+a_2^2 \\ b=\sqrt{h^2+a_2^2}=\sqrt{13.2^2+6.4^2}\doteq 14.6697\text{ cm } \\ {c}_1=\sqrt{v^2-h^2}=\sqrt{26^2-13.2^2}={\displaystyle \frac{112}{5}}=22.4\text{ cm } \\ {c}_2=a-c_1=24-22.4={\displaystyle \frac{8}{5}}=1.6\text{ cm } \\ {d}^2=h^2+c_2^2 \\ d=\sqrt{h^2+c_2^2}=\sqrt{13.2^2+1.6^2}\doteq 13.2966\text{ cm } \\ o=a+b+c+d=24+14.6697+16+13.2966=67.9663\text{ cm} \end{gathered}$$

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