Diagonal

he rectangular ABCD trapeze, whose AD arm is perpendicular to the AB and CD bases, has area 15cm square. Bases have lengths AB = 6cm, CD = 4cm. Calculate the length of the AC diagonal.

Correct result:

u =  3.6056 cm

Solution:

$$\begin{gathered} S=15{\text{cm}}^2 \\ a=6\text{ cm } \\ c=4\text{ cm } \\ S=(a+c)d\mathrm{/}2 \\ d=2\cdot S\mathrm{/}(a+c)=2\cdot 15\mathrm{/}(6+4)=3\text{ cm } \\ u=\sqrt{d^2+(a-c{)}^2}=\sqrt{3^2+(6-4{)}^2}=\sqrt{13}=3.6056\text{ cm} \end{gathered}$$

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