Isosceles triangle

The leg of the isosceles triangle is 5 dm, its height is 20 cm longer than the base. Calculate base length z.

Correct answer:

z =  2.8 dm

Step-by-step explanation:

$$\begin{gathered} r=5\text{ dm } \\ 20cm=2dm \\ v=2+z \\ {r}^2=v^2+(z\mathrm{/}2{)}^2 \\ {5}^2=(2+z{)}^2+(z\mathrm{/}2{)}^2 \\ -1.25z^2-4z+21=0 \\ 1.25z^2+4z-21=0 \\ a=1.25;b=4;c=-21 \\ D=b^2-4ac=4^2-4\cdot 1.25\cdot (-21)=121 \\ D>0 \\ {z}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-4\pm \sqrt{121}}{2.5}} \\ {z}_{1,2}={\displaystyle \frac{-4\pm 11}{2.5}} \\ {z}_{1,2}=-1.6\pm 4.4 \\ {z}_1=2.8 \\ {z}_2=-6 \\ \text{ Factored form of the equation: } \\ 1.25(z-2.8)(z+6)=0 \\ z>0 \\ z=z_1=2.8={\displaystyle \frac{14}{5}} \\ v=2+z=2+2.8={\displaystyle \frac{24}{5}}=4.8\text{ dm } \\ z=r_2=\sqrt{v^2+(z\mathrm{/}2{)}^2}=\sqrt{4.8^2+(2.8\mathrm{/}2{)}^2}=5\text{ dm}=2.8\text{ dm} \end{gathered}$$

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