Isosceles triangle

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

Result

z =  2.8 dm

Solution:

r=5 dm 20 cm=2 dm  v=2+z r2=v2+(z/2)2  52=(2+z)2+(z/2)2  1.25z24z+21=0 1.25z2+4z21=0  a=1.25;b=4;c=21 D=b24ac=4241.25(21)=121 D>0  z1,2=b±D2a=4±1212.5 z1,2=4±112.5 z1,2=1.6±4.4 z1=2.8 z2=6   Factored form of the equation:  1.25(z2.8)(z+6)=0  z>0 z=z1=2.8=145=2.8 dm v=2+z=2+2.8=245=4.8 dm r2=v2+(z/2)2=4.82+(2.8/2)2=5 dmr=5 \ \text{dm} \ \\ 20 \ cm=2 \ dm \ \\ \ \\ v=2 + z \ \\ r^2=v^2 + (z/2)^2 \ \\ \ \\ 5^2=(2+z)^2 + (z/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}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -4 \pm \sqrt{ 121 } }{ 2.5 } \ \\ z_{1,2}=\dfrac{ -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=\dfrac{ 14 }{ 5 }=2.8 \ \text{dm} \ \\ v=2 + z=2 + 2.8=\dfrac{ 24 }{ 5 }=4.8 \ \text{dm} \ \\ r_{2}=\sqrt{ v^2 + (z/2)^2 }=\sqrt{ 4.8^2 + (2.8/2)^2 }=5 \ \text{dm}

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