Respuesta :

Solution :

Let's solve by using Pythagoras theorem,

[tex] \hookrightarrow \: (10) {}^{2} = {(x})^{2} + (x + 2) {}^{2} [/tex]

[tex] \hookrightarrow \: 100 = {x}^{2} + {x}^{2} + 4x + 4[/tex]

[tex] \hookrightarrow \: 100 - 4 = 2 {x}^{2} + 4x[/tex]

[tex] \hookrightarrow \: 96 = 2 {x}^{2} + 4x[/tex]

[tex] \hookrightarrow \: 2 {x}^{2} + 4x - 96 = 0[/tex]

[tex] \hookrightarrow \: 2( {x}^{2} + 2x - 48) = 0[/tex]

[tex] \hookrightarrow \: {x}^{2} + 2x - 48 = 0[/tex]

[tex] \hookrightarrow \: {x}^{2} + 8x - 6x - 48 = 0[/tex]

[tex] \hookrightarrow \: x(x + 8) - 6(x + 8) = 0[/tex]

[tex] \hookrightarrow \: (x + 8)(x - 6) = 0[/tex]

now there are two cases

Case 1 : when (x + 8) = 0

[tex] \hookrightarrow \: x + 8 = 0[/tex]

[tex] \hookrightarrow \: x = - 8[/tex]

but the value of x can't be negative, since side of a triangle isn't a negative value .

Case 2 : when (x - 6) = 0

[tex] \hookrightarrow \: x - 6 = 0[/tex]

[tex] \hookrightarrow \: x = 6[/tex]

therefore the measure of other two sides are :

[tex] \hookrightarrow \: x = 6 \: units[/tex]

[tex] \hookrightarrow \: x + 2 = 6 + 2 = 8 \: units[/tex]

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{TeeNForeveR } [/tex]

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