1 Answer

Matthewb · Answer 1 · 2022-02-11T08:10:47+0000

Solution :

Let's solve by using Pythagoras theorem,

$\hookrightarrow \: (10) {}^(2) = {(x})^(2) + (x + 2) {}^(2)$

$\hookrightarrow \: 100 = {x}^(2) + {x}^(2) + 4x + 4$

$\hookrightarrow \: 100 - 4 = 2 {x}^(2) + 4x$

$\hookrightarrow \: 96 = 2 {x}^(2) + 4x$

$\hookrightarrow \: 2 {x}^(2) + 4x - 96 = 0$

$\hookrightarrow \: 2( {x}^(2) + 2x - 48) = 0$

$\hookrightarrow \: {x}^(2) + 2x - 48 = 0$

$\hookrightarrow \: {x}^(2) + 8x - 6x - 48 = 0$

$\hookrightarrow \: x(x + 8) - 6(x + 8) = 0$

$\hookrightarrow \: (x + 8)(x - 6) = 0$

now there are two cases

Case 1 : when (x + 8) = 0

$\hookrightarrow \: x + 8 = 0$

$\hookrightarrow \: x = - 8$

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

$\hookrightarrow \: x - 6 = 0$

$\hookrightarrow \: x = 6$

therefore the measure of other two sides are :

$\hookrightarrow \: x = 6 \: units$

$\hookrightarrow \: x + 2 = 6 + 2 = 8 \: units$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{TeeNForeveR }$