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solve for x? please help​

Solve For X Please Help class=

Sagot :

Answer:

x = [tex]\sqrt{10}[/tex]

Step-by-step explanation:

The 2 right triangles and the outer right triangle are similar, then ratios of corresponding sides are equal, that is

[tex]\frac{x}{5}[/tex] = [tex]\frac{2}{x}[/tex] ( cross- multiply )

x² = 10 ( take the square root of both sides )

x = [tex]\sqrt{10}[/tex]

Answer:

√10

Step-by-step explanation:

taking the smallest triangle;

y as the hypotenuse, x as the adjacent side, 2 as the opposite side

then: y²=x²+2²

x²=y²-2²= y²-4

taking the other triangle;

z as the hypotenuse, x as the adjacent side, 5 as the opposite side

then: z²=x²+5²

x²=z²-5² = z²-25

take both equations,

x²= y²-4

x²= z²-25

y²-4= z²-25

y²= z²-25+4 = z²-21

taking the biggest triangle (whole triangle);

(5+2)= 7 as the hypotenuse, z as the adjacent side, y as the opposite side

then; 7²=z²+y²

z²+y²=7² =49

but y²= z²-21;

z²+y² =49

z²+ z²-21 =49

2z²-21=49

2z²=49+21= 70

z²= 70/2 = 35

take x²= z²-25

x²= 35-25 = 10

[tex]x = \sqrt{10} [/tex]