Question

Given CS = 2x, SB = 4x + 1, CR = 6, and RA = 13.5. What must the value of x be in order to prove SR II BA?

-2
-4
-5
-3

Answer 1

Answer:

x= -2

Step-by-step explanation:

x = -2

if sr parallel to ba then the sides bc and ac are in proportion, that is

$\frac{sb}{cs}$ = $\frac{ra}{cr}$ ← substitute values

$\frac{4x+1}{2x}$ = $\frac{13.5}{6}$ ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = -2

Answer 2

x= -2

Thales theorem in triangle-

"The line drawn parallel to one side of a triangle and cutting the other two sides divides the other two sides in equal proportion".

Given -

       CS = 2x

       SB = 4x +1

       CR = 6

       RA = 13.5

if sr parallel to ba then the sides bc and ac are in proportion, that is

$\frac{SB}{CS} =\frac{RA}{CR}$    ← substitute values

$\frac{4x + 1}{2x} = \frac{13.5}{6}$

cross- multiply with both sides

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = -2

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