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andriy [413]
3 years ago
11

Someone please help? 25 points!!!!​

Mathematics
2 answers:
solmaris [256]3 years ago
8 0

Use basic proportionality theorem

\\ \sf\longmapsto \dfrac{AD}{DC}=\dfrac{BE}{AC}

\\ \sf\longmapsto \dfrac{2x}{x+3}=\dfrac{2x-1}{x}

  • Use cross multiplication

\\ \sf\longmapsto 2x^2=(2x-1)(x+3)

\\ \sf\longmapsto 2x^2=2x(x+3)-1(x+3)

\\ \sf\longmapsto 2x^2=2x^2+6x-x-3

\\ \sf\longmapsto 2x^2-2x^2=5x-3

\\ \sf\longmapsto 5x-3=0

\\ \sf\longmapsto 5x=3

\\ \sf\longmapsto x=\dfrac{3}{5}

sergij07 [2.7K]3 years ago
6 0

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • Find the value of x.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

Given,

  • DE || AB
  • AD = 2x
  • CD = x + 3
  • BE = 2x - 1
  • CE = x

Value of x (CE) = ?

__________________

We can solve this question using the Basic<u> Proportionality Theorem (BPT).</u> According to BPT, <em>if a line is drawn </em><em>parallel </em><em>(</em><em>|</em><em>|</em><em>)</em><em> </em><em>to</em><em> </em><em>1</em><em> </em><em>side of a triangle to intersect the other </em><em>2</em><em> </em><em>sides </em><em>in </em><em>2</em><em> distinct points then the other </em><em>2</em><em> </em><em>sides </em><em>will </em><em>be</em><em> divided in the same ratio. </em>

__________________

So in this triangle,

\tt\frac{CD}{AD} =  \frac{CE}{BE} \\

Then,

\tt \frac{x + 3}{2x}  =  \frac{x}{2x - 1}  \\  \\  \sf \: Using \: cross \: multiplication.... \\  \\  \tt \: x + 3(2x - 1) = x(2x) \\   \tt \: x(2x - 1) + 3(2x - 1) =  {2x}^{2}  \\   \tt \:  {2x}^{2}  - x + 6x - 3 =  {2x}^{2}  \\  \\ \sf Cancel \: out \: 2 {x}^{2}  \: from \: both \: the \: sides \\  \\  \tt \:  - x + 6x - 3 = 0 \\  \tt \:    - x + 6x = 3 \\  \tt \: 5x = 3 \\     \large\boxed{\boxed{\bf \: x =  \frac{3}{5} }}

__________________

  • The value of x will be <u>3/5 (option b.)</u>
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