Question

In AABC, if DE is parallel to BC, AD= x+1, DB=x-1, AE = x+3 and
EC = x, then the value of x is:

​

Answer 1

Given :-

• $\sf{In\: ΔABC, \: DE || BC.}$
• $\sf{AD= x+1, \:DB=x-1,\: AE = x+3\: and\: EC = x.}$

To Find :-

• $\sf{The \:value\: of \:x. }$

Solution :-

$\sf{In \:ΔABC, \: DE || BC,}$

According to Basic proportionally theorem,

$↪ \sf \: \red{ \frac{AD}{DB} = \frac{AE}{EC}}$

$\sf{[ Putting \: values ]}$

$↪ \sf \: \frac{x+1}{x-1}=\frac{x+3}{x}$

$↪ \sf \: x(x + 1) = (x + 3)(x + 1)$

$↪ \sf \: x {}^{2} + x = x {}^{2} - x + 3x - 3$

$↪ \sf \: x = - x + 3x - 3$

$↪ \sf \: x = 2x - 3$

$↪ \sf \: x - 2x = - 3$

$↪ \sf \: - x = - 3$

$↪ \sf \: \blue{ x = 3}$

Hence,

• The value of x is 3.

Answer 2

Answer:

◦•●◉✿Given :-

\sf{In\: ΔABC, \: DE || BC.}InΔABC,DE∣∣BC.

\sf{AD= x+1, \:DB=x-1,\: AE = x+3\: and\: EC = x.}AD=x+1,DB=x−1,AE=x+3andEC=x.

To Find :-

\sf{The \:value\: of \:x. }Thevalueofx.

Solution :-

\sf{In \:ΔABC, \: DE || BC,}InΔABC,DE∣∣BC,

According to Basic proportionally theorem,

↪ \sf \: \red{ \frac{AD}{DB} = \frac{AE}{EC}}↪

DB

AD

=

EC

AE

\sf{[ Putting \: values ]}[Puttingvalues]

↪ \sf \: \frac{x+1}{x-1}=\frac{x+3}{x}↪

x−1

x+1

=

x

x+3

↪ \sf \: x(x + 1) = (x + 3)(x + 1)↪x(x+1)=(x+3)(x+1)

↪ \sf \: x {}^{2} + x = x {}^{2} - x + 3x - 3↪x

2

+x=x

2

−x+3x−3

↪ \sf \: x = - x + 3x - 3↪x=−x+3x−3

↪ \sf \: x = 2x - 3↪x=2x−3

↪ \sf \: x - 2x = - 3↪x−2x=−3

↪ \sf \: - x = - 3↪−x=−3

↪ \sf \: \blue{ x = 3}↪x=3

Hence,

The value of x is 3.✿◉●•◦

Step-by-step explanation:

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