In the given figure DE || BC, then find the value of x.
Answers
Answer:
given DE ║ Bc
In ΔADE and ΔABC
∠ADE = ∠ABC [as BC║DE ∠ADE and ∠ABC are corresponding angles . nd they are equal ]
∠AED = ∠ ACB
∠DAE=∠BAC (common angle)
△ADE ∼ △ABC by AAA similarity criterion
here AB = 5cm and AD =2cm
and DE = 4cm and BC = x
AB and AD are corresponding sides and corresponding sides are proportional in similar triangles.
therefore AD/AB = DE/BC
2 / 5 = x / 4
2x = 20 [on cross multipling]
x = 10
Given :-
AD = x
DB = 3x + 4
AE = x + 3
EC = 3x + 19
To Find :-
Value of 'x'
Solution :-
Since DE || BC
∴ \dfrac{AD}{DB}= \dfrac{AE}{EC}
DB
AD
=
EC
AE
(By Basic proportionality theorem )
Substituting the value,
➙ \dfrac{x}{3x+4}= \dfrac{x + 3}{3x+19}
3x+4
x
=
3x+19
x+3
➙ x ( 3x+19 ) = ( x+3 ) ( 3x+4 )
➙ 3x² + 19x = 3x² + 4x + 9x + 12
➙ 3x² + 19x = 3x² + 13x + 12
➙ 3x² - 3x² + 19x - 13x - 12 = 0
➙ 6x - 12 = 0
➙ 6x = 12
➙ x = \dfrac{12}{6}
6
12
➙ x = \dfrac{\cancel{12}}{\cancel{6}}
6
12
\huge{\boxed{\mathtt{\red{x\:=\:12}}}}
x=12
Now,
Putting value of 'x' :-
AD = x = 12
DB = 3x + 4 = 3(12) + 4 = 36 + 4 = 40
AE = x + 3 = 12 + 3 = 15
EC = 3x + 19 = 3(12) + 19 = 36 + 19 = 55
Hope it helps..