Question

10. If the figure, DE || BC
AD = x + 5, AB = 3x + 5
AE = x + 1, AC = 3x - 2, find x​

Answer 1

$\large\underline\mathfrak{Answer-}$

$\bold{\sf{\blue{Value\:of\:x\:is\:3}}}$.

$\large\underline\mathfrak{Step\:-\:by\:-\:step\:explanation:}$

Given :

• DE || BC

• AD = x + 5

• AB = 3x + 5

• AE = x + 1

• AC = 3x - 2

To find :

• Value of x.

Solution :

We know that, if a line is drawn parallel to one side of a triangle, then is bisects the other two sides in same ratio.

We can write it as :

$\therefore$ $\sf{\dfrac{AD}{AB}\:=\:\dfrac{AE}{AC}}$

: $\implies$ $\sf{\dfrac{x\:+\:5}{3x\:+\:5}\:=\:\dfrac{x\:+\:1}{3x\:-\:2}}$

: $\implies$ $\sf{(x+5)(3x-2)\:=\:(3x+5)(x+1)}$

: $\implies$ $\sf{\cancel{3x^2}\:-\:2x\:+15x\:-10\:=\:\cancel{3x^2}\:+\:3x\:+\:5x\:+5}$

: $\implies$ $\sf{13x\:-\:8x\:=\:5\:+\:10}$

: $\implies$ $\sf{5x\:=\:15}$

: $\implies$ $\sf{x\:=\:{\cancel{\dfrac{15}{5}}}}$ $\small{\sf{3}}$

: $\implies$ $\sf{\red{x\:=\:3}}$

$\therefore$ $\bold{\sf{\blue{Value\:of\:x\:is\:3}}}$.

Answer 2

$\huge\bold\green{Question}$

If the figure, DE || BC

AD = x + 5, AB = 3x + 5

AE = x + 1, AC = 3x - 2, find x

$\huge\bold\green{Answer}$

According to the question we have given :-

→ AB = 3x + 5

→ AE = x + 1

→ DE || BC

→ AD = x + 5

→ AC = 3x - 2

So, as said in question we have to find out the correct value of " x "

As we know that parallel line bisects the other two sides in same ratio of the triangle

So , Simply We can write it as :-

$\tt{\dfrac{AD}{AB}\:=\:\dfrac{AE}{AC}}$

$\tt{\dfrac{x\:+\:5}{3x\:+\:5}\:=\: \dfrac{x\:+\:1}{3x-2}}$

$\tt{(x+5)(3x-2)\:=\:(3x+5)(x+1)}$

$\tt {\cancel{3x^2}\:-\:2x\:+15x\:-10\:=\:\cancel{3x^2}\+\:3x\:+\:5x\:+5}}$

$\tt{13x\:-\:8x\:=\:5\:+\:10}$

$\tt{5x\:=\:15}$

$\tt{x\:=\:{\cancel{\dfrac{15}{5}}} = 3}$

Hence the required value of " x " is 3