Question

4. In the figure, for
the figure, for what value of x will seg DE be
parallel to AB?
x + 10/
x+6
4x+10)
* + 10
(A) 2
(B) 3
(C) 20
(D) 2 and 20​

Answer 1

Answer:

The value of x will be 2 for $\sf\:seg\:DE\:\parallel\:seg\:AB$.

Option A) 2

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

We have given that

$\sf\:seg\:DE\:\parallel\:seg\:AB$

We have to find the value of x.

We know that,

$\pink{\sf\:\dfrac{AD}{DB}\:=\:\dfrac{AE}{EC}}\sf\:\:\:-\:-\:[\:Basic\:Proportionality\:Theorem\:]\\\\\\\implies\sf\:\dfrac{(\:x\:+\:10\:)}{(\:4x\:+\:10\:)}\:=\:\dfrac{(\:x\:+\:6\:)}{(\:x\:+\:10\:)}\\\\\\\implies\sf\:(\:x\:+\:10\:)\:\times\:(\:x\:+\:10\:)\:=\:(\:x\:+\:6\:)\:\times\:(\:4x\:+\:10\:)\\\\\\\implies\sf\:(\:x\:+\:10\:)^2\:=\:4x^2\:+\:10x\:+\:24x\:+\:60\\\\\\\implies\sf\:x^2\:+\:2\:\times\:x\:\times\:10\:+\:(\:10\:)^2\:=\:4x^2\:+\:34x\:+\:60\\\\\\\implies\sf\:x^2\:+\:20x\:+\:100\:=\:4x^2\:+\:34x\:+\:60\\\\\\\implies\sf\:4x^2\:+\:34x\:+\:60\:-\:x^2\:-\:20x\:-\:100\:=\:0\\\\\\\implies\sf\:4x^2\:-\:x^2\:+\:34x\:-\:20x\:-\:100\:+\:60\:=\:0\\\\\\\implies\sf\:3x^2\:+\:14x\:-\:40\:=\:0$

$\\\\\\\implies\sf\:3x^2\:+\:20x\:-\:6x\:-\:40\:=\:0\\\\\\\implies\sf\:3x^2\:-\:6x\:+\:20x\:-\:40\:=\:0\\\\\\\implies\sf\:3x\:(\:x\:-\:2\:)\:+\:20\:(\:x\:-\:2\:)\:=\:0\\\\\\\implies\sf\:(\:x\:-\:2\:)\:(\:3x\:+\:20\:)\:=\:0\\\\\\\implies\sf\:x\:-\:2\:=\:0\:\:\:or\:\:\:3x\:+\:20\:=\:0\\\\\\\implies\sf\:x\:=\:2\:\:\:or\:\:\:3x\:=\:-\:20\\\\\\\implies\boxed{\red{\sf\:x\:=\:2}}\sf\:\:\:or\:\:\:\boxed{\red{\sf\:x\:=\:-\:\dfrac{20}{3}}}$

But, the length cannot be negative.

$\therefore\sf\:x\:=\:-\:\dfrac{20} {3}\:is\:not\:acceptable$

Hence,

The value of x will be 2 for the given condition.

Answer 2

⭐Given:

• AD= x+10

• BD=4x+10

• AE=x+6

• CE=x+10

• Seg DE || Seg BC

⭐To Find:

Value of x=?

⭐Solution:

It is given that,

Seg DE || Seg BC

Therefore, By basic proportionality therom,

AD/DB=AE/EC

→ x+10/4x+10=x+6/x+10

→(x+10) (x+10) =(4+10) (x+6)

→ x(x+10) +10(x+10) =x(4x+10) +6(4x+10)

→x²+10x+10x+100=4x²+10x+24x+60

→x²+20x+100=4x²+34x+60

→ 4x²-x²+34x-20x+60-100=0

→ 3x²+14x-40=0

Now, Let's solve the above quadratic equation.

3x²+14x-40=0.

→3x²+20x-6x-40=0

→ x(3x+20)-2(3x+20)=0

→ (3x+20)(x-2) =0

→ (3x+20)=0 OR (x-2) =0

→ x= -20/3 OR x=2

But, length can't be negative.

So, x= -20/3 is unacceptable.

Therefore,x= 2

Hence, 2 is the correct value for which Seg DE can be parallel to BC.

For More Information:

Basic Proportionality therom:

⭐ Statement:

If a line parallel to a side of a triangle and intersects the remaining sides in two distinct points,then the line divides the sides in the same proportion.

Converse of Basic proportionality therom:

⭐ Statement:

If a line divides any two sides of a triangle in the same ratio,then the line is parallel to the third side.