Question

In figure line BC parallel line DE, AB =2, BD =3, AC =4 and CE =x, then find the value of x.​

Answer 1

Given :- In figure line BC parallel line DE, AB =2, BD =3, AC =4 and CE =x, then find the value of x. ?

Solution :-

we know that,

• Basic proportionality theorem says that, If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

given that, BC || DE ,

so,

→ AB / BD = AC / CE { By BPT .}

putting given values we get,

→ 2 / 3 = 4 / x

→ 2 * x = 4 * 3

→ 2x = 12

→ x = 6 (Ans.)

Hence, the value of x will be 6.

Learn more :-

*जर △ ABC ~ △ DEF असून AB = 12 सेमी and DE = 14 सेमी. तर △ ABC आणि △ DEF यांच्या क्षेत्रफळाचे गुणोत्तर किती?.*

1️⃣ 49/9...

brainly.in/question/37096819

*In triangle ABC, DE || AB. If CD = 3 cm, EC = 4 cm, BE = 6 cm, then DA is equal to …………….*

1️⃣ 7.5 cm

2️⃣ 3 cm

3️⃣ 4.5...

brainly.in/question/37020783

In the figure , Line PQ || Side BC AP = 6 , PB = 8 , AQ = x and QC = 12 , then write the value of x.

https://brainly.in/question/36586072

Answer 2

Step-by-step explanation:

If DE∣∣AB, then ΔCDE∼ΔABC

By property of similar triangles:

AC

CD

=

BC

CE

⟹

4x+22

x+3

=

4x+4

x

Cross-multiplying:

(x+3)(4x+4)=x(4x+22)

⟹4x

2

+16x+12=4x

2

+22x

12=6x⟹

x=2