Question

In a triangle ABC, FG and DE are two line segments parallel to BC such that AG=GE=EB. Find the combined length of DE and FG, if BC=12 cm

Answer 1

Answer:

The combined length of DE and FG is = 12 cm.

Step-by-step explanation:

In the given question,

In triangle ABC,

FG and DE are two lines parallel to BC such that AG = GE = EB = x (say)

also,

BC = 12 cm (given)

So,

In triangle AED and ABC, as DE || BC,

So,

∠EAD = ∠BAC = ∠A (common)

∠AED = ∠ABC (alternate interior angles)

Therefore,

Triangle AED is similar to Triangle ABC.

So,

$\frac{AE}{AB}=\frac{ED}{BC}=\frac{AD}{AC}\\So,\\\frac{AE}{AB} =\frac{x+x}{x+x+x}= \frac{2}{3}=\frac{ED}{12}\\So,\\ED=8cm$

Similarly,

In triangle AGF and ABC, using similar triangles concept we can say that,

$\frac{AG}{AB}=\frac{GF}{BC}\\\frac{x}{x+x+x}=\frac{1}{3}=\frac{GF}{12}\\So,\\GF =4 cm$

Therefore, DE + FG = 8 + 4 = 12 cm

Hence, the combined length of DE and FG are 12 cm.

Answer 2

Answer:

Given:

BC∥ED∥FG

AG=GE=EB and BC=12cm

Now triangle ΔABC∼ΔAFG...[by AAA criteria]

So,

AG/AA = GF/BC

⟹ 1x/3x = GF/12

⟹GF=4.

Similarly, triangle ΔABC∼ΔAED...[by AAA criteria]

So,

AE/AB= ED/BC

⟹ 2x/ 3x = ED/12

⟹ED=8.

Hence, DE+FG=4+8=12 cm.