The given figure shows an equilateral triangle ABC with each side 15 cm. Also DE // BC, DF // AC and EG //AB. If DE + DF + EG = 20 cm, find FG.

Given,

ABC is an equilateral triangle.

∴ AB = BC = AC = 15 cm and ∠A = ∠B = ∠C = 60°.

In △ ADE,

⇒ ∠ADE = ∠ABC = 60° (Corresponding angles are equal)

⇒ ∠AED = ∠ACB = 60° (Corresponding angles are equal)

By angle sum property of triangle,

⇒ ∠AED + ∠ADE + ∠DAE = 180°

⇒ 60° + 60° + ∠DAE = 180°

⇒ ∠DAE + 120° = 180°

⇒ ∠DAE = 180° - 120° = 60°.

∴ △ ADE is an equilateral triangle with each side equal to x cm.

∴ AD = DE = EA = x cm.

In △ BDF,

⇒ ∠BFD = ∠BCA = 60° (Corresponding angles are equal)

⇒ ∠DBF = ∠B = 60°

By angle sum property of triangle,

⇒ ∠BFD + ∠DBF + ∠BDF = 180°

⇒ 60° + 60° + ∠BDF = 180°

⇒ ∠BDF + 120° = 180°

⇒ ∠BDF = 180° - 120° = 60°.

∴ △ BDF is an equilateral triangle with each side equal to y cm.

∴ DB = BF = FD = y cm.

In △ EGC,

⇒ ∠EGC = ∠ABC = 60° (Corresponding angles are equal)

⇒ ∠ECG = ∠C = 60°

By angle sum property of triangle,

⇒ ∠EGC + ∠ECG + ∠GEC = 180°

⇒ 60° + 60° + ∠GEC = 180°

⇒ ∠GEC + 120° = 180°

⇒ ∠GEC = 180° - 120° = 60°.

∴ △ EGC is an equilateral triangle with each side equal to z cm.

∴ EG = GC = CE = z cm.

Given,

⇒ AB = 15

⇒ AD + BD = 15

⇒ x + y = 15 ……….(1)

⇒ AC = 15

⇒ AE + EC = 15

⇒ x + z = 15 ……….(2)

Given,

⇒ DE + DF + EG = 20

⇒ x + y + z = 20

⇒ 15 + z = 20 [From equation (1)]

⇒ z = 20 - 15 = 5 cm.

Substituting value of z in equation (2), we get :

⇒ x + 5 = 15

⇒ x = 15 - 5 = 10 cm.

Substituting value of x in equation (1), we get :

⇒ 10 + y = 15

⇒ y = 15 - 10 = 5 cm.

From figure,

⇒ BC = 15 cm

⇒ BF + FG + GC = 15

⇒ y + FG + z = 15

⇒ 5 + FG + 5 = 15

⇒ FG + 10 = 15

⇒ FG = 15 - 10 = 5 cm.

Hence, FG = 5 cm.