Question

If a line is drawn parallel to the side of a triangle, the other two sides are divided in the same ratio . Apply this theorem in ABC , where ABC ∽ AXY
ABAX = 53
Also XY = 4cm and BY bisects ∠ XYC

Answer the below questions from the above information
Length BC =
(i) 5/3 cm (ii) 4/3 cm (iii) 20/3 cm (iv) 4/13 cm
Length CY =
(i) 5/3 cm (ii) 4/3 cm (iii) 10cm (iv) 20/3 cm
Length AY =
(i) 20 m (ii) 2 cm (iii) 30 cm (iv ) 10 cm
AY : CY =
(i) 2 : 3 (ii) 3 : 2 (iii) 1 : 5 (iv) 2 : 5
XY : BC =
(i) 5 : 3 (ii) 3 : 5 (iii) 2 : 3 (iv ) 3 : 2

Answer 1

Solution :-

given that,

→ ∆ABC ~ ∆AXY

→ AB/AX = 5/3 .

So,

→ AC/AY = BC/XY = 5/3 ---------- Eqn.(1)

(i)

→ BC/XY = 5/3

→ BC/4 = 5/3

→ BC = 20/3 cm

(ii)

→ ∆ABC ~ ∆AXY (given)

So,

→ XY || BC

then,

→ ∠XYB = ∠CBY { Alternate angles. }

and,

→ ∠XYB = ∠CYB { given that, BY Bisects ∠XYC .}

therefore,

→ ∠CBY = ∠CYB

hence, in Δ BCY ,

→ BC = CY { Sides opposite to equal angles is equal in measure. }

→ CY = 20/3 cm

(iii)

→ AC/AY = 5/3 { from Eqn.(1) }

→ (AY + CY)/AY = 5/3

→ 1 + (CY/AY) = 5/3

→ CY/AY = (5/3) - 1

→ CY/AY = (2/3)

→ (20/3)/AY = (2/3)

→ (20/3AY) = (2/3)

→ AY = 10 cm

(iv)

→ AY : CY = 10 : 20/3

→ AY : CY = 30 : 20

→ AY : CY = 3 : 2

(v)

→ BC/XY = 5/3

→ XY : BC = 3 : 5

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