Question

suppose that AD,BC,AC and BD are the line segments with AD parallel to BC as shown in the figure on the right if AD= 3 units BC= 1 unit and the distance from AD to BC is 5 units find the altitude of the smaller triangle

Answer 1

the altitude of the smaller triangle = 1.25 cm

Step-by-step explanation:

let say intersection point of AC & BD = X

Δ AXD  & ΔCXB

∠AXD = ∠CXB  ( vertically opposite angles)

∠DAX = ∠BCX   ( as AD ║ BC)

∠ADX = ∠CBX    ( as AD ║ BC)

=> Δ AXD  ≈ ΔCXB

Lets draw altitude XM  & XN on smaller & bigger triangle respectively

MN = XM + XN  = 5 cm

as Δ AXD  ≈ ΔCXB

=> XN/XM = AD/BC

=> XN/XM  = 3/1

=> XN = 3XM

XM + XN  = 5 cm

=> XM + 3XM = 5

=> 4XM = 5

=> XM = 1.25

the altitude of the smaller triangle = 1.25 cm

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