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
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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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