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प्रश्न
In the following figure, AE = EF = AF = BE = CF = a, AT ⊥ BC. Show that AB = AC = `sqrt3xxa`
उत्तर
Given: AE = EF = AF = BE = CF,
AT ⊥ EF
ΔAEF is equilateral triangle.
ET = TF =a/2
BT = CT = a +a/2....(1)
In right triangles, ΔATB and ΔATC,
AT = AT … (Side common to both triangles)
∠ATB = ∠ATC … (Right angles)
BT = CT …. (from 1)
∴ ΔATB ≅ ΔATC …..(by SAS)
∴ AB = AC
In ΔAEF, AE = AF = EF …(Given)
∴ ΔAEF is an equilateral triangle.
`AT =sqrt(3)/2 a` ...(Altitude of equilateral triangle)
`In ΔATB, (AB)^2=(AT)^2+(BT)^2`
`(AB)^2=(sqrt(3)/2a)2+(a+a/2)^2=(3a^2)/4+(9a^2)/4=(12a^2)/4=3a^2`
`AB=sqrt(3)a`
`i.e.,AB = AC =sqrt(3)a`
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