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प्रश्न
Prove that in a right-angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
उत्तर
Given: A right triangle ABC in which ∠B = 90°
To Prove: (Hypotenuse)2 = (Base)2 + (perpendicular)2
i.e AC2 = AB2 + BC2
Construction: From B, Draw BD ⊥ AC
In ΔABC and ΔADB
∠BAC = ∠DAB [Common]
∠ABC = ∠ADB [each 90°]
∴ ΔABC ∼ ΔADB [By AA similarity]
`=> "AB"/"AC" = "AD"/"AB"`
`=> AB2 - AD x AC ...(i)
Similarity, ΔABC ∼ ΔBDC
`=> "BC"/"DC" = "AC"/"BC"`
`=> "BC"^2 - "AC" xx "DC"` ...(ii)
On adding (i) and (iii) we get
AB2 + BC2 = AD x AC + AC x DC
`=>` AB2 + BC2 = AC(AD + DC)
`=>` AB2 + BC2 = AC x AC
`=>` AC2 = AB2 + BC2
Notes
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