State and prove Puthagoras Theroem and its Converse part.
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we need to prove ac^2=ab^2+bc^2
let us draw Bd perpendicular AC see figure..
now.. ∆ADB ~∆ABC ( theorem 6.7)
AD/AN =AB/AC (sides are proportional)......... (1)
. AC= AB^2.
also, ∆bdc~∆abc
CD/BC=BC/AC...........(2)
CD. AC=BC^2
adding one and two.....
AD.AC+CD.AC=AB^2+BC^2
AC(AD+CD) =ab^2 +bc^2
AC. AC=ab^2+ bc^2
ac^2 =ab^2+ bc^2...
hence proved.
let us draw Bd perpendicular AC see figure..
now.. ∆ADB ~∆ABC ( theorem 6.7)
AD/AN =AB/AC (sides are proportional)......... (1)
. AC= AB^2.
also, ∆bdc~∆abc
CD/BC=BC/AC...........(2)
CD. AC=BC^2
adding one and two.....
AD.AC+CD.AC=AB^2+BC^2
AC(AD+CD) =ab^2 +bc^2
AC. AC=ab^2+ bc^2
ac^2 =ab^2+ bc^2...
hence proved.
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