Q. 254.8( 8 Votes )

In Fig. 4.145 (a) is right angled at C and . Prove that and hence find the lengths of AE and DE.


Answer :

In ΔABC, by Pythagoras theorem


AB2 = AC2 + BC2


Or, AB2 = 52 + 122


Or, AB2 = 25 + 144


Or, AB2 = = 169


Or AB = 13 (Square root both side)


In Δ AED and Δ ACB


<A = <A (Common)


<AED = <ACB (Each 90°)


Then, Δ AED ~ Δ ACB(Corresponding parts of similar triangle are proportional)


So, AE/AC = DE/ CB =AD/ AB


Or, AE/5 = DE/12 = 3/13


Or, AE/5 = 3/13 and DE/12 = 3/13


Or, AE = 15/13cm and DE = 36/13cm


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