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

AB^{2} = AC^{2} + BC^{2}

Or, AB^{2} = 5^{2} + 12^{2}

Or, AB^{2} = 25 + 144

Or, AB^{2} = = 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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