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AAA Similarity

In the given ...

Question

In the given figure, $Δ$ ABC and $Δ$ AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm. (i) Prove that : (i) $Δ$ ABC ~ $Δ$ AMP
(ii) Find : AB and BC.

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Solution

(i) In ∆ ABC and ∆ AMP,

$∠$ BAC= $∠$ PAM [Common]

$∠$ ABC= $∠$ PMA [Each = 90°]

∆ ABC ~ ∆ AMP [AA Similarity]

(ii)

$A M equals square root of A P squared minus P M squared end root equals square root of 15 squared minus 12 squared end root equals 11 sin c e space increment A B C space tilde increment A M P fraction numerator A B over denominator A M end fraction equals fraction numerator B C over denominator P M end fraction equals fraction numerator A C over denominator A P end fraction fraction numerator A B over denominator 11 end fraction equals fraction numerator B C over denominator 12 end fraction equals 10 over 15 f r o m e space t h i s space w e space c a n space w r i t e fraction numerator A B over denominator 11 end fraction equals 10 over 15 A B equals fraction numerator 10 cross times 11 over denominator 15 end fraction equals 7.33 fraction numerator B C over denominator 12 end fraction equals 10 over 15 B C equals 8 space c m$

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In the given figure, Δ ABC and Δ AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm. (i) Prove that : (i) Δ ABC ~ Δ AMP (ii) Find : AB and BC.

(i) In ∆ ABC and ∆ AMP, ∠BAC= ∠PAM [Common] ∠ABC= ​​​​​​​∠PMA [Each = 90°] ∆ ABC ~ ∆ AMP [AA Similarity] (ii)

In the given figure, $Δ$ ABC and $Δ$ AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm. (i) Prove that : (i) $Δ$ ABC ~ $Δ$ AMP
(ii) Find : AB and BC.