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प्रश्न
In the adjacent figure, ∆ABC is right angled at C and DE ⊥ AB. Prove that ∆ABC ~ ∆ADE and hence find the lengths of AE and DE
उत्तर
In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A ...(common)
∴ ∆ABC ~ ∆ADE ...(By AA similarity)
`"BC"/"DE" = "AB"/"AD" = "AC"/"AE"`
`12/"DE" = 13/3 = 5/"AE"`
In ∆ABC, AB2 = BC2 + AC2
= 122 + 52
= 144 + 25
= 169
AB = `sqrt(169)` = 13
Consider, `13/3 = 5/"AE"`
∴ AE = `(5 xx 3)/13 = 15/13`
AE = `15/13` and DE = `36/13`
Consider, `12/"DE" = 13/3`
DE = `(12 xx 3)/13 = 36/13`
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