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प्रश्न
In ∆ABC, AP ⊥ BC, BQ ⊥ AC B– P–C, A–Q – C then prove that, ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12 then Find AC.
उत्तर
n ∆CPA and ∆CQB,
∠CPA ≅ ∠CQB [Each angle is of measure 90°]
∠ACP ≅ ∠BCQ [Common angle]
∴ ∆CPA ~ ∆CQB [AA test of similarity]
`therefore "AC"/"BC" = "AP"/"BQ"` [Corresponding sides of similar triangle]
`therefore "AC"/12 = 7/8`
`therefore "AC" = "x" = (12 xx 7)/8`
∴ AC = 10.5 units.
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