Sum

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.

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#### Solution

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.

Concept: Similarity of Triangles

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