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In the adjoining figure, seg PS ⊥ seg RQ, seg QT ⊥ seg PR.

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In the adjoining figure, seg PS ⊥ seg RQ, seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.

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In ∆PQR, PR is the base and QT is the corresponding height.

Also, RQ is the base and PS is the corresponding height.

[Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]

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