Question

In ΔABC, AP ⊥ BC and BQ ⊥ AC, B−P−C, A−Q−C, then show that ΔCPA ~ ΔCQB. If AP = 7, BQ = 8, BC = 12, then AC = ?

In ΔCPA and ΔCQB

∠CPA ≅ [∠ ______]    ...[each 90°]

∠ACP ≅ [∠ ______]   ...[common angle]

ΔCPA ~ ΔCQB     ......[______ similarity test]

`"AP"/"BQ" = (["______"])/"BC"`    .......[corresponding sides of similar triangles]

`7/8 = (["______"])/12`

AC × [______] = 7 × 12

AC = 10.5

Answer 1

In ΔCPA and ΔCQB

∠CPA ≅ ∠CQB    ...[each 90°]

∠ACP ≅ ∠BCQ   ...[common angle]

ΔCPA ~ ΔCQB     ...[AA similarity test]

`"AP"/"BQ"` = `"AC"/"BC"`   ...[corresponding sides of similar triangles]

`7/8` = `"AC"/12`

AC × 8 = 7 × 12

AC = `(7 xx 12)/8`

AC = `(7 xx 3)/2`

AC = `21/2`

AC = 10.5.