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 ≅ ∠`square`   ...[each 90°]

∠ACP ≅ ∠`square`   ...[common angle]

ΔCPA ~ ΔCQB   ...[`square` similarity test]

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

`7/8 = (square)/12`

AC × `square` = 7 × 12

AC = 10.5

Answer 1

In ΔCPA and ΔCQB

∠CPA ≅ ∠\[\boxed{\text{CQB}}\]   ...[each 90°]

∠ACP ≅ ∠\[\boxed{\text{BCQ}}\]   ...[common angle]

ΔCPA ~ ΔCQB     ...[\[\boxed{\text{AA}}\] similarity test]

`"AP"/"BQ"` = \[\frac{{\boxed{\text{AC}}}}{{{\text{BC}}}}\]   ...[corresponding sides of similar triangles]

`7/8` = \[\frac{{\boxed{\text{AC}}}}{{{12}}}\]

AC × \[\boxed{8}\] = 7 × 12

AC = `(7 xx 12)/8`

AC = `(7 xx 3)/2`

AC = `21/2`

AC = 10.5