Given:

∠ABC = 90°

∠ACD = 90°

CD = 8cm

AB = 9cm

AD = 17cm

Consider ΔACD

Here, By Pythagoras theorem : AD² = CD² + AC²

17² = 8² + AC²

⇒ AC² = 17²—8²

⇒ AC² = 289 – 64 = 225

⇒ AC = 15

Now, Consider ΔABC

Here, By Pythagoras theorem : AC² = AB² + BC²

15² = 9² + AC²

⇒ BC² = 15²—9²

⇒ BC² = 225 – 81 = 144

⇒ BC = 12

Here,

Area (quad.ABCD) = Area (ΔABC) + Area (ΔACD)

Area (quad.ABCD) = 1/2×AB×BC + 1/2×AC×CD

Area (quad.ABCD) = 1/2×9×12 + 1/2×15×8 = 54 + 60 = 104cm²

∴ Area (quad.ABCD) = 114cm²