The area of quadrilateral ABCD in the given figure is
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