Question

In the given figure ABCD is a quadrilateral in which ∠ABC = 90°, ∠BDC, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quadrilateral ABCD is

(a) 102cm²

(b) 114cm²

(c) 95cm²

(d) 57cm²

Answer 1

Correct answer is (b) 114 cm²

Using Pythagoras theorem in \(\triangle ABC,\) we get:

AC² = AB² + BC²

\(\Rightarrow AB =\sqrt{AC^2-BC^2}\)

= \(\sqrt{17^2-15^2}\)

= 8 cm

∴ Area of quadrilateral ABCD = Ar(\(\triangle ABC\)) + Ar(\(\triangle BCD\)) = 54 + 60 = 114 cm²