In the given figure ABCD is quadrilateral in which ∠ABC = 90°, ∠BDC = 90°, AC = 17 cm, BC = 15 cm, BD = 12 cm and CD = 9 cm. The area of quad ABCD is
Given:
AC = 17 cm
BC = 15 cm
BD = 12 cm
CD = 9 cm.
∠ABC = 90°
∠BDC = 90°
In ∆ABC,
Using Pythagoras theorem,
AB2 + BC2 = AC2
⇒ AB2 = AC2- BC2
⇒ AB = √( AC2- BC2)
⇒ AB = √( 172- 152)
⇒ AB = √(289-225)
⇒ AB = √64
⇒ AB = 8 cm
Therefore,
Area of ∆ABC = 1/2 × AB × BC
= 1/2 × 8 × 15
= 60 cm2
And,
In ∆BDC,
Area of ∆BDC = 1/2 × BD × DC
= 1/2 × × 12 × 9
= 54 cm2
Therefore,
Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆BDC
= 60 cm2 + 54 cm2
= 114 cm2