Question

Find the perimeter and area of the quadrilateral ABCD in which AB = 17 cm, AD = 9 cm, CD = 12 cm, ACB = 90° and AC = 15 cm.

Answer 1

In right ∆ACB using pythagorous theorem: 

(AB)² = (AC)² + (BC)² 

(17)² = (15)² + (BC)²

^{\(\sqrt{289-225}\) = BC}

BC = 8 cm

Perimeter of quad ABCD = AB+BC+CD+DA = 17+8+12+9 = 46 cm

Area of right angled ∆ACB = \(\frac{1}{2}\) (Base x Altitude)

⇒ \(\frac{1}{2}\)(15 x 8) = 60 cm² 

Now in equilateral ∆ACD 

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

Hence, area of quad ABCD = 60+54 = 114 cm²