Question

in trapezium ABCD, AB=AD=BC=13 AND CD=23cm. Find the area of trapezium

Answer 1

From B draw BP perpendicular DC 

Therefore, AB = DP = 20 cm 

So, PC = DC - DP 

= (25 - 20) cm 

= 5 cm 

Now, area of trapezium ABCD = Area of rectangle ABPD + Area of △ BPC 

△BPC is right angled at ∠BPC 

Therefore, using Pythagoras theorem, 

      BC² = BP² + PC²

     13² = BP² + 5²

⇒ 169 = BP² + 25

⇒ 169 - 25 = BP² 

⇒ 144 = BP²

⇒ BP = 12 

Now, area of trapezium ABCD = Area of rectangle ABPD + Area of ∆BPC 

                                                = AB × BP + 1/2 × PC × BP 

                                                = 20 × 12 + 1/2 × 5 × 12 

                                                = 240 + 30 

                                                = 270 cm²

Answer 2

Area of trapezium= 1/2 × ( sum of sides) × height
= 1/2 × (13+13) × 23
=299cm sq
hope it helps.