Question

ABCD is a trapezium in which AB ∥ DC, AB = 78 cm, CD = 52 cm, AD = 28 cm and BC = 30 cm. Find the area of the trapezium. 

Answer 1

$<u><b><huge>HELLO<u><b><huge>$


Draw CE ∥ AD and CF ⊥ AB. 

Now, EB = (AB - AE) = (AB - DC) = (78 - 52) cm = 26 cm, 



CE = AD = 28 cm and BC = 30 cm. 

Now, in ∆CEB, we have 

S = ¹/₂ (28 + 26 + 30) cm = 42 cm. 

(s - a) = (42 - 28) cm = 14 cm, 

(s - b) = (42 - 26) cm = 16 cm, and 

(s - c) = (42 - 30) cm = 12 cm. 

area of ∆CEB = √{s(s - a)(s - b)(s - c)} 

                      = √(42 × 14 × 16 × 12) cm² 

                      = 336 cm²

Also, area of ∆CEB = ¹/₂ × EB × CF 

                               = (¹/₂ × 26 × CF) cm² 

                               = (13 × CF) cm²

Therefore, 13 × CF = 336 

⇒ CF = 336/13 cm

Area of a trapezium ABCD

                    = {¹/₂ × (AB + CD) × CF} square units 

                    = {¹/₂ × (78 + 52) × ³³⁶/₁₃} cm²

                    = 1680 cm²

Answer 2

Answer:

$1680 {cm}^{2}$

Step-by-step explanation:

explanation are in the image