Question

ABCD is a trapezium in which
AB || CD and CE I AB. Also AB = 10 cm, CD = 6 cm and BC 75 cm. Find the area of the trapezium

Answer 1

Answer:

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

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Answer 2

Answer:

wrong answer ok hello bye