Question

Hey everyone pls solve this!!!
QUESTION :
ABCD is a Trapezium in which
AB ll DC, AB = 29 cm, DC = 20 cm, AD = 15 cm and BC = 12 cm. Find the area of Trapezium.

Answer 1

Answer:

294 cm²

Step-by-step explanation:

Given, AB = 29 cm, DC = 20 cm, AD = 15 cm and BC = 12 cm.

(i)

Draw a line segment BE || AD.

∴ BE = AD = 15 cm, DE = 29 cm, EC = 9 cm

We know that Area of triangle s = a + b + c/2

                                                     = 15 + 9 + 12/2

                                                     = 18 cm.

Area of ΔBEC = √s(s - a)(s - b)(s - c)

                       = √18(18 - 15)(18 - 9)(18 - 12)

                       = √18 * 3 * 9 * 6

                       = √2916

                       = 54 cm²

(ii)

Area of ΔBEC = (1/2) * height * base

⇒ 54 = (1/2) * height * 9

⇒ height = 12 cm

(iii)

Area of trapezium ABCD = (1/2) * (Sum of parallel sides) * height

⇒ (1/2) * (29 + 20) * 12

⇒ (29 + 20) * 6

⇒ 294 cm².

Therefore, Area of trapezium = 294 cm².

Hope it helps!