Question

plz answer the question....​

Answer 1

ΔABD is a right triangle.

∴ AB² = BD² + AD²

BD² = AB² - AD²

BD² = 25² - 15²

BD² = 625 - 225

BD² = 400

BD = √400

BD = 20

Draw a perpendicular DM from D to AB.

Let AM be x. So BM becomes 25 - x.

ΔDAM is a right triangle.

∴ AD² = AM² + DM²

DM² = AD² - AM²           →          (1)

ΔBDM is also a right triangle.

∴ BD² = DM² + BM²

DM² = BD² - BM²           →          (2)

From (1) and (2),

AD² - AM² = BD² - BM²

15² - x² = 20² - (25 - x)²

225 - x² = 400 - (625 - 50x + x²)

225 - x² = 400 - 625 + 50x - x²

225 = 400 - 625 + 50x

50x = 225 - 400 + 625

50x = 450

x = 450 / 50

x = 9

From (1),

DM² = AD² - AM²

DM² = 15² - x²

DM² = 15² - 9²

DM² = 225 - 81

DM² = 144

DM = √144

DM = 12

Area of ABCD =

1/2 × DM(AB + DC)

1/2 × 12(25 + 7)

1/2 × 12 × 32

1/2 × 384

192

So the area of the trapezium is 192 unit².

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