In a quadrilateral ABCD,AB is perpendicular to BC and CD is perpendicular BC AB=9 CD=7.PQ is perpendicular bisector of AD. 1. Find

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1. Answer:

   ✧Given Find the area of the quadrilateral ABCD in which AB=9cm , BC=40cm , CD=28cm, DA =15cm and ∆ABC = 90°

   Now ABCD is a quadrilateral.

   Triangle ABC = 90

   In right angle triangle ABC

   AC^2 = AB^2 + BC^2

   ➡ = 9^2 + 40^2

   ➡ = 81 + 1600

   ➡ = 1681

   So AC = 41 cm

   ✧Now area of triangle = 1/2 x b x h

   = 1/2 x 9 x 40 = 180 sq cm

   ✧Now to find the other side of triangle using heron’s formula we get

   s = a + b + c / 2

   s = 15 + 41 + 28 / 2

   s = 42 cm

   ✧Now area of triangle ACD

   ➡ = √s (s – a)(s – b)(s – c)

   ➡= √42(42 – 15)(42 – 41)(42 – 28)

   ➡ = √42 x 27 x 1 x 14

   ➡ = 126 sq cm

   ✧So Area of quadrilateral will be 126 + 180 = 306 sq cm

   The ans is 306cm square

   ✧Find the area of the quadrilateral ABCD in which AB=9cm , BC=40cm , CD=28cm, DA =15cm and ∆ABC = 90°

   if we draw diagonal AC

   ✧then we can divide quadrilateral abcd into two triangle abc & adc

   area of abc = (1/2)×9×40= 180 sqcm

   ac^2 = ab^2 + bc^2

   ac^2 = 9^2 + 40^2

   ac^2 = 81 + 1600

   ac^2 = 1681

   ac = 41 cm

   triangle adc

   ➡ac = 41 cm cd = 28cm da = 15 cm

   s= (41 + 28 + 15)/2 = 42

   using heron formula

   ✧area of triangle^2 = (42)(42-41)(42-28)(42-15)

   ➡= 42 × 1 × 14 × 27

   ➡= 14 × 3 × 14 × 3 × 9

   ➡= (42 × 3)^2

   area = 126 cm^2

   total area = 180 + 126 = 306 cm^2

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