Question

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

Answer 1

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

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${\huge{\underline{\underline{\bold{\red{Solution:-}}}}}}$

✧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 hero 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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$\huge\boxed\blue{Thanks}$