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 area of ABPQ . 2. the area of triangle AQP.
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Answers
Step-by-step explanation:
✧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