Given:
In quadrilateral ABCD, angle B=90° , AB=9m, BC=40m, CD=15m, DA=28m.
To find:
The area of the quadrilateral ABCD.
Solution:
In quadrilateral ABCD, draw a diagonal AC.
Using Pythagoras theorem in triangle ABC, we get
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=9^2+40^2[/tex]
[tex]AC^2=81+1600[/tex]
[tex]AC^2=1681[/tex]
Taking square root on both sides, we get
[tex]AC=\sqrt{1681}[/tex]
[tex]AC=41[/tex]
Area of the triangle ABC is:
[tex]A_1=\dfrac{1}{2}\times base\times height[/tex]
[tex]A_1=\dfrac{1}{2}\times BC\times AB[/tex]
[tex]A_1=\dfrac{1}{2}\times 40\times 9[/tex]
[tex]A_1=180[/tex]
So, the area of the triangle ABC is 180 square m.
According to the Heron's formula, the area of a triangle is
[tex]Area=\sqrt{s(s-a)(s-b)(s-c)}[/tex]
where,
[tex]s=\dfrac{a+b+c}{2}[/tex]
In triangle ACD,
[tex]s=\dfrac{28+15+41}{2}[/tex]
[tex]s=\dfrac{84}{2}[/tex]
[tex]s=42[/tex]
Using Heron's formula, the area of the triangle ACD, we get
[tex]A_2=\sqrt{42(42-28)(42-15)(42-41)}[/tex]
[tex]A_2=\sqrt{42(14)(27)(1)}[/tex]
[tex]A_2=\sqrt{15876}[/tex]
[tex]A_2=126[/tex]
Now, the area of the quadrilateral is the sum of area of the triangle ABC and triangle ACD.
[tex]A=A_1+A_2[/tex]
[tex]A=180+126[/tex]
[tex]A=306[/tex]
Therefore, the area of the quadrilateral ABCD is 306 square meter.
