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
![AC^2=AB^2+BC^2](https://tex.z-dn.net/?f=AC%5E2%3DAB%5E2%2BBC%5E2)
![AC^2=9^2+40^2](https://tex.z-dn.net/?f=AC%5E2%3D9%5E2%2B40%5E2)
![AC^2=81+1600](https://tex.z-dn.net/?f=AC%5E2%3D81%2B1600)
![AC^2=1681](https://tex.z-dn.net/?f=AC%5E2%3D1681)
Taking square root on both sides, we get
![AC=\sqrt{1681}](https://tex.z-dn.net/?f=AC%3D%5Csqrt%7B1681%7D)
![AC=41](https://tex.z-dn.net/?f=AC%3D41)
Area of the triangle ABC is:
![A_1=\dfrac{1}{2}\times base\times height](https://tex.z-dn.net/?f=A_1%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20base%5Ctimes%20height)
![A_1=\dfrac{1}{2}\times BC\times AB](https://tex.z-dn.net/?f=A_1%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20BC%5Ctimes%20AB)
![A_1=\dfrac{1}{2}\times 40\times 9](https://tex.z-dn.net/?f=A_1%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2040%5Ctimes%209)
![A_1=180](https://tex.z-dn.net/?f=A_1%3D180)
So, the area of the triangle ABC is 180 square m.
According to the Heron's formula, the area of a triangle is
![Area=\sqrt{s(s-a)(s-b)(s-c)}](https://tex.z-dn.net/?f=Area%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D)
where,
![s=\dfrac{a+b+c}{2}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7Ba%2Bb%2Bc%7D%7B2%7D)
In triangle ACD,
![s=\dfrac{28+15+41}{2}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B28%2B15%2B41%7D%7B2%7D)
![s=\dfrac{84}{2}](https://tex.z-dn.net/?f=s%3D%5Cdfrac%7B84%7D%7B2%7D)
![s=42](https://tex.z-dn.net/?f=s%3D42)
Using Heron's formula, the area of the triangle ACD, we get
![A_2=\sqrt{42(42-28)(42-15)(42-41)}](https://tex.z-dn.net/?f=A_2%3D%5Csqrt%7B42%2842-28%29%2842-15%29%2842-41%29%7D)
![A_2=\sqrt{42(14)(27)(1)}](https://tex.z-dn.net/?f=A_2%3D%5Csqrt%7B42%2814%29%2827%29%281%29%7D)
![A_2=\sqrt{15876}](https://tex.z-dn.net/?f=A_2%3D%5Csqrt%7B15876%7D)
![A_2=126](https://tex.z-dn.net/?f=A_2%3D126)
Now, the area of the quadrilateral is the sum of area of the triangle ABC and triangle ACD.
![A=A_1+A_2](https://tex.z-dn.net/?f=A%3DA_1%2BA_2)
![A=180+126](https://tex.z-dn.net/?f=A%3D180%2B126)
![A=306](https://tex.z-dn.net/?f=A%3D306)
Therefore, the area of the quadrilateral ABCD is 306 square meter.