Question

the sides of a quadrangular field taken in order are 26m ,27m , 7m and 24m respectively. the angle contained by the last two sides is a right angle.find it's area.

Answer 1

Hope this answer helps you...

Answer 2

Area of quadrilateral ABCD is $=375.85\ m^{2}$.

To find the area of the quadrangular field given in diagram, the area of right angle triangle and equilateral triangle should be  

From the figure, $A C=\sqrt{(A B)^{2}+(B C)^{2}}=\sqrt{24^{2}+7^{2}}=\sqrt{576+49}=\sqrt{625}=25 m$

$Area\ of\ \Delta A B C=\frac{1}{2} \times A B \times B C=\frac{1}{2} \times 24 m \times 7 m=84 m^{2}$

$\Delta A D C,=\frac{25+26+27}{2}=\frac{78}{2}=39$

$Area\ of \begin{aligned} \Delta A D C &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=\sqrt{39(39-25)(39-26)(39-27)} m^{2} \\ &=\sqrt{39 \times 14 \times 13 \times 12} m^{2} \\ &=\sqrt{85176} m^{2} \\ &=291.85 m^{2} \end{aligned}$

Area of quadrilateral ABCD $=84\ m^{2}+291.85\ m^{2}=375.85\ m^{2}$